This is page i
Printer: Opaque thisElementary Number Theory:
Primes, Congruences, and Secrets
William Stein
January 23, 2017

v

To my wife Clarita LefthandviThis is page vii
Printer: Opaque thisContents

Prefaceix

1 Prime Numbers
1.1 Prime Factorization . . . . . . . . . . . . . . . . . . . . . .
1.2 The Sequence of Prime Numbers . . . . . . . . . . . . . . .
1.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1
2
10
19

2 The Ring of Integers Modulo n
2.1 Congruences Modulo n . . . . . . . . . . . . . . . . . . . . .
2.2 The Chinese Remainder Theorem . . . . . . . . . . . . . . .
2.3 Quickly Computing Inverses and Huge Powers . . . . . . . .
2.4 Primality Testing . . . . . . . . . . . . . . . . . . . . . . . .
2.5 The Structure of (Z/pZ)∗ . . . . . . . . . . . . . . . . . . .
2.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . .

21
22
29
31
36
39
44

3 Public-key Cryptography
3.1 Playing with Fire . . . . . . . . . . . . . . . . . . . . . . . .
3.2 The Diffie-Hellman Key Exchange . . . . . . . . . . . . . .
3.3 The RSA Cryptosystem . . . . . . . . . . . . . . . . . . . .
3.4 Attacking RSA . . . . . . . . . . . . . . . . . . . . . . . . .
3.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . .

49
49
51
56
61
67

4 Quadratic Reciprocity
4.1 Statement of the Quadratic Reciprocity Law

69
70. . . . . . . .

viii

Contents

4.2
4.3
4.4
4.5
4.6

Euler’s Criterion . . . . . . . . . . . . . . . . . . . . . . . .
First Proof of Quadratic Reciprocity . . . . . . . . . . . . .
A Proof of Quadratic Reciprocity Using Gauss Sums . . . .
Finding Square Roots . . . . . . . . . . . . . . . . . . . . .
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . .

73
75
81
86
89

5 Continued Fractions
5.1 The Definition . . . . . . . . . . . . . . . . . . . . . . . . .
5.2 Finite Continued Fractions . . . . . . . . . . . . . . . . . .
5.3 Infinite Continued Fractions . . . . . . . . . . . . . . . . . .
5.4 The Continued Fraction of e . . . . . . . . . . . . . . . . . .
5.5 Quadratic Irrationals . . . . . . . . . . . . . . . . . . . . . .
5.6 Recognizing Rational Numbers . . . . . . . . . . . . . . . .
5.7 Sums of Two Squares . . . . . . . . . . . . . . . . . . . . .
5.8 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . .

93
94
95
101
107
110
115
117
121

6 Elliptic Curves
123
6.1 The Definition . . . . . . . . . . . . . . . . . . . . . . . . . 124
6.2 The Group Structure on an Elliptic Curve . . . . . . . . . . 125
6.3 Integer Factorization Using Elliptic Curves . . . . . . . . . 129
6.4 Elliptic Curve Cryptography . . . . . . . . . . . . . . . . . 135
6.5 Elliptic Curves Over the Rational Numbers . . . . . . . . . 140
6.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146
Answers and Hints

149

References

155

Index

160

This is page ix
Printer: Opaque thisPrefaceThis is a book about prime numbers, congruences, secret messages, andelliptic curves that you can read cover to cover. It grew out of undergraduate courses that the author taught at Harvard, UC San Diego, and the
University of Washington.
The systematic study of number theory was initiated around 300B.C.
when Euclid proved that there are infinitely many prime numbers, andalso cleverly deduced the fundamental theorem of arithmetic, which assertsthat every positive integer factors uniquely as a product of primes. Over a
thousand years later (around 972A.D.) Arab mathematicians formulated
the congruent number problem that asks for a way to decide whether or not
a given positive integer n is the area of a right triangle, all three of whosesides are rational numbers. Then another thousand years later (in 1976),
Diffie and Hellman introduced the first ever public-key cryptosystem, whichenabled two people to communicate secretely over a public communications
channel with no predetermined secret; this invention and the ones that
followed it revolutionized the world of digital communication. In the 1980s
and 1990s, elliptic curves revolutionized number theory, providing striking
new insights into the congruent number problem, primality testing, publickey cryptography, attacks on public-key systems, and playing a central rolein Andrew Wiles’ resolution of Fermat’s Last Theorem.
Today, pure and applied number theory is an exciting mix of simultaneously broad and deep theory, which is constantly informed and motivatedby algorithms and explicit computation. Active research is underway thatpromises to resolve the congruent number problem, deepen our understanding into the structure of prime numbers, and both challenge and improvex

Prefaceour ability to communicate securely. The goal of this book is to bring thereader closer to this world.
The reader is strongly encouraged to do every exercise in this book,checking their answers in the back (where many, but not all, solutions
are given). Also, throughout the text there, are examples of calculations
done using the powerful free open source mathematical software system
Sage (http://www.sagemath.org), and the reader should try every such
example and experiment with similar examples.
Background. The reader should know how to read and write mathematical proofs and must know the basics of groups, rings, and fields. Thus, theprerequisites for this book are more than the prerequisites for most elementary number theory books, while still being aimed at undergraduates.
Notation and Conventions. We let N = {1, 2, 3, . . .} denote the natural
numbers, and use the standard notation Z, Q, R, and C for the rings ofinteger, rational, real, and complex numbers, respectively. In this book, wewill use the words proposition, theorem, lemma, and corollary as follows.
Usually a proposition is a less important or less fundamental assertion, atheorem is a deeper culmination of ideas, a lemma is something that we will
use later in this book to prove a proposition or theorem, and a corollary
is an easy consequence of a proposition, theorem, or lemma. More difficult
exercises are marked with a (*).
Acknowledgements. I would like to thank Brian Conrad, Carl Pomerance, and Ken Ribet for many clarifying comments and suggestions. Baurzhan Bektemirov, Lawrence Cabusora, and Keith Conrad read drafts ofthis book and made many comments, and Carl Witty commented extensively on the first two chapters. Frank Calegari used the course whenteaching Math 124 at Harvard, and he and his students provided much
feedback. Noam Elkies made comments and suggested Exercise 4.6. Seth
Kleinerman wrote a version of Section 5.4 as a class project. Hendrik
Lenstra made helpful remarks about how to present his factorization algorithm. Michael Abshoff, Sabmit Dasgupta, David Joyner, Arthur Patterson, George Stephanides, Kevin Stern, Eve Thompson, Ting-You Wang,and Heidi Williams all suggested corrections. I also benefited from conversations with Henry Cohn and David Savitt. I used Sage ([Sag08]), emacs,
and LATEX in the preparation of this book.

This is page 1
Printer: Opaque this1
Prime Numbers

Every positive integer can be written uniquely as a product of prime numbers, e.g., 100 = 22 · 52 . This is surprisingly difficult to prove, as we willsee below. Even more astounding is that actually finding a way to write
certain 1,000-digit numbers as a product of primes seems out of the reach ofpresent technology, an observation that is used by millions of people every
day when they buy things online.
Since prime numbers are the building blocks of integers, it is natural towonder how the primes are distributed among the integers.
“There are two facts about the distribution of prime numbers.
The first is that, [they are] the most arbitrary and ornery objects studied by mathematicians: they grow like weeds among
the natural numbers, seeming to obey no other law than that ofchance, and nobody can predict where the next one will sprout.
The second fact is even more astonishing, for it states just the
opposite: that the prime numbers exhibit stunning regularity,that there are laws governing their behavior, and that they obey
these laws with almost military precision.”— Don Zagier [Zag75]
The Riemann Hypothesis, which is the most famous unsolved problem in
number theory, postulates a very precise answer to the question of how theprime numbers are distributed.
This chapter lays the foundations for our study of the theory of numbers
by weaving together the themes of prime numbers, integer factorization,and the distribution of primes. In Section 1.1, we rigorously prove that the2

1. Prime Numbers

every positive integer is a product of primes, and give examples of specificintegers for which finding such a decomposition would win one a large cash
bounty. In Section 1.2, we discuss theorems about the set of prime numbers,
starting with Euclid’s proof that this set is infinite, and discuss the largest
known prime. Finally we discuss the distribution of primes via the prime
number theorem and the Riemann Hypothesis.

1.1 Prime Factorization
1.1.1

Primes

The set of natural numbers is
N = {1, 2, 3, 4, . . .},and the set of integers is
Z = {. . . , −2, −1, 0, 1, 2, . . .}.
Definition 1.1.1 (Divides). If a, b ∈ Z we say that a divides b, written
a | b, if ac = b for some c ∈ Z. In this case, we say a is a divisor of b. We
say that a does not divide b, written a - b, if there is no c ∈ Z such thatac = b.
For example, we have 2 | 6 and −3 | 15. Also, all integers divide 0, and 0
divides only 0. However, 3 does not divide 7 in Z.
.
Remark 1.1.2. The notation b : a for “b is divisible by a” is common in
Russian literature on number theory.
Definition 1.1.3 (Prime and Composite). An integer n > 1 is prime if
the only positive divisors of n are 1 and n. We call n composite if n is notprime.
The number 1 is neither prime nor composite. The first few primes of N
are2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, . . . ,and the first few composites are
4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 25, 26, 27, 28, 30, 32, 33, 34, . . . .
Remark 1.1.4. J. H. Conway argues in [Con97, viii] that −1 should beconsidered a prime, and in the 1914 table [Leh14], Lehmer considers 1 tobe a prime. In this book, we consider neither −1 nor 1 to be prime.
SAGE Example 1.1.5. We use Sage to compute all prime numbers between
a and b − 1.

1.1 Prime Factorization

3

sage: prime_range(10,50)[11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47]
We can also compute the composites in an interval.
sage: [n for n in range(10,30) if not is_prime(n)][10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 25, 26, 27, 28]
Every natural number is built, in a unique way, out of prime numbers:
Theorem 1.1.6 (Fundamental Theorem of Arithmetic). Every natural
number can be written as a product of primes uniquely up to order.
Note that primes are the products with only one factor and 1 is theempty product.
Remark 1.1.7. Theorem 1.1.6, which we will prove in Section 1.1.4, is trickier to prove than you might first think. For example, unique factorization
fails in the ring√√Z[ −5] = {a + b −5 : a, b ∈ Z} ⊂ C,
where 6 factors in two different ways:√√
6 = 2 · 3 = (1 + −5) · (1 − −5).

1.1.2

The Greatest Common Divisor

We will use the notion of the greatest common divisor of two integers toprove that if p is a prime and p | ab, then p | a or p | b. Proving this is the
key step in our proof of Theorem 1.1.6.
Definition 1.1.8 (Greatest Common Divisor). Let
gcd(a, b) = max {d ∈ Z : d | a and d | b} ,unless both a and b are 0 in which case gcd(0, 0) = 0.
For example, gcd(1, 2) = 1, gcd(6, 27) = 3, and for any a, gcd(0, a) =gcd(a, 0) = a.
If a 6= 0, the greatest common divisor exists because if d | a then d ≤ |a|,and there are only |a| positive integers ≤ |a|. Similarly, the gcd exists whenb 6= 0.
Lemma 1.1.9. For any integers a and b, we havegcd(a, b) = gcd(b, a) = gcd(±a, ±b) = gcd(a, b − a) = gcd(a, b + a).
Proof. We only prove that gcd(a, b) = gcd(a, b − a), since the other cases
are proved in a similar way. Suppose d | a and d | b, so there exist integers
c1 and c2 such that dc1 = a and dc2 = b. Then b−a = dc2 −dc1 = d(c2 −c1 ),4

1. Prime Numbers

so d | b − a. Thus gcd(a, b) ≤ gcd(a, b − a), since the set over which we aretaking the max for gcd(a, b) is a subset of the set for gcd(a, b − a). The
same argument with a replaced by −a and b replaced by b − a, shows thatgcd(a, b − a) = gcd(−a, b − a) ≤ gcd(−a, b) = gcd(a, b), which proves thatgcd(a, b) = gcd(a, b − a).
Lemma 1.1.10. Suppose a, b, n ∈ Z. Then gcd(a, b) = gcd(a, b − an).
Proof. By repeated application of Lemma 1.1.9, we havegcd(a, b) = gcd(a, b − a) = gcd(a, b − 2a) = · · · = gcd(a, b − an).

Assume for the moment that we have already proved Theorem 1.1.6. A
naive way to compute gcd(a, b) is to factor a and b as a product of primes
using Theorem 1.1.6; then the prime factorization of gcd(a, b) can be readoff from that of a and b. For example, if a = 2261 and b = 1275, thena = 7 · 17 · 19 and b = 3 · 52 · 17, so gcd(a, b) = 17. It turns out thatthe greatest common divisor of two integers, even huge numbers (millions
of digits), is surprisingly easy to compute using Algorithm 1.1.13 below,
which computes gcd(a, b) without factoring a or b.
To motivate Algorithm 1.1.13, we compute gcd(2261, 1275) in a different
way. First, we recall a helpful fact.
Proposition 1.1.11. Suppose that a and b are integers with b 6= 0. Then
there exists unique integers q and r such that 0 ≤ r < |b| and a = bq + r.
Proof. For simplicity, assume that both a and b are positive (we leave thegeneral case to the reader). Let Q be the set of all nonnegative integers nsuch that a − bn is nonnegative. Then Q is nonempty because 0 ∈ Q and Q
is bounded because a − bn < 0 for all n > a/b. Let q be the largest elementof Q. Then r = a − bq < b, otherwise q + 1 would also be in Q. Thus q
and r satisfy the existence conclusion.
To prove uniqueness, suppose that q 0 and r0 also satisfy the conclusion.
Then q 0 ∈ Q since r0 = a − bq 0 ≥ 0, so q 0 ≤ q, and we can write q 0 = q − mfor some m ≥ 0. If q 0 6= q, then m ≥ 1 so
r0 = a − bq 0 = a − b(q − m) = a − bq + bm = r + bm ≥ bsince r ≥ 0, a contradiction. Thus q = q 0 and r0 = a − bq 0 = a − bq = r, as
claimed.
For us, an algorithm is a finite sequence of instructions that can be followed to perform a specific task, such as a sequence of instructions in acomputer program, which must terminate on any valid input. The word “algorithm” is sometimes used more loosely (and sometimes more precisely)
than defined here, but this definition will suffice for us.

1.1 Prime Factorization

5

Algorithm 1.1.12 (Division Algorithm). Suppose a and b are integers
with b 6= 0. This algorithm computes integers q and r such that 0 ≤ r < |b|and a = bq + r.
We will not describe the actual steps of Algorithm 1.1.12, since it is justthe familiar long division algorithm. Note that it might not be exactly thesame as the standard long division algorithm you learned in school, because
we make the remainder positive even when dividing a negative number by
a positive number.
We use the division algorithm repeatedly to compute gcd(2261, 1275).
Dividing 2261 by 1275 we find that
2261 = 1 · 1275 + 986,
so q = 1 and r = 986. Notice that if a natural number d divides both 2261
and 1275, then d divides their difference 986 and d still divides 1275. On
the other hand, if d divides both 1275 and 986, then it has to divide theirsum 2261 as well! We have made progress:
gcd(2261, 1275) = gcd(1275, 986).
This equality also follows by applying Lemma 1.1.9. Repeating, we have
1275 = 1 · 986 + 289,
so gcd(1275, 986) = gcd(986, 289). Keep going:
986 = 3 · 289 + 119
289 = 2 · 119 + 51
119 = 2 · 51 + 17.
Thus gcd(2261, 1275) = · · · = gcd(51, 17), which is 17 because 17 | 51. Thus
gcd(2261, 1275) = 17.
Aside from some tedious arithmetic, that computation was systematic, andit was not necessary to factor any integers (which is something we do notknow how to do quickly if the numbers involved have hundreds of digits).
Algorithm 1.1.13 (Greatest Common Division). Given integers a, b, thisalgorithm computes gcd(a, b).
1. [Assume a > b > 0] We have gcd(a, b) = gcd(|a|, |b|) = gcd(|b|, |a|),so we may replace a and b by their absolute values and hence assumea, b ≥ 0. If a = b, output a and terminate. Swapping if necessary, we
assume a > b. If b = 0, we output a.
2. [Quotient and Remainder] Using Algorithm 1.1.12, write a = bq + r,
with 0 ≤ r < b and q ∈ Z.

6

1. Prime Numbers

3. [Finished?] If r = 0, then b | a, so we output b and terminate.
4. [Shift and Repeat] Set a ← b and b ← r, then go to Step 2.
Proof. Lemmas 1.1.9–1.1.10 imply that gcd(a, b) = gcd(b, r) so the gcd does
not change in Step 4. Since the remainders form a decreasing sequence ofnonnegative integers, the algorithm terminates.
Example 1.1.14. Set a = 15 and b = 6.
15

=

6·2+3

gcd(15, 6) = gcd(6, 3)

6

=

3·2+0

gcd(6, 3) = gcd(3, 0) = 3

Note that we can just as easily do an example that is ten times as big, anobservation that will be important in the proof of Theorem 1.1.19 below.
Example 1.1.15. Set a = 150 and b = 60.
150=

60 · 2 + 30

gcd(150, 60) = gcd(60, 30)60=

30 · 2 + 0

gcd(60, 30) = gcd(30, 0) = 30

SAGE Example 1.1.16. Sage uses the gcd command to compute the greatest common divisor of two integers. For example,
sage: gcd(97,100)
1
sage: gcd(97 * 10^15, 19^20 * 97^2)
97
Lemma 1.1.17. For any integers a, b, n, we havegcd(an, bn) = gcd(a, b) · |n|.
Proof. The idea is to follow Example 1.1.15; we step through Euclid’s algorithm for gcd(an, bn) and note that at every step the equation is theequation from Euclid’s algorithm for gcd(a, b) but multiplied through by n.
For simplicity, assume that both a and b are positive. We will prove thelemma by induction on a + b. The statement is true in the base case whena + b = 2, since then a = b = 1. Now assume a, b are arbitrary with a ≥ b.
Let q and r be such that a = bq + r and 0 ≤ r < b. Then by Lemmas 1.1.9–
1.1.10, we have gcd(a, b) = gcd(b, r). Multiplying a = bq + r by n we seethat an = bnq + rn, so gcd(an, bn) = gcd(bn, rn). Thenb + r = b + (a − bq) = a − b(q − 1) ≤ a < a + b,
so by induction gcd(bn, rn) = gcd(b, r) · |n|. Since gcd(a, b) = gcd(b, r), thisproves the lemma.
Lemma 1.1.18. Suppose a, b, n ∈ Z are such that n | a and n | b. Thenn | gcd(a, b).

1.1 Prime Factorization

7

Proof. Since n | a and n | b, there are integers c1 and c2 , such that a = nc1
and b = nc2 . By Lemma 1.1.17, gcd(a, b) = gcd(nc1 , nc2 ) = n gcd(c1 , c2 ),so n divides gcd(a, b).
With Algorithm 1.1.13, we can prove that if a prime divides the productof two numbers, then it has got to divide one of them. This result is the
key to proving that prime factorization is unique.
Theorem 1.1.19 (Euclid). Let p be a prime and a, b ∈ N. If p | ab thenp | a or p | b.
You might think this theorem is “intuitively obvious,” but that might bebecause the fundamental theorem of arithmetic (Theorem 1.1.6) is deeply
ingrained in your intuition. Yet Theorem 1.1.19 will be needed in our proofof the fundamental theorem of arithmetic.
Proof of Theorem 1.1.19. If p | a we are done. If p - a then gcd(p, a) = 1,
since only 1 and p divide p. By Lemma 1.1.17, gcd(pb, ab) = b. Since p | pb
and, by hypothesis, p | ab, it follows (using Lemma 1.1.17) thatp | gcd(pb, ab) = b gcd(p, a) = b · 1 = b.

1.1.3

Numbers Factor as Products of Primes

In this section, we prove that every natural number factors as a productof primes. Then we discuss the difficulty of finding such a decomposition
in practice. We will wait until Section 1.1.4 to prove that factorization is
unique.
As a first example, let n = 1275. The sum of the digits of n is divisible
by 3, so n is divisible by 3 (see Proposition 2.1.9), and we have n = 3 · 425.
The number 425 is divisible by 5, since its last digit is 5, and we have
1275 = 3 · 5 · 85. Again, dividing 85 by 5, we have 1275 = 3 · 52 · 17,
which is the prime factorization of 1275. Generalizing this process proves
the following proposition.
Proposition 1.1.20. Every natural number is a product of primes.
Proof. Let n be a natural number. If n = 1, then n is the empty productof primes. If n is prime, we are done. If n is composite, then n = ab witha, b < n. By induction, a and b are products of primes, so n is also a productof primes.
Two questions immediately arise: (1) is this factorization unique, and(2) how quickly can we find such a factorization? Addressing (1), what if
we had done something differently when breaking apart 1275 as a productof primes? Could the primes that show up be different? Let’s try: we have8

1. Prime Numbers

1275 = 5 · 255. Now 255 = 5 · 51 and 51 = 17 · 3, and again the factorizationis the same, as asserted by Theorem 1.1.6. We will prove the uniqueness ofthe prime factorization of any integer in Section 1.1.4.
SAGE Example 1.1.21. The factor command in Sage factors an integer
as a product of primes with multiplicities. For example,
sage: factor(1275)
3 * 5^2 * 17
sage: factor(2007)3^2 * 223
sage: factor(31415926535898)
2 * 3 * 53 * 73 * 2531 * 534697
Regarding (2), there are algorithms for integer factorization. It is a major
open problem to decide how fast integer factorization algorithms can be. We
say that an algorithm to factor n is polynomial time if there is a polynomial
f (x) such that for any n the number of steps needed by the algorithm tofactor n is less than f (log10 (n)). Note that log10 (n) is an approximation
for the number of digits of the input n to the algorithm.
Open Problem 1.1.22. Is there an algorithm that can factor any integer nin polynomial time?
Peter Shor [Sho97] devised a polynomial time algorithm for factoring
integers on quantum computers. We will not discuss his algorithm further,
except to note that in 2001 IBM researchers built a quantum computerthat used Shor’s algorithm to factor 15 (see [LMG+ 01, IBM01]). Building
much larger quantum computers appears to be extremely difficult.
You can earn money by factoring certain large integers. Many cryptosystems would be easily broken if factoring certain large integers was easy.
Since nobody has proven that factoring integers is difficult, one way toincrease confidence that factoring is difficult is to offer cash prizes for factoring certain integers. For example, until recently there was a $10,000
bounty on factoring the following 174-digit integer (see [RSA]):
1881988129206079638386972394616504398071635633794173827007
6335642298885971523466548531906060650474304531738801130339
6716199692321205734031879550656996221305168759307650257059
This number is known as RSA-576 since it has 576 digits when written inbinary (see Section 2.3.2 for more on binary numbers). It was factored at theGerman Federal Agency for Information Technology Security in December
2003 (see [Wei03]):
398075086424064937397125500550386491199064362342526708406
385189575946388957261768583317
×472772146107435302536223071973048224632914695302097116459
852171130520711256363590397527

1.1 Prime Factorization

9

The previous RSA challenge was the 155-digit number
1094173864157052742180970732204035761200373294544920599091
3842131476349984288934784717997257891267332497625752899781
833797076537244027146743531593354333897.
It was factored on 22 August 1999 by a group of sixteen researchers in fourmonths on a cluster of 292 computers (see [ACD+ 99]). They found that
RSA-155 is the product of the following two 78-digit primes:
p = 10263959282974110577205419657399167590071656780803806
6803341933521790711307779
q = 10660348838016845482092722036001287867920795857598929
1522270608237193062808643.
The next RSA challenge is RSA-640:
31074182404900437213507500358885679300373460228427275457201619
48823206440518081504556346829671723286782437916272838033415471
07310850191954852900733772482278352574238645401469173660247765
2346609,
and its factorization was worth $20,000 until November 2005 when it was
factored by F. Bahr, M. Boehm, J. Franke, and T. Kleinjun. This factorization took five months. Here is one of the prime factors (you can find theother):
16347336458092538484431338838650908598417836700330923121811108
52389333100104508151212118167511579.
(This team also factored a 663-bit RSA challenge integer.)
The smallest currently open challenge is RSA-704, worth $30,000:
74037563479561712828046796097429573142593188889231289084936232
63897276503402826627689199641962511784399589433050212758537011
89680982867331732731089309005525051168770632990723963807867100
86096962537934650563796359
SAGE Example 1.1.23. Using Sage, we see that the above number has 212
decimal digits and is definitely composite:
sage: n = 7403756347956171282804679609742957314259318888\...9231289084936232638972765034028266276891996419625117\...8439958943305021275853701189680982867331732731089309\...0055250511687706329907239638078671008609696253793465\...0563796359
sage: len(n.str(2))

10

1. Prime Numbers

704
sage: len(n.str(10))
212
sage: n.is_prime()
False# this is instant

These RSA numbers were factored using an algorithm called the number
field sieve (see [LL93]), which is the best-known general purpose factorization algorithm. A description of how the number field sieve works is beyond
the scope of this book. However, the number field sieve makes extensive useof the elliptic curve factorization method, which we will describe in Section 6.3.

1.1.4

The Fundamental Theorem of Arithmetic

We are ready to prove Theorem 1.1.6 using the following idea. Suppose
we have two factorizations of n. Using Theorem 1.1.19, we cancel common
primes from each factorization, one prime at a time. At the end, we discover that the factorizations must consist of exactly the same primes. The
technical details are given below.
Proof. If n = 1, then the only factorization is the empty product of primes,
so suppose n > 1.
By Proposition 1.1.20, there exist primes p1 , . . . , pd such thatn = p1 p2 · · · pd .
Suppose thatn = q1 q2 · · · qm
is another expression of n as a product of primes. Since
p1 | n = q1 (q2 · · · qm ),Euclid’s theorem implies that p1 = q1 or p1 | q2 · · · qm . By induction, we
see that p1 = qi for some i.
Now cancel p1 and qi , and repeat the above argument. Eventually, we
find that, up to order, the two factorizations are the same.

1.2 The Sequence of Prime Numbers
This section is concerned with three questions:
1. Are there infinitely many primes?
2. Given a, b ∈ Z, are there infinitely many primes of the form ax + b?

1.2 The Sequence of Prime Numbers

11

3. How are the primes spaced along the number line?
We first show that there are infinitely many primes, then state Dirichlet’s
theorem that if gcd(a, b) = 1, then ax + b is a prime for infinitely many
values of x. Finally, we discuss the Prime Number Theorem which assertsthat there are asymptotically x/ log(x) primes less than x, and we make aconnection between this asymptotic formula and the Riemann Hypothesis.

1.2.1

There Are Infinitely Many Primes

Each number on the left in the following table is prime. We will see soonthat this pattern does not continue indefinitely, but something similarworks.
3=2+1
7=2·3+1
31 = 2 · 3 · 5 + 1
211 = 2 · 3 · 5 · 7 + 1
2311 = 2 · 3 · 5 · 7 · 11 + 1
Theorem 1.2.1 (Euclid). There are infinitely many primes.
Proof. Suppose that p1 , p2 , . . . , pn are n distinct primes. We construct a
prime pn+1 not equal to any of p1 , . . . , pn , as follows. If
N = p1 p2 p3 · · · pn + 1,

(1.2.1)then by Proposition 1.1.20 there is a factorization
N = q1 q2 · · · qm
with each qi prime and m ≥ 1. If q1 = pi for some i, then pi | N . Becauseof (1.2.1), we also have pi | N − 1, so pi | 1 = N − (N − 1), which is acontradiction. Thus the prime pn+1 = q1 is not in the list p1 , . . . , pn , andwe have constructed our new prime.
For example,
2 · 3 · 5 · 7 · 11 · 13 + 1 = 30031 = 59 · 509.
Multiplying together the first six primes and adding 1 doesn’t produce aprime, but it produces an integer that is merely divisible by a new prime.
Joke 1.2.2 (Hendrik Lenstra). There are infinitely many composite numbers. Proof. To obtain a new composite number, multiply together the
first n composite numbers and don’t add 1.

12

1. Prime Numbers

1.2.2

Enumerating Primes

In this section we describe a sieving process that allows us to enumerate
all primes up to n. The sieve works by first writing down all numbers up
to n, noting that 2 is prime, and crossing off all multiples of 2. Next, notethat the first number not crossed off is 3, which is prime, and cross off all
multiples of 3, etc. Repeating this process, we obtain a list of the primes
up to n. Formally, the algorithm is as follows:
Algorithm 1.2.3 (Prime Sieve). Given a positive integer n, this algorithm
computes a list of the primes up to n.
1. [Initialize] Let X = [3, 5, . . .] be the list of all odd integers between 3
and n. Let P = [2] be the list of primes found so far.
√
2. [Finished?] Let p be the first element of X. If p > n, append eachelement of X to P and terminate. Otherwise append p to P .
3. [Cross Off] Set X equal to the sublist of elements in X that are notdivisible by p. Go to Step 2.
For example, to list the primes ≤ 40 using the sieve, we proceed asfollows. First P = [2] andX = [3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39].
We append 3 to P and cross off all multiples of 3 to obtain the new list
X = [5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37].
Next we append 5 to P , obtaining P = [2, 3, 5], and cross off the multiplesof 5, to obtain X = [7, 11, 13, 17, 19, 23, 29, 31, 37]. Because 72 ≥ 40, we
append X to P and find that the primes less than 40 are
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37.
Proof of Algorithm 1.2.3. The part of the algorithm
that is not clear is√that when the first element a of X satisfies a ≥√ n, then each element of
X is prime. To see this, suppose√m is in X, so Q
n ≤ m ≤ n and that m isdivisible by no prime that is ≤ n. Write m =√ pei i with the pi distinct
primes ordered so that p1 < p2 < . . .. If pi > n for each i and there ismorethan one pi , then m > n, a contradiction. Thus some pi is less than√n, which also contradicts our assumptions on m.

1.2.3

The Largest Known Prime

Though Theorem 1.2.1 implies that there are infinitely many primes, it still
makes sense to ask the question “What is the largest known prime?”

1.2 The Sequence of Prime Numbers

13

A Mersenne prime is a prime of the form 2q − 1. According to [Cal] thelargest known prime as of March 2007 is the 44th known Mersenne prime
p = 232582657 − 1,which has 9,808,358 decimal digits1 . This would take over 2000 pages toprint, assuming a page contains 60 lines with 80 characters per line. The
Electronic Frontier Foundation has offered a $100,000 prize to the first
person who finds a 10,000,000 digit prime.
Euclid’s theorem implies that there definitely are infinitely many primes
bigger than p. Deciding whether or not a number is prime is interesting, asa theoretical problem, and as a problem with applications to cryptography,
as we will see in Section 2.4 and Chapter 3.
SAGE Example 1.2.4. We can compute the decimal expansion of p in Sage,although watch out as this is a serious computation that may take arounda minute on your computer. Also, do not print out p or s below, because
both would take a very long time to scroll by.
sage: p = 2^32582657 - 1
sage: p.ndigits()
9808358
Next we convert p to a decimal string and look at some of the digits.
sage: s = p.str(10) # this takes a long time
sage: len(s)# s is a very long string (long time)
9808358
sage: s[:20]# the first 20 digits of p (long time)’12457502601536945540’
sage: s[-20:]# the last 20 digits(long time)’11752880154053967871’

1.2.4

Primes of the Form ax + b

Next we turn to primes of the form ax + b, where a and b are fixed integers
with a > 1 and x varies over the natural numbers N. We assume thatgcd(a, b) = 1, because otherwise there is no hope that ax + b is prime
infinitely often. For example, 2x + 2 = 2(x + 1) is only prime if x = 0, andis not prime for any x ∈ N.
Proposition 1.2.5. There are infinitely many primes of the form 4x − 1.
Why might this be true? We list numbers of the form 4x−1 and underline
those that are prime.
3, 7, 11, 15, 19, 23, 27, 31, 35, 39, 43, 47, . . .
1 The 49th known Mersenne prime may have been found on January 7, 2016.

14

1. Prime Numbers

Not only is it plausible that underlined numbers will continue to appear
indefinitely, it is something we can easily prove.
Proof. Suppose p1 , p2 , . . . , pn are distinct primes of the form 4x − 1. Consider the number
N = 4p1 p2 · · · pn − 1.
Then pi - N for any i. Moreover, not every prime p | N is of the form
4x + 1; if they all were, then N would be of the form 4x + 1. Since N isodd, each prime divisor pi is odd so there is a p | N that is of the form
4x − 1. Since p 6= pi for any i, we have found a new prime of the form
4x − 1. We can repeat this process indefinitely, so the set of primes of theform 4x − 1 cannot be finite.
Note that this proof does not work if 4x − 1 is replaced by 4x + 1, sincea product of primes of the form 4x − 1 can be of the form 4x + 1.
Example 1.2.6. Set p1 = 3, p2 = 7. Then
N = 4 · 3 · 7 − 1 = 83
is a prime of the form 4x − 1. Next
N = 4 · 3 · 7 · 83 − 1 = 6971,
which is again a prime of the form 4x − 1. Again,
N = 4 · 3 · 7 · 83 · 6971 − 1 = 48601811 = 61 · 796751.
This time 61 is a prime, but it is of the form 4x + 1 = 4 · 15 + 1. However,
796751 is prime and 796751 = 4 · 199188 − 1. We are unstoppable.
N = 4 · 3 · 7 · 83 · 6971 · 796751 − 1 = 5591 · 6926049421.
This time the small prime, 5591, is of the form 4x − 1 and the large one isof the form 4x + 1.
Theorem 1.2.7 (Dirichlet). Let a and b be integers with gcd(a, b) = 1.
Then there are infinitely many primes of the form ax + b.
Proofs of this theorem typically use tools from advanced number theory,
and are beyond the scope of this book (see e.g., [FT93, §VIII.4]).

1.2.5

How Many Primes are There?

We saw in Section 1.2.1 that there are infinitely many primes. In order to
get a sense of just how many primes there are, we consider a few warmup questions. Then we consider some numerical evidence and state theprime number theorem, which gives an asymptotic answer to our question,

1.2 The Sequence of Prime Numbers

15

and connect this theorem with a form of the famous Riemann Hypothesis.
Our discussion of counting primes in this section is very cursory; for moredetails, read Crandall and Pomerance’s excellent book [CP01, §1.1.5].
The following vague discussion is meant to motivate a precise way tomeasure the number (or percentage) of primes. What percentage of natural numbers are even? Answer: Half of them. What percentage of natural
numbers are of the form 4x − 1? Answer: One fourth of them. What percentage of natural numbers are perfect squares? Answer: Zero percent of
all natural numbers, in the sense that the limit of the proportion of perfectsquares to all natural numbers converges to 0. More precisely,
lim

x→∞

#{n ∈ N : n ≤ x and n is a perfect square}
= 0,
x

√
√
since the numerator is roughly x and limx→∞ xx = 0. Likewise, it isan easy consequence of Theorem 1.2.10 that zero percent of all natural
numbers are prime (see Exercise 1.4).
We are thus led to ask another question:√ How many positive integers ≤ x
are perfect squares? Answer: Roughly x. In the context of primes, we ask,

Question 1.2.8. How many natural numbers ≤ x are prime?
Letπ(x) = #{p ∈ N : p ≤ x is a prime}.
For example,
π(6) = #{2, 3, 5} = 3.
Some values of π(x) are given in Table 1.1, and Figures 1.1 and 1.2 containgraphs of π(x). These graphs look like straight lines, which maybe bend
down slightly.
SAGE Example 1.2.9. To compute π(x) in Sage use the prime pi(x) command:sage: prime_pi(6)3sage: prime_pi(100)25sage: prime_pi(3000000)
216816
We can also draw a plot of π(x) using the plot command:sage: plot(prime_pi, 1,1000, rgbcolor=(0,0,1))
Gauss was an inveterate computer: he wrote in an 1849 letter that there
are 216, 745 primes less than 3, 000, 000 (this is wrong but close; the correct
count is 216, 816).

16

1. Prime Numbers
TABLE 1.1. Values of π(x)

x
π(x)

100
25

200
46

300
62

400
78

500
95

600
109

700
125

800
139

200

1000
168

(1000, 168)
(900, 154)150
(500, 95)

100
50

900
154

(100, 25)

(200, 46)

250

500

750

1000

FIGURE 1.1. Graph of π(x) for x < 1000

Gauss conjectured the following asymptotic formula for π(x), which waslater proved independently by Hadamard and Vallée Poussin in 1896 (butwill not be proved in this book).
Theorem 1.2.10 (Prime Number Theorem). The function π(x) is asymptotic to x/ log(x), in the sense that
lim

π(x)

x→∞ x/ log(x)= 1.

We do nothing more here than motivate this deep theorem with a few
further observations. The theorem implies that
lim

x→∞

π(x)
1
= lim
= 0,
x→∞ log(x)xso for any a,
lim

π(x)

x→∞ x/(log(x) − a)

= lim

π(x)

x→∞ x/ log(x)

−

aπ(x)
= 1.
x

Thus x/(log(x) − a) is also asymptotic to π(x) for any a. See [CP01, §1.1.5]
for a discussion of why a = 1 is the best choice. Table 1.2 compares π(x)and x/(log(x) − 1) for several x < 10000.
The record for counting primes isπ(1023 ) = 1925320391606803968923.
Note that such computations are very difficult to get exactly right, so theabove might be slightly wrong.
For the reader familiar with complex analysis, we mention a connection
between π(x) and the Riemann Hypothesis. The Riemann zeta function
ζ(s) is a complex analytic function on C \ {1} that extends the function

1.2 The Sequence of Prime Numbers

TABLE 1.2. Comparison of π(x) and x/(log(x) − 1)

x
1000
2000
3000
4000
5000
6000
7000
8000
9000
10000

π(x)
168
303
430
550
669
783
900
1007
1117
1229

x/(log(x) − 1) (approx)
169.2690290604408165186256278
302.9888734545463878029800994
428.1819317975237043747385740
548.3922097278253264133400985
665.1418784486502172369455815
779.2698885854778626863677374
891.3035657223339974352567759
1001.602962794770080754784281
1110.428422963188172310675011
1217.976301461550279200775705

1250

10000

1000

7500

750

5000

500
2500

250
2500

5000

7500

10000

25000

50000

75000

FIGURE 1.2. Graphs of π(x) for x < 10000 and x < 100000

10000017

18

1. Prime Numbers

P∞defined on a right half plane by n=1 n−s . The Riemann Hypothesis is
the conjecture that the zeros in C of ζ(s) with positive real part lie on theline Re(s) = 1/2. This conjecture is one of the Clay Math Institute million
dollar millennium prize problems [Cla].
According to [CP01, §1.4.1], the Riemann Hypothesis is equivalent to theconjecture that
Z x
1
Li(x) =dtlog(t)
2
is a “good” approximation to π(x), in the following precise sense.
Conjecture 1.2.11 (Equivalent to the Riemann Hypothesis).
For all x ≥ 2.01,√|π(x) − Li(x)| ≤ x log(x).
√If x = 2, then π(2) = 1 and Li(2) = 0, but 2 log(2) = 0.9802 . . ., so theinequality is not true for x ≥ 2, but 2.01 is big enough. We will do nothing
more to explain this conjecture, and settle for one numerical example.
Example 1.2.12. Let x = 4 · 1022 . Thenπ(x) = 783964159847056303858,
Li(x) = 783964159852157952242.7155276025801473 . . . ,|π(x) − Li(x)| = 5101648384.71552760258014 . . . ,√x log(x) = 10408633281397.77913344605 . . . ,x/(log(x) − 1) = 783650443647303761503.5237113087392967 . . . .
SAGE Example 1.2.13. We use Sage to graph π(x), Li(x), and√x log(x).

sage: P = plot(Li, 2,10000, rgbcolor=’purple’)sage: Q = plot(prime_pi, 2,10000, rgbcolor=’black’)sage: R = plot(sqrt(x)*log(x),2,10000,rgbcolor=’red’)sage: show(P+Q+R,xmin=0, figsize=[8,3])
1250
1000
750
500
250
2500

5000

7500

10000

The
√ topmost line is Li(x), the next line is π(x), and the bottom line isx log(x).

1.3 Exercises

19

For more on the prime number theorem and the Riemann hypothesis see[Zag75] and [MS08].

1.3 Exercises
1.1 Compute the greatest common divisor gcd(455, 1235) by hand.
1.2 Use the prime enumeration sieve to make a list of all primes up to100.
1.3 Prove that there are infinitely many primes of the form 6x − 1.
1.4 Use Theorem 1.2.10 to deduce that lim

x→∞

π(x)
= 0.
x

1.5 Let ψ(x) be the number of primes of the form 4k −1 that are ≤ x. Usea computer to make a conjectural guess about limx→∞ ψ(x)/π(x).
1.6 So far 49 Mersenne primes 2p − 1 have been discovered. Give a guess,backed up by an argument, about when the next Mersenne prime
might be discovered (you will have to do some online research).
1.7 (a) Let y = 10000. Compute π(y) = #{primes p ≤ y}.
x(b) The prime number theorem implies π(x) is asymptotic to log(x).
How close is π(y) to y/ log(y), where y is as in (a)?

1.8 Let a, b, c, n be integers. Prove that(a) if a | n and b | n with gcd(a, b) = 1, then ab | n.
(b) if a | bc and gcd(a, b) = 1, then a | c.
1.9 Let a, b, c, d, and m be integers. Prove that(a) if a | b and b | c then a | c.
(b) if a | b and c | d then ac | bd.
(c) if m 6= 0, then a | b if and only if ma | mb.
(d) if d | a and a 6= 0, then |d| ≤ |a|.
1.10 In each of the following, apply the division algorithm to find q and rsuch that a = bq + r and 0 ≤ r < |b|:a = 300, b = 17, a = 729, b = 31, a = 300, b = −17, a = 389, b = 4.
1.11 (a) (Do this part by hand.) Compute the greatest common divisor of
323 and 437 using the algorithm described in class that involves
quotients and remainders (i.e., do not just factor a and b).

20

1. Prime Numbers(b) Compute by any means the greatest common divisor of
314159265358979323846264338
and
271828182845904523536028747.
1.12 (a) Suppose a, b and n are positive integers. Prove that if an | bn ,then a | b.
(b) Suppose p is a prime and a and k are positive integers. Provethat if p | ak , then pk | ak .
1.13 (a) Prove that if a positive integer n is a perfect square, then ncannot be written in the form 4k + 3 for k an integer. (Hint:
Compute the remainder upon division by 4 of each of (4m)2 ,(4m + 1)2 , (4m + 2)2 , and (4m + 3)2 .)(b) Prove that no integer in the sequence11, 111, 1111, 11111, 111111, . . .
is a perfect square. (Hint: 111 · · · 111 = 111 · · · 108 + 3 = 4k + 3.)
1.14 Prove that a positive integer n√is prime if and only if n is not divisible
by any prime p with 1 < p ≤ n.

This is page 21
Printer: Opaque this2
The Ring of Integers Modulo n

A startling fact about numbers is that it takes less than a second to decidewith near certainty whether or not any given 1,000 digit number n is aprime, without actually factoring n. The algorithm for this involves doingsome arithmetic with n that works differently depending on whether n isprime or composite. In particular, we do arithmetic with the set (in fact,“ring”) of integers {0, 1, . . . , n − 1} using an innovative rule for addition
and multiplication, where the sum and product of two elements of that set
is again in that set.
Another surprising fact is that one can almost instantly compute the last
1,000 digits of a massive multi-billion digit number like n = 12341234567890
without explicitly writing down all the digits of n. Again, this calculation
involves arithmetic with the ring {0, 1, . . . , n − 1}.
This chapter is about the ring Z/nZ of integers modulo n, the beautiful structure this ring has, and how to apply it to the above mentioned
problems, among others. It is foundational for the rest of this book. In Section 2.1, we discuss when linear equations modulo n have a solution, thenintroduce the Euler ϕ function and prove Euler’s Theorem and Wilson’s
theorem. In Section 2.2, we prove the Chinese Remainer Theorem, which
addresses simultaneous solubility of several linear equations modulo coprime moduli. With these theoretical foundations in place, in Section 2.3,
we introduce algorithms for doing powerful computations modulo n, including computing large powers quickly, and solving linear equations. We
finish in Section 2.4 with a discussion of recognizing prime numbers using
arithmetic modulo n.

22

2. The Ring of Integers Modulo n

2.1 Congruences Modulo n
Definition 2.1.1 (Group). A group is a set G equipped with a binary
operation G × G → G (denoted by multiplication below) and an identity
element 1 ∈ G such that:
1. For all a, b, c ∈ G, we have (ab)c = a(bc).
2. For each a ∈ G, we have 1a = a1 = a, and there exists b ∈ G suchthat ab = 1.
Definition 2.1.2 (Abelian Group). An abelian group is a group G suchthat ab = ba for every a, b ∈ G.
Definition 2.1.3 (Ring). A ring R is a set equipped with binary operations+ and × and elements 0, 1 ∈ R such that R is an abelian group under +,and for all a, b, c ∈ R we have
• 1a = a1 = a• (ab)c = a(bc)• a(b + c) = ab + ac.
If, in addition, ab = ba for all a, b ∈ R, then we call R a commutative ring.
In this section, we define the ring Z/nZ of integers modulo n, introduce
the Euler ϕ-function, and relate it to the multiplicative order of certain
elements of Z/nZ.
If a, b ∈ Z and n ∈ N, we say that a is congruent to b modulo n if n | a−b,
and write a ≡ b (mod n). Let nZ = (n) be the subset of Z consisting of all
multiples of n (this is called the “ideal of Z generated by n”).
Definition 2.1.4 (Integers Modulo n). The ring Z/nZ of integers modulo n is the set of equivalence classes of integers modulo n. It is equippedwith its natural ring structure:(a + nZ) + (b + nZ) = (a + b) + nZ
(a + nZ) · (b + nZ) = (a · b) + nZ.
Example 2.1.5. For example,
Z/3Z = {{. . . , −3, 0, 3, . . .}, {. . . , −2, 1, 4, . . .}, {. . . , −1, 2, 5, . . .}}SAGE Example 2.1.6. In Sage, we list the elements of Z/nZ as follows:
sage: R = Integers(3)sage: list(R)[0, 1, 2]

2.1 Congruences Modulo n

23

We use the notation Z/nZ because Z/nZ is the quotient of the ring Z
by the “ideal” nZ of multiples of n. Because Z/nZ is the quotient of a ringby an ideal, the ring structure on Z induces a ring structure on Z/nZ. We
often let a or a (mod n) denote the equivalence class a + nZ of a.
Definition 2.1.7 (Field). A field K is a ring such that for every nonzero
element a ∈ K there is an element b ∈ K such that ab = 1.
For example, if p is a prime, then Z/pZ is a field (see Exercise 2.12).
Definition 2.1.8 (Reduction Map and Lift). We call the natural reduction
map Z → Z/nZ, which sends a to a + nZ, reduction modulo n. We also
say that a is a lift of a + nZ. Thus, e.g., 7 is a lift of 1 mod 3, since
7 + 3Z = 1 + 3Z.
We can use that arithmetic in Z/nZ is well defined to derive tests fordivisibility by n (see Exercise 2.8).
Proposition 2.1.9. A number n ∈ Z is divisible by 3 if and only if thesum of the digits of n is divisible by 3.
Proof. Writen = a + 10b + 100c + · · · ,where the digits of n are a, b, c, etc. Since 10 ≡ 1 (mod 3),
n = a + 10b + 100c + · · · ≡ a + b + c + · · ·(mod 3),from which the proposition follows.

2.1.1

Linear Equations Modulo n

In this section, we are concerned with how to decide whether or not a linear
equation of the form ax ≡ b (mod n) has a solution modulo n. Algorithms
for computing solutions to ax ≡ b (mod n) are the topic of Section 2.3.
First, we prove a proposition that gives a criterion under which one cancancel a quantity from both sides of a congruence.
Proposition 2.1.10 (Cancellation). If gcd(c, n) = 1 andac ≡ bc (mod n),then a ≡ b (mod n).
Proof. By definitionn | ac − bc = (a − b)c.
Since gcd(n, c) = 1, it follows from Theorem 1.1.6 that n | a − b, soa ≡ b (mod n),
as claimed.

24

2. The Ring of Integers Modulo n

When a has a multiplicative inverse a0 in Z/nZ (i.e., aa0 ≡ 1 (mod n))then the equation ax ≡ b (mod n) has a unique solution x ≡ a0 b (mod n).
Thus, it is of interest to determine the units in Z/nZ, i.e., the elements
which have a multiplicative inverse.
We will use complete sets of residues to prove that the units in Z/nZ
are exactly the a ∈ Z/nZ such that gcd(ã, n) = 1 for any lift ã of a to Z
(it doesn’t matter which lift).
Definition 2.1.11 (Complete Set of Residues). We call a subset R ⊂ Z
of size n whose reductions modulo n are pairwise distinct a complete set ofresidues modulo n. In other words, a complete set of residues is a choice ofrepresentative for each equivalence class in Z/nZ.
For example,
R = {0, 1, 2, . . . , n − 1}
is a complete set of residues modulo n. When n = 5, R = {0, 1, −1, 2, −2}
is a complete set of residues.
Lemma 2.1.12. If R is a complete set of residues modulo n and a ∈ Z
with gcd(a, n) = 1, then aR = {ax : x ∈ R} is also a complete set ofresidues modulo n.
Proof. If ax ≡ ax0 (mod n) with x, x0 ∈ R, then Proposition 2.1.10 implies
that x ≡ x0 (mod n). Because R is a complete set of residues, this implies
that x = x0 . Thus the elements of aR have distinct reductions modulo n. It
follows, since #aR = n, that aR is a complete set of residues modulo n.
Proposition 2.1.13 (Units). If gcd(a, n) = 1, then the equation ax ≡ b(mod n) has a solution, and that solution is unique modulo n.
Proof. Let R be a complete set of residues modulo n, so there is a uniqueelement of R that is congruent to b modulo n. By Lemma 2.1.12, aR is also
a complete set of residues modulo n, so there is a unique element ax ∈ aR
that is congruent to b modulo n, and we have ax ≡ b (mod n).
Algebraically, this proposition asserts that if gcd(a, n) = 1, then the map
Z/nZ → Z/nZ given by left multiplication by a is a bijection.
Example 2.1.14. Consider the equation 2x ≡ 3 (mod 7), and the complete
set R = {0, 1, 2, 3, 4, 5, 6} of coset representatives. We have
2R = {0, 2, 4, 6, 8 ≡ 1, 10 ≡ 3, 12 ≡ 5},
so 2 · 5 ≡ 3 (mod 7).
When gcd(a, n) 6= 1, then the equation ax ≡ b (mod n) may or may
not have a solution. For example, 2x ≡ 1 (mod 4) has no solution, but
2x ≡ 2 (mod 4) does, and in fact it has more than one mod 4 (x = 1
and x = 3). Generalizing Proposition 2.1.13, we obtain the following moregeneral criterion for solvability.

2.1 Congruences Modulo n

25

Proposition 2.1.15 (Solvability). The equation ax ≡ b (mod n) has asolution if and only if gcd(a, n) divides b.
Proof. Let g = gcd(a, n). If there is a solution x to the equation ax ≡ b(mod n), then n | (ax − b). Since g | n and g | a, it follows that g | b.
Conversely, suppose that g | b. Then n | (ax − b) if and only if
n
|
g



a
b
x−
g
g


.

Thus ax ≡ b (mod n) has a solution if and only if ag x ≡ gb (mod ng ) has
a solution. Since gcd(a/g, n/g) = 1, Proposition 2.1.13 implies this latter
equation does have a solution.
In Chapter 4, we will study quadratic reciprocity, which gives a nicecriterion for whether or not a quadratic equation modulo n has a solution.

2.1.2

Euler’s Theorem

Let (Z/nZ)∗ denote the set of elements [x] ∈ Z/nZ such that gcd(x, n) = 1.
The set (Z/nZ)∗ is a group, called the group of units of the ring Z/nZ;
it will be of great interest to us. Each element of this group has an order,and Lagrange’s theorem from group theory implies that each element of(Z/nZ)∗ has an order that divides the order of (Z/nZ)∗ . In elementary
number theory, this fact goes by the monicker “Fermat’s Little Theorem”
when n is prime and “Euler’s Theorem” in general, and we reprove it frombasic principles in this section.
Definition 2.1.16 (Order of an Element). Let n ∈ N and x ∈ Z andsuppose that gcd(x, n) = 1. The order of x modulo n is the smallest m ∈ Nsuch thatxm ≡ 1 (mod n).
To show that the definition makes sense, we verify that such an m exists.
Consider x, x2 , x3 , . . . modulo n. There are only finitely many residue classes
modulo n, so we must eventually find two integers i, j with i < j such thatxj ≡ xi(mod n).

Since gcd(x, n) = 1, Proposition 2.1.10 implies that we can cancel x’s andconclude thatxj−i ≡ 1 (mod n).
SAGE Example 2.1.17. Use x.multiplicative order() to compute theorder of an element of Z/nZ in Sage.

26

2. The Ring of Integers Modulo nsage: R = Integers(10)sage: a = R(3)# create an element of Z/10Z
sage: a.multiplicative_order()4
Notice that the powers of a are periodic with period 4, i.e., there are four
powers and they repeat:sage: [a^i for i in range(15)][1, 3, 9, 7, 1, 3, 9, 7, 1, 3, 9, 7, 1, 3, 9]
The command range(n) we use above returns the list of integers between
0 and n − 1, inclusive.
Definition 2.1.18 (Euler’s ϕ-function). For n ∈ N, let
ϕ(n) = #{a ∈ N : a ≤ n and gcd(a, n) = 1}.
For example,
ϕ(1) = #{1} = 1,
ϕ(2) = #{1} = 1,
ϕ(5) = #{1, 2, 3, 4} = 4,
ϕ(12) = #{1, 5, 7, 11} = 4.

Also, if p is any prime number thenϕ(p) = #{1, 2, . . . , p − 1} = p − 1.
In Section 2.2.1, we prove that if gcd(m, r) = 1, then ϕ(mr) = ϕ(m)ϕ(r).
This will yield an easy way to compute ϕ(n) in terms of the prime factorization of n.
SAGE Example 2.1.19. Use the euler phi(n) command to compute ϕ(n)in Sage:sage: euler_phi(2007)
1332
Theorem 2.1.20 (Euler’s Theorem). If gcd(x, n) = 1, thenxϕ(n) ≡ 1

(mod n).

Proof. As mentioned above, Euler’s Theorem has the following grouptheoretic interpretation. The set of units in Z/nZ is a group(Z/nZ)∗ = {a ∈ Z/nZ : gcd(a, n) = 1}
that has order ϕ(n). The theorem then asserts that the order of an elementof (Z/nZ)∗ divides the order ϕ(n) of (Z/nZ)∗ . This is a special case of2.1 Congruences Modulo n

27

the more general fact (Lagrange’s Theorem) that if G is a finite group andg ∈ G, then the order of g divides the cardinality of G.
We now give an elementary proof of the theorem. Let
P = {a : 1 ≤ a ≤ n and gcd(a, n) = 1}.
In the same way that we proved Lemma 2.1.12, we see that the reductions
modulo n of the elements of xP are the same as the reductions of theelements of P . ThusYY(xa) ≡a (mod n),
a∈P

a∈P

since the products are over the same numbers modulo n. Now cancel thea’s on both sides to getx#P ≡ 1(mod n),as claimed.
SAGE Example 2.1.21. We illustrate Euler’s Theorem using Sage. The
Mod(x,n) command returns the equivalence class of x in Z/nZ.
sage: n = 20
sage: k = euler_phi(n); k
8
sage: [Mod(x,n)^k for x in range(n) if gcd(x,n) == 1][1, 1, 1, 1, 1, 1, 1, 1]

2.1.3

Wilson’s Theorem

The following characterization of prime numbers, from the 1770s, is called“Wilson’s Theorem,” though it was first proved by Lagrange.
Proposition 2.1.22 (Wilson’s Theorem). An integer p > 1 is prime if
and only if (p − 1)! ≡ −1 (mod p).
For example, if p = 3, then (p − 1)! = 2 ≡ −1 (mod 3). If p = 17, then(p − 1)! = 20922789888000 ≡ −1(mod 17).

But if p = 15, then(p − 1)! = 87178291200 ≡ 0(mod 15),

so 15 is composite. Thus Wilson’s theorem could be viewed as a primality
test, though, from a computational point of view, it is probably one of theworld’s least efficient primality tests since computing (n−1)! takes so many
steps.

28

2. The Ring of Integers Modulo n

Proof. The statement is clear when p = 2, so henceforth we assume thatp > 2. We first assume that p is prime and prove that (p − 1)! ≡ −1(mod p). If a ∈ {1, 2, . . . , p − 1}, then the equationax ≡ 1(mod p)has a unique solution a0 ∈ {1, 2, . . . , p − 1}. If a = a0 , then a2 ≡ 1 (mod p),
so p | a2 −1 = (a−1)(a+1), so p | (a−1) or p | (a+1), so a ∈ {1, p−1}. We
can thus pair off the elements of {2, 3, . . . , p − 2}, each with their inverse.
Thus
2 · 3 · · · · · (p − 2) ≡ 1 (mod p).
Multiplying both sides by p − 1 proves that (p − 1)! ≡ −1 (mod p).
Next, we assume that (p − 1)! ≡ −1 (mod p) and prove that p must be
prime. Suppose not, so that p ≥ 4 is a composite number. Let ` be a prime
divisor of p. Then ` < p, so ` | (p − 1)!. Also, by assumption,` | p | ((p − 1)! + 1).
This is a contradiction, because a prime can not divide a number a andalso divide a + 1, since it would then have to divide (a + 1) − a = 1.
Example 2.1.23. We illustrate the key step in the above proof in the casep = 17. We have
2·3 · · · 15 = (2·9)·(3·6)·(4·13)·(5·7)·(8·15)·(10·12)·(14·11) ≡ 1(mod 17),

where we have paired up the numbers a, b for which ab ≡ 1 (mod 17).
SAGE Example 2.1.24. We use Sage to create a table of triples; the firstcolumn contains n, the second column contains (n − 1)! modulo n, and thethird contains −1 modulo n. Notice that the first columns contains a primeprecisely when the second and third columns are equal. (The ... notation
indicates a multi-line command in Sage; you should not type the dots in
explicitly.)
sage: for n in range(1,10):...
print n, factorial(n-1) % n, -1 % n
1 0 0
2 1 1
3 2 2
4 2 3
5 4 4
6 0 5
7 6 6
8 0 7
9 0 8

2.2 The Chinese Remainder Theorem

29

2.2 The Chinese Remainder Theorem
In this section, we prove the Chinese Remainder Theorem, which gives
conditions under which a system of linear equations is guaranteed to havea solution. In the 4th century a Chinese mathematician asked the following:
Question 2.2.1. There is a quantity whose number is unknown. Repeatedly divided by 3, the remainder is 2; by 5 the remainder is 3; and by 7 theremainder is 2. What is the quantity?
In modern notation, Question 2.2.1 asks us to find a positive integer
solution to the following system of three equations:x≡2(mod 3)x≡3(mod 5)x≡2(mod 7)The Chinese Remainder Theorem asserts that a solution exists, and theproof gives a method to find one. (See Section 2.3 for the necessary algorithms.)
Theorem 2.2.2 (Chinese Remainder Theorem). Let a, b ∈ Z and n, m ∈N such that gcd(n, m) = 1. Then there exists x ∈ Z such thatx ≡ a (mod m),x ≡ b (mod n).
Moreover x is unique modulo mn.
Proof. If we can solve for t in the equationa + tm ≡ b (mod n),then x = a + tm will satisfy both congruences. To see that we can solve,
subtract a from both sides and use Proposition 2.1.13 together with ourassumption that gcd(n, m) = 1 to see that there is a solution.
For uniqueness, suppose that x and y solve both congruences. Then z =x − y satisfies z ≡ 0 (mod m) and z ≡ 0 (mod n), so m | z and n | z. Since
gcd(n, m) = 1, it follows that nm | z, so x ≡ y (mod nm).
Algorithm 2.2.3 (Chinese Remainder Theorem). Given coprime integers
m and n and integers a and b, this algorithm finds an integer x such thatx ≡ a (mod m) and x ≡ b (mod n).
1. [Extended GCD] Use Algorithm 2.3.7 below to find integers c, d suchthat cm + dn = 1.
2. [Answer] Output x = a + (b − a)cm and terminate.

30

2. The Ring of Integers Modulo n

Proof. Since c ∈ Z, we have x ≡ a (mod m), and using that cm + dn = 1,
we have a + (b − a)cm ≡ a + (b − a) ≡ b (mod n).
Now we can answer Question 2.2.1. First, we use Theorem 2.2.2 to finda solution to the pair of equationsx≡2(mod 3),x≡3(mod 5).

Set a = 2, b = 3, m = 3, n = 5. Step 1 is to find a solution to t · 3 ≡ 3 − 2(mod 5). A solution is t = 2. Then x = a + tm = 2 + 2 · 3 = 8. Since any x0
with x0 ≡ x (mod 15) is also a solution to those two equations, we cansolve all three equations by finding a solution to the pair of equations
x≡8(mod 15)x≡2(mod 7).

Again, we find a solution to t · 15 ≡ 2 − 8 (mod 7). A solution is t = 1, so
x = a + tm = 8 + 15 = 23.
Note that there are other solutions. Any x0 ≡ x (mod 3 · 5 · 7) is also asolution; e.g., 23 + 3 · 5 · 7 = 128.
SAGE Example 2.2.4. The CRT(a,b,m,n) command in Sage computes aninteger x such that x ≡ a (mod m) and x ≡ b (mod n). For example,
sage: CRT(2,3, 3, 5)
8
The CRT list command computes a number that reduces to several numbers modulo coprime moduli. We use it to answer Question 2.2.1:
sage: CRT_list([2,3,2], [3,5,7])
23

2.2.1

Multiplicative Functions

Recall from Definition 2.1.18 that the Euler ϕ-function isϕ(n) = #{a : 1 ≤ a ≤ n and gcd(a, n) = 1}.
Lemma 2.2.5. Suppose that m, n ∈ N and gcd(m, n) = 1. Then the mapψ : (Z/mnZ)∗ → (Z/mZ)∗ × (Z/nZ)∗ .
defined byψ(c) = (c mod m, c mod n)
is a bijection.

(2.2.1)

2.3 Quickly Computing Inverses and Huge Powers

31

Proof. We first show that ψ is injective. If ψ(c) = ψ(c0 ), then m | c − c0 andn | c − c0 , so nm | c − c0 because gcd(n, m) = 1. Thus c = c0 as elements of(Z/mnZ)∗ .
Next we show that ψ is surjective, i.e., that every element of (Z/mZ)∗ ×(Z/nZ)∗ is of the form ψ(c) for some c. Given a and b with gcd(a, m) = 1
and gcd(b, n) = 1, Theorem 2.2.2 implies that there exists c with c ≡ a(mod m) and c ≡ b (mod n). We may assume that 1 ≤ c ≤ nm, andsince gcd(a, m) = 1 and gcd(b, n) = 1, we must have gcd(c, nm) = 1. Thus
ψ(c) = (a, b).
Definition 2.2.6 (Multiplicative Function). A function f : N → C ismultiplicative if, whenever m, n ∈ N and gcd(m, n) = 1, we havef (mn) = f (m) · f (n).
Proposition 2.2.7 (Multiplicativity of ϕ). The function ϕ is multiplicative.
Proof. The map ψ of Lemma 2.2.5 is a bijection, so the set on the left in(2.2.1) has the same size as the product set on the right in (2.2.1). Thus
ϕ(mn) = ϕ(m) · ϕ(n).

The proposition is helpful in computing ϕ(n), at least if we assume we cancompute the factorization of n (see Section 3.4.1 for a connection between
factoring n and computing ϕ(n)). For example,
ϕ(12) = ϕ(22 ) · ϕ(3) = 2 · 2 = 4.
Also, for n ≥ 1, we haveϕ(pn ) = pn −pn
= pn − pn−1 = pn−1 (p − 1),p(2.2.2)since ϕ(pn ) is the number of numbers less than pn minus the number of
those that are divisible by p. Thus, e.g.,ϕ(389 · 112 ) = 388 · (112 − 11) = 388 · 110 = 42680.

2.3 Quickly Computing Inverses and Huge Powers
This section is about how to solve the equation ax ≡ 1 (mod n) when
we know it has a solution, and how to efficiently compute am (mod n).
We also discuss a simple probabilistic primality test that relies on ourability to compute am (mod n) quickly. All three of these algorithms areof fundamental importance to the cryptography algorithms of Chapter 3.

32

2.3.1

2. The Ring of Integers Modulo n

How to Solve ax ≡ 1 (mod n)

Suppose a, n ∈ N with gcd(a, n) = 1. Then by Proposition 2.1.13 theequation ax ≡ 1 (mod n) has a unique solution. How can we find it?
Proposition 2.3.1 (Extended Euclidean Representation). Suppose a, b ∈Z and let g = gcd(a, b). Then there exists x, y ∈ Z such thatax + by = g.
Remark 2.3.2. If e = cg is a multiple of g, then cax + cby = cg = e, so
e = (cx)a + (cy)b can also be written in terms of a and b.
Proof of Proposition 2.3.1. Let g = gcd(a, b). Then gcd(a/g, b/g) = 1, so
by Proposition 2.1.15, the equationab· x ≡ 1 mod(2.3.1)gg
has a solution x ∈ Z. Multiplying (2.3.1) through by g yields ax ≡ g(mod b), so there exists y such that b · (−y) = ax − g. Then ax + by = g,
as required.
Given a, b and g = gcd(a, b), our proof of Proposition 2.3.1 gives a way to
explicitly find x, y such that ax + by = g, assuming one knows an algorithm
to solve linear equations modulo n. Since we do not know such an algorithm,
we now discuss a way to explicitly find x and y. This algorithm will in fact
enable us to solve linear equations modulo n. To solve ax ≡ 1 (mod n)when gcd(a, n) = 1, use the Algorithm 2.3.7 to find x and y such thatax + ny = 1. Then ax ≡ 1 (mod n).
Example 2.3.3. Suppose a = 5 and b = 7. The steps of Algorithm 1.1.13
to compute gcd(5, 7) are as follows. Here we underline certain numbers,
because it clarifies the subsequent back substitution we will use to find x
and y.
7=1·5+2

so 2 = 7 − 5

5=2·2+1

so 1 = 5 − 2 · 2 = 5 − 2(7 − 5) = 3 · 5 − 2 · 7

On the right, we have back-substituted in order to write each partial remainder as a linear combination of a and b. In the last step, we obtaingcd(a, b) as a linear combination of a and b, as desired.
Example 2.3.4. That example was not too complicated, so we try anotherone. Let a = 130 and b = 61. We have
130 = 2 · 61 + 8

8 = 130 − 2 · 61

61 = 7 · 8 + 5

5 = −7 · 130 + 15 · 61

8=1·5+3

3 = 8 · 130 − 17 · 61

5=1·3+2

2 = −15 · 130 + 32 · 61

3=1·2+1

1 = 23 · 130 − 49 · 61

2.3 Quickly Computing Inverses and Huge Powers

33

Thus x = 23 and y = −49 is a solution to 130x + 61y = 1.
Example 2.3.5. This example is just like Example 2.3.4 above, except wemake the notation on the right more compact.
130 = 2 · 61 + 8

8 = (1, −2)

61 = 7 · 8 + 5

5 = (−7, 15) = (0, 1) − 7(1, −2)

8=1·5+3

3 = (8, −17) = (1, −2) − (−7, 15)

5=1·3+2

2 = (−15, 32) = (−7, 15) − (8, −17)

3=1·2+1

1 = (23, −49) = (8, −17) − (−15, 32)

Notice at each step that the vector on the right is just the vector fromtwo steps ago minus a multiple of the vector from one step ago, where themultiple is the cofficient of what we divide by.
SAGE Example 2.3.6. The xgcd(a,b) command computes the greatest
common divisor g of a and b along with x, y such that ax + by = g.
sage: xgcd(5,7)(1, 3, -2)
sage: xgcd(130,61)(1, 23, -49)
Algorithm 2.3.7 (Extended Euclidean Algorithm). Suppose a and b are
integers and let g = gcd(a, b). This algorithm finds g, x and y such thatax + by = g. We describe only the steps when a > b ≥ 0, since one caneasily reduce to this case.
1. [Initialize] Set x = 1, y = 0, r = 0, s = 1.
2. [Finished?] If b = 0, set g = a and terminate.
3. [Quotient and Remainder] Use Algorithm 1.1.12 to write a = qb + cwith 0 ≤ c < b.
4. [Shift] Set (a, b, r, s, x, y) = (b, c, x − qr, y − qs, r, s) and go to Step 2.
(This shift step is nicely illustrated in Example 2.3.5.)
Proof. This algorithm is the same as Algorithm 1.1.13, except that we keep
track of extra variables x, y, r, s, so it terminates and when it terminatesg = gcd(a, b). We omit the rest of the inductive proof that the algorithm
is correct, and instead refer the reader to [Knu97, §1.2.1].
Algorithm 2.3.8 (Inverse Modulo n). Suppose a and n are integers andgcd(a, n) = 1. This algorithm finds an x such that ax ≡ 1 (mod n).
1. [Compute Extended GCD] Use Algorithm 2.3.7 to compute integers
x, y such that ax + ny = gcd(a, n) = 1.
2. [Finished] Output x.

34

2. The Ring of Integers Modulo n

Proof. Reduce ax+ny = 1 modulo n to see that x satisfies ax ≡ 1 (mod n).
Example 2.3.9. Solve 17x ≡ 1 (mod 61). First, we use Algorithm 2.3.7 tofind x, y such that 17x + 61y = 1:
61 = 3 · 17 + 10

10 = 61 − 3 · 17

17 = 1 · 10 + 7

7 = −61 + 4 · 17

10 = 1 · 7 + 3

3 = 2 · 61 − 7 · 17

3=2·3+1

1 = −5 · 61 + 18 · 17

Thus 17 · 18 + 61 · (−5) = 1 so x = 18 is a solution to 17x ≡ 1 (mod 61).
SAGE Example 2.3.10. Sage implements the above algorithm for quickly
computing inverses modulo n. For example,
sage: a = Mod(17, 61)
sage: a^(-1)
18

2.3.2

How to Compute am (mod n)

Let a and n be integers, and m a nonnegative integer. In this section, we describe an efficient algorithm to compute am (mod n). For the cryptography
applications in Chapter 3, m will have hundreds of digits.
The naive approach to computing am (mod n) is to simply computema = a · a · · · a (mod n) by repeatedly multiplying by a and reducing modulo m. Note that after each arithmetic operation is completed, we reduce
the result modulo n so that the sizes of the numbers involved do not get
too large. Nonetheless, this algorithm is horribly inefficient because it takesm − 1 multiplications, which is huge if m has hundreds of digits.
A much more efficient algorithm for computing am (mod n) involvesiwriting m in binary, then expressing am as a product of expressions a2 , forvarious i. These latter expressions can be computed by repeatedly squaringia2 . This more clever algorithm is not “simpler,” but it is vastly more
efficient since the number of operations needed grows with the numberof binary digits of m, whereas with the naive algorithm in the previous
paragraph, the number of operations is m − 1.
Algorithm 2.3.11 (Write a Number in Binary). Let m be a nonnegative
integer.
PrThis algorithm writes m in binary, so it finds εi ∈ {0, 1} such thatm = i=0 εi 2i with each εi ∈ {0, 1}.
1. [Initialize] Set i = 0.
2. [Finished?] If m = 0, terminate.
3. [Digit] If m is odd, set εi = 1, otherwise εi = 0. Increment i.

2.3 Quickly Computing Inverses and Huge Powers

4. [Divide by 2] Set m =35

m
2 , the greatest integer ≤ m/2. Go to Step 2.

SAGE Example 2.3.12. To write a number in binary using Sage, use thestr command:sage: 100.str(2)’1100100’
Notice the above is the correct binary expansion:
sage: 0*2^0 + 0*2^1 + 1*2^2 + 0*2^3 + 0*2^4 + 1*2^5 + 1*2^6100
Algorithm 2.3.13 (Compute Power). Let a and n be integers and m anonnegative integer. This algorithm computes am modulo n.
1. [Write in Binary] Write m in binary using Algorithm 2.3.11, so am =Q
2i(mod n).
εi =1 a2

32

2. [Compute Powers] Compute a, a2 , a2 = (a2 )2 , a2 = (a2 )2 , etc., upr
to a2 , where r + 1 is the number of binary digits of m.
i

3. [Multiply Powers] Multiply together the a2 such that εi = 1, always
working modulo n.
Example 2.3.14. We can compute the last 2 digits of 791 , by finding 791(mod 100). First, because gcd(7, 100) = 1, we have by Theorem 2.1.20 that
7ϕ(100) ≡ 1 (mod 100). Because ϕ is multiplicative,ϕ(100) = ϕ(22 · 52 ) = (22 − 2) · (52 − 5) = 40.
Thus 740 ≡ 1 (mod 100), hence
791 ≡ 740+40+11 ≡ 711(mod 100).

We now compute 711 (mod 100) using the above algorithm. First, write 11
in binary by repeatedly dividing by 2.
11 = 5 · 2 + 1
5=2·2+12=1·2+01=0·2+1

So in binary, (11)2 = 1011, which we check:
11 = 1 · 8 + 1 · 2 + 1.

36

2. The Ring of Integers Modulo n

Next, compute a, a2 , a4 , a8 and output a8 · a2 · a. We havea=7
a2 ≡ 49
a4 ≡ 492 ≡ 1
a8 ≡ 12 ≡ 1

Note: it is easiest to square 49 by working modulo 4 and 25 and using the
Chinese Remainder Theorem. Finally,
791 ≡ 711 ≡ a8 · a2 · a ≡ 1 · 49 · 7 ≡ 43

(mod 100).

SAGE Example 2.3.15. Sage implements the above algorithm for computing powers efficiently. For example,
sage: Mod(7,100)^91
43
We can also, of course, directly compute 791 in Sage, though we would notwant to do this by hand:sage: 7^9180153343160247310515380886994816022539378033762994852
007501964604841680190743

2.4 Primality Testing
Theorem 2.4.1 (Pseudoprimality). An integer p > 1 is prime if and only
if for every a 6≡ 0 (mod p),
ap−1 ≡ 1(mod p).

Proof. If p is prime, then the statement follows from Theorem 2.1.20. If p
is composite, then there is a divisor a of p with 2 ≤ a < p. If ap−1 ≡ 1(mod p), then p | ap−1 − 1. Since a | p, we have a | ap−1 − 1, hence there
exists an integer k such that ak = ap−1 − 1. Subtracting, we see thatap−1 − ak = 1, so a(ap−2 − k) = 1. This implies that a | 1, which is acontradiction since a ≥ 2.
Suppose n ∈ N. Using Theorem 2.4.1 and Algorithm 2.3.13, we can either
quickly prove that n is not prime, or convince ourselves that n is likely prime(but not quickly prove that n is prime). For example, if 2n−1 6≡ 1 (mod n),then we have proved that n is not prime. On the other hand, if an−1 ≡ 1(mod n) for a few a, it “seems likely” that n is prime, and we loosely referto such a number that seems prime for several bases as a pseudoprime.

2.4 Primality Testing

37

There are composite numbers n (called Carmichael numbers) with theamazing property that an−1 ≡ 1 (mod n) for all a with gcd(a, n) = 1. The
first Carmichael number is 561, and it is a theorem that there are infinitely
many such numbers ([AGP94]).
Example 2.4.2. Is p = 323 prime? We compute 2322 (mod 323). Making atable as above, we have
i
0
1
2
3
4
5
6
7
8

m
322
161
80
40
20
10
5
2
1

εi
0
1
0
0
0
0
1
0
1

i

22 mod 323
2
4
16
256
290
120
188
137
35

Thus
2322 ≡ 4 · 188 · 35 ≡ 157(mod 323),

so 323 is not prime, though this computation gives no information about
how 323 factors as a product of primes. In fact, one finds that 323 = 17 · 19.
SAGE Example 2.4.3. It’s possible to easily prove that a large number iscomposite, but the proof does not easily yield a factorization. For example
ifn = 95468093486093450983409583409850934850938459083,
then 2n−1 6≡ 1 (mod n), so n is composite.
sage: n = 95468093486093450983409583409850934850938459083
sage: Mod(2,n)^(n-1)
34173444139265553870830266378598407069248687241
Note that factoring n actually takes much longer than the above computation (which was essentially instant).
sage: factor(n)# takes up to a few seconds.
1610302526747 * 59285812386415488446397191791023889
Another practical primality test is the Miller-Rabin test, which has theproperty that each time it is run on a number n it either correctly assertsthat the number is definitely not prime, or that it is probably prime, andthe probability of correctness goes up with each successive call. If MillerRabin is called m times on n and in each case claims that n is probably
prime, then one can in a precise sense bound the probability that n iscomposite in terms of m.

38

2. The Ring of Integers Modulo n

We state the Miller-Rabin algorithm precisely, but do not prove anything
about the probability that it will succeed.
Algorithm 2.4.4 (Miller-Rabin Primality Test). Given an integer n ≥ 5this algorithm outputs either true or false. If it outputs true, then n is“probably prime,” and if it outputs false, then n is definitely composite.
1. [Split Off Power of 2] Compute the unique integers m and k suchthat m is odd and n − 1 = 2k · m.
2. [Random Base] Choose a random integer a with 1 < a < n.
3. [Odd Power] Set b = am (mod n). If b ≡ ±1 (mod n) output true
and terminate.
r

4. [Even Powers] If b2 ≡ −1 (mod n) for any r with 1 ≤ r ≤ k − 1,
output true and terminate. Otherwise output false.
If Miller-Rabin outputs true for n, we can call it again with n and if itagain outputs true then the probability that we have incorrectly determinedthat n is prime (when n is actually composite) decreases.
Proof. We will prove that the algorithm is correct, but will prove nothing
about how likely the algorithm is to assert that a composite is prime.
We must prove that if the algorithm pronounces an integer n composite,then n really is composite. Thus suppose n is prime, yet the algorithm
pronounces n composite. Then am 6≡ ±1 (mod n), and for all r with 1 ≤rr ≤ k − 1 we have a2 m 6≡ −1 (mod n). Since n is prime and 2k−1 m =k−1(n − 1)/2, Proposition 4.2.1 implies that a2 m ≡ ±1 (mod n), so by
k−1
k−2
our hypothesis a2 m ≡ 1 (mod n). But then (a2 m )2 ≡ 1 (mod n), so
by Proposition 2.5.3 (which is proved right after it is stated, and whosek−2
proof does not depend on this argument), we have a2 m ≡ ±1 (mod n).
k−2
Again, by our hypothesis, this implies a2 m ≡ 1 (mod n). Repeating thisargument inductively, we see that am ≡ ±1 (mod n), which contradicts
our hypothesis on a.
Until recently it was an open problem to give an algorithm (with proof)that decides whether or not any integer is prime in time bounded by a polynomial in the number of digits of the integer. Agrawal, Kayal, and Saxena
recently found the first polynomial-time primality test (see [AKS02]). We
will not discuss their algorithm further, because for our applications tocryptography Miller-Rabin or pseudoprimality tests will be sufficient. See[Sho05, Ch. 21] for a book that gives a detailed exposition of this algorithm.
SAGE Example 2.4.5. The is prime command uses a combination of techniques to determine (provably correctly!) whether or not an integer is prime.
sage: n = 95468093486093450983409583409850934850938459083
sage: is_prime(n)False2.5 The Structure of (Z/pZ)∗

39

We use the is prime function to make a table of the first few Mersenne
primes (see Section 1.2.3).
sage: for p in primes(100):...
if is_prime(2^p - 1):...
print p, 2^p - 1
2 3
3 7
5 31
7 127
13 8191
17 131071
19 524287
31 2147483647
61 2305843009213693951
89 618970019642690137449562111
There is a specialized test for primality of Mersenne numbers called theLucas-Lehmer test. This remarkably simple algorithm determines provably
correctly whether or not a number 2p − 1 is prime. We implement it in a
few lines of code and use the Lucas-Lehmer test to check for primality of
two Mersenne numbers:
sage: def is_prime_lucas_lehmer(p):...
s = Mod(4, 2^p - 1)...
for i in range(3, p+1):...
s = s^2 - 2...
return s == 0
sage: # Check primality of 2^9941 - 1
sage: is_prime_lucas_lehmer(9941)
True
sage: # Check primality of 2^next_prime(1000)-1
sage: is_prime_lucas_lehmer(next_prime(1000))
False
For more on Mersenne primes, see the Great Internet Mersenne Prime
Search (GIMPS) project at http://www.mersenne.org/.

2.5 The Structure of (Z/pZ)∗
This section is about the structure of the group (Z/pZ)∗ of units modulo
a prime number p. The main result is that this group is always cyclic. We
will use this result later in Chapter 4 in our proof of quadratic reciprocity.
Definition 2.5.1 (Primitive Root). A primitive root modulo an integer n
is an element of (Z/nZ)∗ of order ϕ(n).

40

2. The Ring of Integers Modulo n

We will prove that there is a primitive root modulo every prime p. Since
the unit group (Z/pZ)∗ has order p−1, this implies that (Z/pZ)∗ is a cyclic
group, a fact that will be extremely useful, since it completely determines
the structure of (Z/pZ)∗ as a group.
If n is an odd prime power, then there is a primitive root modulo n (seeExercise 2.28), but there is no primitive root modulo the prime power 23 ,and hence none mod 2n for n ≥ 3 (see Exercise 2.27).
Section 2.5.1 is the key input to our proof that (Z/pZ)∗ is cyclic; here
we show that for every divisor d of p − 1 there are exactly d elements of(Z/pZ)∗ whose order divides d. We then use this result in Section 2.5.2 toproduce an element of (Z/pZ)∗ of order q r when q r is a prime power thatexactly divides p − 1 (i.e., q r divides p − 1, but q r+1 does not divide p − 1),and multiply together these elements to obtain an element of (Z/pZ)∗ oforder p − 1.
SAGE Example 2.5.2. Use the primitive root command to compute thesmallest positive integer that is a primitive root modulo n. For example,
below we compute primitive roots modulo p for each prime p < 20.

sage: for p in primes(20):...
print p, primitive_root(p)
2 1
3 2
5 2
7 3
11 2
13 2
17 3
19 2

2.5.1

Polynomials over Z/pZ

The polynomials x2 − 1 has four roots in Z/8Z, namely 1, 3, 5, and 7.
In contrast, the following proposition shows that a polynomial of degree dover a field, such as Z/pZ, can have at most d roots.

Proposition 2.5.3 (Root Bound). Let f ∈ k[x] be a nonzero polynomial
over a field k. Then there are at most deg(f ) elements α ∈ k such thatf (α) = 0.

2.5 The Structure of (Z/pZ)∗

41

Proof. We prove the proposition by induction on deg(f ). The cases in whichdeg(f ) ≤ 1 are clear. Write f = an xn + · · · a1 x + a0 . If f (α) = 0, thenf (x) = f (x) − f (α)= an (xn − αn ) + · · · + a1 (x − α) + a0 (1 − 1)
= (x − α)(an (xn−1 + · · · + αn−1 ) + · · · + a2 (x + α) + a1 )= (x − α)g(x),for some polynomial g(x) ∈ k[x]. Next, suppose that f (β) = 0 with β 6= α.
Then (β −α)g(β) = 0, so, since β −α 6= 0 and k is a field, we have g(β) = 0.
By our inductive hypothesis, g has at most n − 1 roots, so there are at mostn − 1 possibilities for β. It follows that f has at most n roots.
SAGE Example 2.5.4. We use Sage to find the roots of a polynomials over
Z/13Z.
sage: R.<x> = PolynomialRing(Integers(13))
sage: f = x^15 + 1
sage: f.roots()[(12, 1), (10, 1), (4, 1)]
sage: f(12)0The output of the roots command above lists each root along with its
multiplicity (which is 1 in each case above).
Proposition 2.5.5. Let p be a prime number and let d be a divisor ofp − 1. Then f = xd − 1 ∈ (Z/pZ)[x] has exactly d roots in Z/pZ.
Proof. Let e = (p − 1)/d. We havexp−1 − 1 = (xd )e − 1= (xd − 1)((xd )e−1 + (xd )e−2 + · · · + 1)
= (xd − 1)g(x),where g ∈ (Z/pZ)[x] and deg(g) = de − d = p − 1 − d. Theorem 2.1.20
implies that xp−1 − 1 has exactly p − 1 roots in Z/pZ, since every nonzero
element of Z/pZ is a root! By Proposition 2.5.3, g has at most p − 1 − droots and xd − 1 has at most d roots. Since a root of (xd − 1)g(x) is a rootof either xd − 1 or g(x) and xp−1 − 1 has p − 1 roots, g must have exactlyp − 1 − d roots and xd − 1 must have exactly d roots, as claimed.
SAGE Example 2.5.6. We use Sage to illustrate the proposition.
sage: R.<x> = PolynomialRing(Integers(13))
sage: f = x^6 + 1
sage: f.roots()[(11, 1), (8, 1), (7, 1), (6, 1), (5, 1), (2, 1)]

42

2. The Ring of Integers Modulo n

We pause to reemphasize that the analog of Proposition 2.5.5 is false
when p is replaced by a composite integer n, since a root mod n of a
product of two polynomials need not be a root of either factor. For example,
f = x2 − 1 = (x − 1)(x + 1) ∈ Z/15Z[x] has the four roots 1, 4, 11, and 14.

2.5.2

Existence of Primitive Roots

Recall from Section 2.1.2 that the order of an element x in a finite groupis the smallest m ≥ 1 such that xm = 1. In this section, we prove that
(Z/pZ)∗ is cyclic by using the results of Section 2.5.1 to produce an elementof (Z/pZ)∗ of order d for each prime power divisor d of p − 1, and then we
multiply these together to obtain an element of order p − 1.
We will use the following lemma to assemble elements of each orderdividing p − 1 to produce an element of order p − 1.
Lemma 2.5.7. Suppose a, b ∈ (Z/nZ)∗ have orders r and s, respectively,
and that gcd(r, s) = 1. Then ab has order rs.
Proof. This is a general fact about commuting elements of any group; ourproof only uses that ab = ba and nothing special about (Z/nZ)∗ . Since(ab)rs = ars brs = 1,
the order of ab is a divisor of rs. Write this divisor as r1 s1 where r1 | r ands1 | s. Raise both sides of the equation
ar1 s1 br1 s1 = (ab)r1 s1 = 1
to the power r2 = r/r1 to obtainar1 r2 s1 br1 r2 s1 = 1.
Since ar1 r2 s1 = (ar1 r2 )s1 = 1, we have
br1 r2 s1 = 1,
so s | r1 r2 s1 . Since gcd(s, r1 r2 ) = gcd(s, r) = 1, it follows that s = s1 .
Similarly r = r1 , so the order of ab is rs.
Theorem 2.5.8 (Primitive Roots). There is a primitive root modulo any
prime p. In particular, the group (Z/pZ)∗ is cyclic.
Proof. The theorem is true if p = 2, since 1 is a primitive root, so we may
assume p > 2. Write p − 1 as a product of distinct prime powers qini :p − 1 = q1n1 q2n2 · · · qrnr .
ni

By Proposition 2.5.5, the polynomial xqi − 1 has exactly qini roots, andni −1
the polynomial xqi− 1 has exactly qini −1 roots. There are qini − qini −1 =

2.5 The Structure of (Z/pZ)∗
ni

43

ni −1

qini −1 (qi − 1) elements a ∈ Z/pZ such that aqi = 1 but aqi
6= 1; eachof these elements has order qini . Thus for each i = 1, . . . , r, we can choose
an ai of order qini . Then, using Lemma 2.5.7 repeatedly, we see thata = a1 a2 · · · ar
has order q1n1 · · · qrnr = p − 1, so a is a primitive root modulo p.
Example 2.5.9. We illustrate the proof of Theorem 2.5.8 when p = 13. We
havep − 1 = 12 = 22 · 3.
The polynomial x4 − 1 has roots {1, 5, 8, 12} and x2 − 1 has roots {1, 12},
so we may take a1 = 5. The polynomial x3 − 1 has roots {1, 3, 9}, and we
set a2 = 3. Then a = 5 · 3 = 15 ≡ 2 is a primitive root. To verify this, notethat the successive powers of 2 (mod 13) are
2, 4, 8, 3, 6, 12, 11, 9, 5, 10, 7, 1.
Example 2.5.10. Theorem 2.5.8 is false if, for example, p is replaced by apower of 2 bigger than 4. For example, the four elements of (Z/8Z)∗ eachhave order dividing 2, but ϕ(8) = 4.
Theorem 2.5.11 (Primitive Roots mod pn ). Let pn be a power of an odd
prime. Then there is a primitive root modulo pn .
The proof is left as Exercise 2.28.
Proposition 2.5.12 (Number of Primitive Roots). If there is a primitiveroot modulo n, then there are exactly ϕ(ϕ(n)) primitive roots modulo n.
Proof. The primitive roots modulo n are the generators of (Z/nZ)∗ , whichby assumption is cyclic of order ϕ(n). Thus they are in bijection with thegenerators of any cyclic group of order ϕ(n). In particular, the number ofprimitive roots modulo n is the same as the number of elements of Z/ϕ(n)Zwith additive order ϕ(n). An element of Z/ϕ(n)Z has additive order ϕ(n)if and only if it is coprime to ϕ(n). There are ϕ(ϕ(n)) such elements, as
claimed.
Example 2.5.13. For example, there are ϕ(ϕ(17)) = ϕ(16) = 24 − 23 =8 primitive roots mod 17, namely 3, 5, 6, 7, 10, 11, 12, 14. The ϕ(ϕ(9)) =ϕ(6) = 2 primitive roots modulo 9 are 2 and 5. There are no primitiveroots modulo 8, even though ϕ(ϕ(8)) = ϕ(4) = 2 > 0.

2.5.3

Artin’s Conjecture

Conjecture 2.5.14 (Emil Artin). Suppose a ∈ Z is not −1 or a perfectsquare. Then there are infinitely many primes p such that a is a primitiveroot modulo p.

44

2. The Ring of Integers Modulo n

There is no single integer a such that Artin’s conjecture is known tobe true. For any given a, Pieter [Mor93] proved that there are infinitely
many p such that the order of a is divisible by the largest prime factor
of p − 1. Hooley [Hoo67] proved that something called the Generalized
Riemann Hypothesis implies Conjecture 2.5.14.
Remark 2.5.15. Artin conjectured more precisely that if N (x, a) is thenumber of primes p ≤ x such that a is a primitive root modulo p, thenN (x, a) is asymptotic to C(a)π(x), where C(a) is a positive constant that
depends only on a and π(x) is the number of primes up to x.

2.5.4

Computing Primitive Roots

Theorem 2.5.8 does not suggest an efficient algorithm for finding primitiveroots. To actually find a primitive root mod p in practice, we try a = 2,
then a = 3, etc., until we find an a that has order p − 1. Computing theorder of an element of (Z/pZ)∗ requires factoring p − 1, which we do notknow how to do quickly in general, so finding a primitive root modulo p
for large p seems to be a difficult problem.
Algorithm 2.5.16 (Primitive Root). Given a prime p, this algorithm computes the smallest positive integer a that generates (Z/pZ)∗ .
1. [p = 2?] If p = 2 output 1 and terminate. Otherwise set a = 2.
2. [Prime Divisors] Compute the prime divisors p1 , . . . , pr of p − 1.
3. [Generator?] If for every pi , we have a(p−1)/pi 6≡ 1 (mod p), then a isa generator of (Z/pZ)∗ , so output a and terminate.
4. [Try next] Set a = a + 1 and go to Step 3.
Proof. Let a ∈ (Z/pZ)∗ . The order of a is a divisor d of the order p − 1 ofthe group (Z/pZ)∗ . Write d = (p − 1)/n, for some divisor n of p − 1. If a is
not a generator of (Z/pZ)∗ , then since n | (p − 1), there is a prime divisor
pi of p − 1 such that pi | n. Thena(p−1)/pi = (a(p−1)/n )n/pi ≡ 1(mod p).

Conversely, if a is a generator, then a(p−1)/pi 6≡ 1 (mod p) for any pi . Thus
the algorithm terminates with Step 3 if and only if the a under consideration
is a primitive root. By Theorem 2.5.8, there is at least one primitive root,
so the algorithm terminates.

2.6 Exercises
2.1 Prove that for any positive integer n, the set (Z/nZ)∗ under multiplication modulo n is a group.

2.6 Exercises

45

2.2 Compute the following gcd’s using Algorithm 1.1.13:
gcd(15, 35)

gcd(247, 299)

gcd(51, 897)

gcd(136, 304)

2.3 Use Algorithm 2.3.7 to find x, y ∈ Z such that 2261x + 1275y = 17.
2.4 Prove that if a and b are integers and p is a prime, then (a + b)p ≡ap + bp (mod p). You may assume that the binomial coefficientp!
r!(p − r)!
is an integer.
2.5 (a) Prove that if x, y is a solution to ax + by = d, with d = gcd(a, b),then for all c ∈ Z,bx0 = x + c · ,dy0 = y − c ·ad(2.6.1)is also a solution to ax + by = d.
(b) Find two distinct solutions to 2261x + 1275y = 17.
(c) Prove that all solutions are of the form (2.6.1) for some c.
2.6 Let f (x) = x2 + ax + b ∈ Z[x] be a quadratic polynomial with integer
coefficients, for example, f (x) = x2 + x + 6. Formulate a conjecture
about when the set{f (n) : n ∈ Z and f (n) is prime}is infinite. Give numerical evidence that supports your conjecture.
2.7 Find four complete sets of residues modulo 7, where the ith set satisfies the ith condition: (1) nonnegative, (2) odd, (3) even, (4) prime.
2.8 Find rules in the spirit of Proposition 2.1.9 for divisibility of an integer
by 5, 9, and 11, and prove each of these rules using arithmetic modulo
a suitable n.
2.9 (*) (The following problem is from the 1998 Putnam Competition.)
Define a sequence of decimal integers an as follows: a1 = 0, a2 =1, and an+2 is obtained by writing the digits of an+1 immediately
followed by those of an . For example, a3 = 10, a4 = 101, and a5 =10110. Determine the n such that an is a multiple of 11, as follows:(a) Find the smallest integer n > 1 such that an is divisible by 11.
(b) Prove that an is divisible by 11 if and only if n ≡ 1 (mod 6).
2.10 Find an integer x such that 37x ≡ 1 (mod 101).

46

2. The Ring of Integers Modulo n

2.11 What is the order of 2 modulo 17?
2.12 Let p be a prime. Prove that Z/pZ is a field.
2.13 Find an x ∈ Z such that x ≡ −4 (mod 17) and x ≡ 3 (mod 23).
2.14 Prove that if n > 4 is composite then(n − 1)! ≡ 0(mod n).

2.15 For what values of n is ϕ(n) odd?
2.16 (a) Prove that ϕ is multiplicative as follows. Suppose m, n are positive integers and gcd(m, n) = 1. Show that the natural map
ψ : Z/mnZ → Z/mZ × Z/nZ is an injective homomorphism ofrings, hence bijective by counting, then look at unit groups.
(b) Prove conversely that if gcd(m, n) > 1, then the natural map
ψ : Z/mnZ → Z/mZ × Z/nZ is not an isomorphism.
2.17 Seven competitive math students try to share a huge hoard of stolen
math books equally between themselves. Unfortunately, six books areleft over, and in the fight over them, one math student is expelled.
The remaining six math students, still unable to share the math books
equally since two are left over, again fight, and another is expelled.
When the remaining five share the books, one book is left over, and
it is only after yet another math student is expelled that an equal
sharing is possible. What is the minimum number of books that allows
this to happen?
2.18 Show that if p is a positive integer such that both p and p2 + 2 areprime, then p = 3.
2.19 Let ϕ : N → N be the Euler ϕ function.
(a) Find all natural numbers n such that ϕ(n) = 1.
(b) Do there exist natural numbers m and n such that ϕ(mn) 6=
ϕ(m) · ϕ(n)?
2.20 Find a formula for ϕ(n) directly in terms of the prime factorizationof n.
2.21 (a) Prove that if ϕ : G → H is a group homomorphism, then ker(ϕ)
is a subgroup of G.
(b) Prove that ker(ϕ) is normal, i.e., if a ∈ G and b ∈ ker(ϕ), thena−1 ba ∈ ker(ϕ).
2.22 Is the set Z/5Z = {0, 1, 2, 3, 4} with binary operation multiplication
modulo 5 a group?

2.6 Exercises

47

2.23 Find all four solutions to the equationx2 − 1 ≡ 0(mod 35).

2.24 Prove that for any positive integer n the fraction (12n + 1)/(30n + 2)
is in reduced form.
2.25 Suppose a and b are positive integers.
(a) Prove that gcd(2a − 1, 2b − 1) = 2gcd(a,b) − 1.
(b) Does it matter if 2 is replaced by an arbitrary prime p?
(c) What if 2 is replaced by an arbitrary positive integer n?
2.26 For every positive integer b, show that there exists a positive integern such that the polynomial x2 − 1 ∈ (Z/nZ)[x] has at least b roots.
2.27 (a) Prove that there is no primitive root modulo 2n for any n ≥ 3.
(b) (*) Prove that (Z/2n Z)∗ is generated by −1 and 5.
2.28 Let p be an odd prime.
(a) (*) Prove that there is a primitive root modulo p2 . (Hint: Usethat if a, b have orders n, m, with gcd(n, m) = 1, then ab hasorder nm.)(b) Prove that for any n, there is a primitive root modulo pn .
(c) Explicitly find a primitive root modulo 125.
2.29 (*) In terms of the prime factorization of n, characterize the integers nsuch that there is a primitive root modulo n.
2.30 Compute the last two digits of 345 .
2.31 Find the integer a such that 0 ≤ a < 113 and10270 + 1 ≡ a37(mod 113).

2.32 Find the proportion of primes p < 1000 such that 2 is a primitiveroot modulo p.
2.33 Find a prime p such that the smallest primitive root modulo p is 37.

48

2. The Ring of Integers Modulo nThis is page 49
Printer: Opaque this3
Public-key Cryptography

In the 1970s, techniques from number theory changed the world foreverby providing, for the first time ever, a way for two people to communicate
secret messages under the assumption that all of their communication isintercepted and read by an adversary. This idea has stood the test of time.
In fact, whenever you buy something online, you use such a system, which
typically involves working in the ring of integers modulo n. This chapter
tells the story of several such systems.

3.1 Playing with Fire
I recently watched a TV show called La Femme Nikita about a woman
named Nikita who is forced to be an agent for a shady anti-terrorist organization called Section One. Nikita has strong feelings for fellow agent
Michael, and she most trusts Walter, Section One’s ex-biker gadgets and explosives expert. Often Nikita’s worst enemies are her superiors and coworkers at Section One. A synopsis for a Season Three episode is as follows:
PLAYING WITH FIRE
On a mission to secure detonation chips from a terrorist organization’s heavily armed base camp, Nikita is captured as a
hostage by the enemy. Or so it is made to look. Michael and
Nikita have actually created the scenario in order to secretly
rendezvous with each other. The ruse works, but when Birkoff

50

3. Public-key Cryptography

FIGURE 3.1. Diffie and Hellman (photos from [Sin99])[Section One’s master hacker] accidentally discovers encrypted
messages between Michael and Nikita sent with Walter’s help,
Birkoff is forced to tell Madeline. Suspecting that Michael and
Nikita may be planning a coup d’état, Operations and Madeline
use a second team of operatives to track Michael and Nikita’s
next secret rendezvous... killing them if necessary.
What sort of encryption might Walter have helped them to use? I let
my imagination run free, and this is what I came up with. After being
captured at the base camp, Nikita is given a phone by her captors in hopesthat she’ll use it and they’ll be able to figure out what she is really up to.
Everyone is eagerly listening in on her calls.
Remark 3.1.1. In this book, we will assume a method is available for producing random integers. Methods for generating random integers are involved and interesting, but we will not discuss them in this book. For an
in-depth treatment of random numbers, see [Knu98, Ch. 3].
Nikita remembers a conversation with Walter about a public-key cryptosystem called the “Diffie-Hellman key exchange.” She remembers that itallows two people to agree on a secret key in the presence of eavesdroppers.
Moreover, Walter mentioned that though Diffie-Hellman was the first everpublic-key exchange system, it is still in common use today (for example,
in OpenSSH protocol version 2, see http://www.openssh.com/).
Nikita pulls out her handheld computer and phone, calls up Michael, andthey do the following, which is wrong (try to figure out what is wrong as
you read it).
1. Together they choose a big prime number p and a number g with
1 < g < p.
2. Nikita secretly chooses an integer n.

3.2 The Diffie-Hellman Key Exchange

51

3. Michael secretly chooses an integer m.
4. Nikita tells Michael ng (mod p).
5. Michael tells mg (mod p) to Nikita.
6. The “secret key” is s = nmg (mod p), which both Nikita and Michael
can easily compute.
Here’s a very simple example with small numbers that illustrates what
Michael and Nikita do. (They really used much larger numbers.)
1. p = 97, g = 5
2. n = 31
3. m = 95
4. ng ≡ 58 (mod 97)
5. mg ≡ 87 (mod 97)
6. s = nmg = 78 (mod 97)
Nikita and Michael are foiled because everyone easily figures out s:
1. Everyone knows p, g, ng (mod p), and mg (mod p).
2. Using Algorithm 2.3.7, anyone can easily find a, b ∈ Z such thatag + bp = 1, which exists because gcd(g, p) = 1.
3. Then, ang ≡ n (mod p), so everyone knows Nikita’s secret key n,and hence can easily compute the shared secret s.
To taunt her, Nikita’s captors give her a paragraph from a review of Diffieand Hellman’s 1976 paper “New Directions in Cryptography” [DH76]:“The authors discuss some recent results in communications
theory [...] The first [method] has the feature that an unauthorized ‘eavesdropper’ will find it computationally infeasible to decipher the message [...] They propose a couple of techniques for
implementing the system, but the reviewer was unconvinced.”

3.2 The Diffie-Hellman Key Exchange
As night darkens Nikita’s cell, she reflects on what has happened. Upon realizing that she mis-remembered how the system works, she phones Michael
and they do the following:52

3. Public-key Cryptography

1. Together Michael and Nikita choose a 200-digit integer p that is likely
to be prime (see Section 2.4), and choose a number g with 1 < g < p.
2. Nikita secretly chooses an integer n.
3. Michael secretly chooses an integer m.
4. Nikita computes g n (mod p) on her handheld computer and tells
Michael the resulting number over the phone.
5. Michael tells Nikita g m (mod p).
6. The shared secret key is thens ≡ (g n )m ≡ (g m )n ≡ g nm

(mod p),which both Nikita and Michael can compute.
Here is a simplified example that illustrates what they did, that involves
only relatively simple arithmetic.
1. p = 97, g = 5
2. n = 31
3. m = 95
4. g n ≡ 7 (mod p)
5. g m ≡ 39 (mod p)
6. s ≡ (g n )m ≡ 14 (mod p)

3.2.1

The Discrete Log Problem

Nikita communicates with Michael by encrypting everything using theiragreed upon secret key (for example, using a standard symmetric cipher
such as AES, Arcfour, Cast128, 3DES, or Blowfish). In order to understandthe conversation, the eavesdropper needs s, but it takes a long time tocompute s given only p, g, g n , and g m . One way would be to compute n fromknowledge of g and g n ; this is possible, but appears to be “computationallyinfeasible,” in the sense that it would take too long to be practical.
Let a, b, and n be real numbers with a, b > 0 and n ≥ 0. Recall that the“log to the base b” function is characterized bylogb (a) = n if and only if a = bn .
We use the logb function in algebra to solve the following problem: Givena base b and a power a of b, find an exponent n such thata = bn .

3.2 The Diffie-Hellman Key Exchange

53

That is, given a = bn and b, find n.
SAGE Example 3.2.1. The number a = 19683 is the nth power of b = 3
for some n. We quickly find thatn = log3 (19683) = log(19683)/ log(3) = 9.
sage: log(19683.0)
9.88751059801299
sage: log(3.0)
1.09861228866811
sage: log(19683.0) / log(3.0)
9.00000000000000
Sage can quickly compute a numerical approximation for log(x), for any x,
by computing a partial sum of an appropriate rapidly-converging infinite
series (at least for x in a certain range).
The discrete log problem is the analog of computing logb (a) but where
both b and a are elements of a finite group.
Problem 3.2.2 (Discrete Log Problem). Let G be a finite group, for example, G = (Z/pZ)∗ . Given b ∈ G and a power a of b, find a positive
integer n such that bn = a.
As far as we know, finding discrete logarithms in (Z/pZ)∗ when p islarge is “very difficult” in practice. Over the years, many people have beenvery motivated to try. For example, if Nikita’s captors could efficiently
solve Problem 3.2.2, then they could read the messages she exchanges withMichael. Unfortunately, we have no formal proof that computing discrete
logarithms on a classical computer is difficult. Also, Peter Shor [Sho97]
showed that if one could build a sufficiently complicated quantum computer, it could solve the discrete logarithm problem in time bounded by apolynomial function of the number of digits of #G.
It is easy to give an inefficient algorithm that solves the discrete log
problem. Simply try b1 , b2 , b3 , etc., until we find an exponent n such thatbn = a. For example, suppose a = 18, b = 5, and p = 23. Working modulo
23, we haveb1 = 5, b2 = 2, b3 = 10, . . . , b12 = 18,
so n = 12. When p is large, computing the discrete log this way soon becomes impractical, because increasing the number of digits of the modulus
makes the computation take vastly longer.
SAGE Example 3.2.3. Perhaps part of the reason that computing discrete
logarithms is difficult, is that the logarithm in the real numbers is continuous, but the (minimum) logarithm of a number mod n bounces around at
random. We illustrate this exotic behavior in Figure 3.2.
This draws the continuous plot.

54

3. Public-key Cryptography
100

2.5
2
1.5
1
0.5
-0.5
-1
-1.5
-2
-2.5

75

2.5

5

7.5

10

50
25

25

50

75

100

FIGURE 3.2. Graphs of the continuous log and of the discrete log modulo 53.
Which picture looks easier to predict?

sage: plot(log, 0.1,10, rgbcolor=(0,0,1))This draws the discrete plot.
sage: p = 53
sage: R = Integers(p)sage: a = R.multiplicative_generator()sage: v = sorted([(a^n, n) for n in range(p-1)])
sage: G = plot(point(v,pointsize=50,rgbcolor=(0,0,1)))
sage: H = plot(line(v,rgbcolor=(0.5,0.5,0.5)))
sage: G + H

3.2.2

Realistic Diffie-Hellman Example

In this section, we present an example that uses bigger numbers. First, we
prove a proposition that we can use to choose a prime p in such a way that
it is easy to find a g ∈ (Z/pZ)∗ with order p − 1. We have already seen in
Section 2.5 that for every prime p there exists an element g of order p − 1,and we gave Algorithm 2.5.16 for finding a primitive root for any prime.
The significance of Proposition 3.2.4 below is that it suggests an algorithm
for finding a primitive root that is easier to use in practice when p is large,
because it does not require factoring p−1. Of course, one could also just use
a random g for Diffie-Hellman; it is not essential that g generates (Z/pZ)∗ .
Proposition 3.2.4. Suppose p is a prime such that (p−1)/2 is also prime.
Then each element of (Z/pZ)∗ has order one of 1, 2, (p − 1)/2, or p − 1.
Proof. Since p is prime, the group (Z/pZ)∗ is of order p−1. By assumption,
the prime factorization of p − 1 is 2 · ((p − 1)/2). Let a ∈ (Z/pZ)∗ . Thenby Theorem 2.1.20, ap−1 = 1, so the order of a is a divisor of p − 1, which
proves the proposition.

3.2 The Diffie-Hellman Key Exchange

55

Given a prime p with (p − 1)/2 prime, find an element of order p − 1 asfollows. If 2 has order p − 1, we are done. If not, 2 has order (p − 1)/2 since
2 does not have order either 1 or 2. Then −2 has order p − 1.
Let p = 93450983094850938450983409611. Then p is prime, but (p −1)/2 is not. So we keep adding 2 to p and testing pseudoprimality using
algorithms from Section 2.4 until we find that the next pseudoprime after pisq = 93450983094850938450983409623.
It turns out that q pseudoprime and (q − 1)/2 is also pseudoprime. We findthat 2 has order (q − 1)/2, so g = −2 has order q − 1 modulo q, and ishence a generator of (Z/qZ)∗ , at least assuming that q is really prime.
The secret random numbers generated by Nikita and Michael aren = 18319922375531859171613379181
andm = 82335836243866695680141440300.
Nikita sends
g n = 45416776270485369791375944998 ∈ (Z/pZ)∗
to Michael, and Michael sends
g m = 15048074151770884271824225393 ∈ (Z/pZ)∗
to Nikita. They agree on the secret key
g nm = 85771409470770521212346739540 ∈ (Z/pZ)∗ .
SAGE Example 3.2.5. We illustrate the above computations using Sage.
sage: q = 93450983094850938450983409623
sage: q.is_prime()True
sage: is_prime((q-1)//2)
True
sage: g = Mod(-2, q)
sage: g.multiplicative_order()93450983094850938450983409622
sage: n = 18319922375531859171613379181
sage: m = 82335836243866695680141440300
sage: g^n45416776270485369791375944998
sage: g^m
15048074151770884271824225393
sage: (g^n)^m
85771409470770521212346739540
sage: (g^m)^n
85771409470770521212346739540

56

3. Public-key Cryptography

3.2.3

The Man in the Middle Attack

Since their first system was broken, instead of talking on the phone, Michael
and Nikita can now only communicate via text messages. One of her captors, The Man, is watching each of the transmissions; moreover, he canintercept messages and send false messages. When Nikita sends a message to Michael announcing g n (mod p), The Man intercepts this message,
and sends his own number g t (mod p) to Michael. Eventually, Michael and
The Man agree on the secret key g tm (mod p), and Nikita and The Man
agree on the key g tn (mod p). When Nikita sends a message to Michael she
unwittingly uses the secret key g tn (mod p); The Man then intercepts it,
decrypts it, changes it, and re-encrypts it using the key g tm (mod p), andsends it on to Michael. This is bad because now The Man can read every
message sent between Michael and Nikita, and moreover, he can change
them in transmission in subtle ways.
One way to get around this attack is to use a digital signature scheme
based on the RSA cryptosystem. We will not discuss digital signatures
further in this book, but will discuss RSA in the next section.

3.3 The RSA Cryptosystem
The Diffie-Hellman key exchange has drawbacks. As discussed in Section
3.2.3, it is susceptible to the man in the middle attack. This section isabout the RSA public-key cryptosystem of Rivest, Shamir, and Adleman[RSA78], which is an alternative to Diffie-Hellman that is more flexible insome ways.
We first describe the RSA cryptosystem, then discuss several ways toattack it. It is important to be aware of such weaknesses, in order to avoid
foolish mistakes when implementing RSA. We barely scratched the surface
here of the many possible attacks on specific implementations of RSA orother cryptosystems.

3.3.1

How RSA works

The fundamental idea behind RSA is to try to construct a trap-door or
one-way function on a set X. This is an invertible function
E:X→X
such that it is easy for Nikita to compute E −1 , but extremely difficult for
anybody else to do so.
Here is how Nikita makes a one-way function E on the set of integers
modulo n.
1. Using a method hinted at in Section 2.4, Nikita picks two large
primes p and q, and lets n = pq.

3.3 The RSA Cryptosystem

57

2. It is then easy for Nikita to computeϕ(n) = ϕ(p) · ϕ(q) = (p − 1) · (q − 1).
3. Nikita next chooses a random integer e with
1 < e < ϕ(n) and gcd(e, ϕ(n)) = 1.
4. Nikita uses the algorithm from Section 2.3.2 to find a solution x = d
to the equationex ≡ 1 (mod ϕ(n)).
5. Finally, Nikita defines a function E : Z/nZ → Z/nZ by
E(x) = xe ∈ Z/nZ.
Note that anybody can compute E fairly quickly using the repeatedsquaring algorithm from Section 2.3.2. Nikita’s public key is the pair of
integers (n, e), which is just enough information for people to easily compute E. Nikita knows a number d such that ed ≡ 1 (mod ϕ(n)), so, as we
will see, she can quickly compute E −1 .
To send Nikita a message, proceed as follows. Encode your message, insome way, as a sequence of numbers modulo n (see Section 3.3.2)
m1 , . . . , mr ∈ Z/nZ,
then sendE(m1 ), . . . , E(mr )to Nikita. (Recall that E(m) = me for m ∈ Z/nZ.)
When Nikita receives E(mi ), she finds each mi by using that E −1 (m) =d
m , a fact that follows from Proposition 3.3.1
Proposition 3.3.1 (Decryption Key). Let n be an integer that is a productof distinct primes and let d, e ∈ N be such that p − 1 | de − 1 for each prime
p | n. Then ade ≡ a (mod n) for all a ∈ Z.
Proof. Since n | ade − a, if and only if p | ade − a for each prime divisor pof n, it suffices to prove that ade ≡ a (mod p) for each prime divisor p of n.
If gcd(a, p) 6= 1, then a ≡ 0 (mod p), so ade ≡ a (mod p). If gcd(a, p) = 1,
then Theorem 2.1.20 asserts that ap−1 ≡ 1 (mod p). Since p − 1 | de − 1,
we have ade−1 ≡ 1 (mod p) as well. Multiplying both sides by a shows thatade ≡ a (mod p).
Thus to decrypt E(mi ) Nikita computes
E(mi )d = (mei )d = mi .

58

3. Public-key Cryptography

SAGE Example 3.3.2. We implement the RSA cryptosystem using Sage.
The rsa function creates a key with (at most) the given number of bits,i.e., if bits equals 20, it creates a key n = pq such that n is approximately
220 . Typical real-life cryptosystems would choose keys that are 512, 1024,
or 2048 bits long. Try generating large keys yourself using Sage; how long
does it take?
sage: def rsa(bits):...
# only prove correctness up to 1024 bits...
proof = (bits <= 1024)...
p = next_prime(ZZ.random_element(2**(bits//2 +1)),...
proof=proof)...
q = next_prime(ZZ.random_element(2**(bits//2 +1)),...
proof=proof)...
n = p * q...
phi_n = (p-1) * (q-1)...
while True:...
e = ZZ.random_element(1,phi_n)...
if gcd(e,phi_n) == 1: break...
d = lift(Mod(e,phi_n)^(-1))...
return e, d, n...
sage: def encrypt(m,e,n):...
return lift(Mod(m,n)^e)...
sage: def decrypt(c,d,n):...
return lift(Mod(c,n)^d)...
sage: e,d,n = rsa(20)sage: c = encrypt(123, e, n)sage: decrypt(c, d, n)
123

3.3.2

Encoding a Phrase in a Number

In order to use the RSA cryptosystem to encrypt messages, it is necessary
to encode them as a sequence of numbers of size less than n = pq. We
now describe a simple way to do this. Note that in any actual deployed
implementation, it is crucial that you add extra random characters (“salt”)
at the beginning of each block of the message, so that the same plain text
encodes differently each time. This helps thwart chosen plain text attacks.
Suppose s is a sequence of capital letters and spaces, and that s does not
begin with a space. We encode s as a number in base 27 as follows: a single
space corresponds to 0, the letter A to 1, B to 2, . . ., Z to 26. Thus “RUN

3.3 The RSA Cryptosystem

59

NIKITA” is a number written in base 27.
RUN NIKITA↔ 279 · 18 + 278 · 21 + 277 · 14 + 276 · 0 + 275 · 14
+ 274 · 9 + 273 · 11 + 272 · 9 + 27 · 20 + 1= 143338425831991 (in decimal).

To recover the letters from the decimal number, repeatedly divide by 27
and read off the letter corresponding to each remainder.
143338425831991
5308830586370
196623355050
7282346483
269716536
9989501
369981
13703
507
18

= 5308830586370 · 27=
196623355050 · 27=
7282346483 · 27=
269716536 · 27=
9989501 · 27=
369981 · 27=
13703 · 27=
507 · 27=
18 · 27=
0 · 27

+
+
+
+
+
+
+
+
+
+

1
20
9
11
9
14
0
14
21
18

“A”
“T”
“I”
“K”
“I”
“N”
“”
“N”
“U”
“R”

If 27k ≤ n, then any sequence of k letters can be encoded as above using
a positive integer ≤ n. Thus if we can encrypt integers of size at most n,then we must break our message up into blocks of size at most log27 (n).
SAGE Example 3.3.3. We use Sage to implement conversion between astring and a number, though in a bit more generally than in the toy illustration above (which used only base 27). The input string s on a computeris stored in a format called ASCII, so each “letter” corresponds to an integer between 0 and 255, inclusive. This number is obtained from the letterusing the ord command.
sage: def encode(s):...
s = str(s)# make input a string...
return sum(ord(s[i])*256^i for i in range(len(s)))
sage: def decode(n):...
n = Integer(n) # make input an integer...
v = []...
while n != 0:...
v.append(chr(n % 256))...
n //= 256
# this replaces n by floor(n/256).
...
return ’’.join(v)sage: m = encode(’Run Nikita!’); m40354769014714649421968722
sage: decode(m)’Run Nikita!’

60

3. Public-key Cryptography

3.3.3

Some Complete Examples

To make the arithmetic easier to follow, we use small prime numbers p and q
and encrypt the single letter “X” using the RSA cryptosystem. First, we
compute the parameters of an RSA cryptosystem.
1. Choose p and q: Let p = 17, q = 19, so n = pq = 323.
2. Compute ϕ(n):ϕ(n) = ϕ(p · q) = ϕ(p) · ϕ(q) = (p − 1)(q − 1)
= pq − p − q + 1 = 323 − 17 − 19 + 1 = 288.
3. Randomly choose an e < 288: We choose e = 95.
4. Solve
95x ≡ 1(mod 288).

Using the GCD algorithm, we find that d = 191 solves the equation.
We have thus computed the parameters of an RSA public key cryptosystem.
The public key is (323, 95), so the encryption function isE(x) = x95 ,and the decryption function is D(x) = x191 .
Next, we encrypt the letter “X”. It is encoded as the number 24, since X
is the 24th letter of the alphabet. We haveE(24) = 2495 = 294 ∈ Z/323Z.
To decrypt, we compute E −1 :E −1 (294) = 294191 = 24 ∈ Z/323Z.
This next example illustrates RSA but with bigger numbers. Let
p = 738873402423833494183027176953, q = 3787776806865662882378273.
Then,n = p · q = 2798687536910915970127263606347911460948554197853542169
andϕ(n) = (p − 1)(q − 1)
= 2798687536910915970127262867470721260308194351943986944.

3.4 Attacking RSA

61

Using a pseudo-random number generator on a computer, the author randomly chose the integere = 1483959194866204179348536010284716655442139024915720699.
Then,
d = 2113367928496305469541348387088632973457802358781610803
Since log27 (n) ≈ 38.04, we can encode then encrypt single blocks of
up to 38 letters. Let’s encrypt the string RUN NIKITA, which encodes as
m = 143338425831991. We haveE(m) = me= 1504554432996568133393088878600948101773726800878873990.
Remark 3.3.4. In practice, one usually choses e to be small, since that does
not seem to reduce the security of RSA, and makes the key size smaller. For
example, in the OpenSSL documentation (see http://www.openssl.org/)
about their implementation of RSA, it states that “The exponent is an odd
number, typically 3, 17 or 65537.”

3.4 Attacking RSA
Suppose Nikita’s public key is (n, e) and her decryption key is d, so ed ≡ 1(mod ϕ(n)). If somehow we compute the factorization n = pq, then we cancompute ϕ(n) = (p−1)(q−1) and hence compute d. Thus, if we can factor nthen we can break the corresponding RSA public-key cryptosystem.

3.4.1

Factoring n Given ϕ(n)Suppose n = pq. Given ϕ(n), it is very easy to compute p and q. We haveϕ(n) = (p − 1)(q − 1) = pq − (p + q) + 1,
so we know both pq = n and p + q = n + 1 − ϕ(n). Thus, we know thepolynomial
x2 − (p + q)x + pq = (x − p)(x − q)whose roots are p and q. These roots can be found using the quadratic
formula.
Example 3.4.1. The number n = pq = 31615577110997599711 is a productof two primes, and ϕ(n) = 31615577098574867424. We havef = x2 − (n + 1 − ϕ(n))x + n= x2 − 12422732288x + 31615577110997599711= (x − 3572144239)(x − 8850588049),

62

3. Public-key Cryptography

where the factorization step is easily accomplished using the quadratic
formula:√−b + b2 − 4ac
2a√12422732288 + 124227322882 − 4 · 31615577110997599711=
2= 8850588049.
We conclude that n = 3572144239 · 8850588049.
SAGE Example 3.4.2. The following Sage function factors n = pq given n
and ϕ(n).
sage: def crack_rsa(n, phi_n):...
R.<x> = PolynomialRing(QQ)...
f = x^2 - (n+1 -phi_n)*x + n...
return [b for b, _ in f.roots()]
sage: crack_rsa(31615577110997599711, 31615577098574867424)[8850588049, 3572144239]

3.4.2

When p and q are Close

Suppose that p and q are “close” to each other. Then it is easy to factor nusing a factorization method of Fermat called the Fermat Factorization
Method.
Suppose n = pq with p > q. Then,
2 
2

p−q
p+q
−
.
n=
2
2
Since p and q are “close,”
s=

p−q
2

is small,

p+q
2
√
2
is only slightly larger than n, and t − n = s2 is a perfect square. So, wejust try√√√t = d ne, t = d ne + 1, t = d ne + 2, . . .
t=until t2 −n is a perfect square s2 . (Here dxe denotes the least integer n ≥ x.)
Then
p = t + s,
q = t − s.
Example 3.4.3. Suppose n = 23360947609. Then√n = 152842.88 . . . .

3.4 Attacking RSA

63√
If t = 152843, then √t2 − n = 187.18 . . ..
If t = 152844, then √t2 − n = 583.71 . . ..
If t = 152845, then t2 − n = 804 ∈ Z.
Thus s = 804. We find that p = t + s = 153649 and q = t − s = 152041.
SAGE Example 3.4.4. We implement the above algorithm√ for factoring an
RSA modulus n = pq, when one of p and q is close to n.
sage: def crack_when_pq_close(n):...
t = Integer(ceil(sqrt(n)))...
while True:...
k = t^2 - n...
if k > 0:...
s = Integer(int(round(sqrt(t^2 - n))))...
if s^2 + n == t^2:...
return t+s, t-s...
...
t += 1...
sage: crack_when_pq_close(23360947609)(153649, 152041)
For example, you might think that choosing a random prime, and thenext prime after would be a good idea, but instead it creates an easy-tocrack cryptosystem.
sage: p = next_prime(2^128); p340282366920938463463374607431768211507
sage: q = next_prime(p)sage: crack_when_pq_close(p*q)(340282366920938463463374607431768211537,
340282366920938463463374607431768211507)

3.4.3

Factoring n Given d

In this section, we show that finding the decryption key d for an RSA
cryptosystem is, in practice, at least as difficult as factoring n. We give aprobabilistic algorithm that given a decryption key determines the factorization of n.
Consider an RSA cryptosystem with modulus n and encryption key e.
Suppose we somehow finding an integer d such thataed ≡ a(mod n)for all a. Then m = ed − 1 satisfies am ≡ 1 (mod n) for all a that are
coprime to n. As we saw in Section 3.4.1, knowing ϕ(n) leads directly to afactorization of n. Unfortunately, knowing d does not seem to lead easily to64

3. Public-key Cryptographya factorization of n. However, there is a probabilistic procedure that, given
an m such that am ≡ 1 (mod n), will find a factorization of n with “highprobability” (we will not analyze the probability here).
Algorithm 3.4.5 (Probabilistic Algorithm to Factor n). Let n = pq be
the product of two distinct odd primes, and suppose m is an integer suchthat am ≡ 1 (mod n) for all a coprime to n. This probabilistic algorithm
factors n with “high probability.” In the steps below, a always denotes aninteger coprime to n = pq.
1. [Divide out powers of 2] If m is even and am/2 ≡ 1 (mod n) for several
randomly chosen a, set m = m/2, and go to Step 1, otherwise let abe such that am/2 6≡ 1 (mod n).
2. [Compute GCD] Choose a random a and compute g = gcd(am/2 −1, n).
3. [Terminate?] If g is a proper divisor of n, output g and terminate.
Otherwise go to Step 2.
Before giving the proof, we introduce some more terminology from algebra.
Definition 3.4.6 (Group Homomorphism). Let G and H be groups. A
map ϕ : G → H is a group homomorphism if for all a, b ∈ G we have ϕ(ab) =ϕ(a)ϕ(b). A group homomorphism is called surjective if for every c ∈ H
there is a ∈ G such that ϕ(a) = c. The kernel of a group homomorphism
ϕ : G → H is the set ker(ϕ) of elements a ∈ G such that ϕ(a) = 1. A grouphomomorphism is injective if ker(ϕ) = {1}.
Definition 3.4.7 (Subgroup). If G is a group and H is a subset of G, thenH is a subgroup if H is a group under the group operation on G.
For example, if ϕ : G → H is a group homomorphism, then ker(ϕ) is a
subgroup of G (see Exercise 2.21).
We now return to discussing Algorithm 3.4.5. In Step 1, note that m iseven since (−1)m ≡ 1 (mod n), so it makes sense to consider m/2. It is notpractical to determine whether or not am/2 ≡ 1 (mod n) for all a, because
it would require doing a computation for too many a. Instead, we try a
few random a; if am/2 ≡ 1 (mod n) for the a we check, we divide m by 2.
Also note that if there exists even a single a such that am/2 6≡ 1 (mod n),then half the a have this property, since then a 7→ am/2 is a surjective
homomorphism (Z/nZ)∗ → {±1} and the kernel has index 2.
Proposition 2.5.3 implies that if x2 ≡ 1 (mod p) then x = ±1 (mod p).
In Step 2, since (am/2 )2 ≡ 1 (mod n), we also have (am/2 )2 ≡ 1 (mod p)and (am/2 )2 ≡ 1 (mod q), so am/2 ≡ ±1 (mod p) and am/2 ≡ ±1 (mod q).
Since am/2 6≡ 1 (mod n), there are three possibilities for these signs, so withpositive probability one of the following two possibilities occurs:1.

am/2 ≡ +1 (mod p)

andam/2 ≡ −1 (mod q)

3.4 Attacking RSA2.

am/2 ≡ −1 (mod p)

and

65

am/2 ≡ +1 (mod q).

The only other possibility is that both signs are −1. In the first case,p | am/2 − 1but

q - am/2 − 1,

so gcd(am/2 − 1, pq) = p, and we have factored n. Similarly, in the second
case, gcd(am/2 − 1, pq) = q, and we again factor n.
Example 3.4.8. Somehow we discover that the RSA cryptosystem withn = 32295194023343

ande = 29468811804857

has decryption key d = 11127763319273. We use this information and Algorithm 3.4.5 to factor n. If
m = ed − 1 = 327921963064646896263108960,
then ϕ(pq) | m, so am ≡ 1 (mod n) for all a coprime to n. For each a ≤ 20
we find that am/2 ≡ 1 (mod n), so we replace m withm
= 163960981532323448131554480.
2
Again, we find with this new m that for each a ≤ 20, am/2 ≡ 1 (mod n), so
we replace m by 81980490766161724065777240. Yet again, for each a ≤ 20,
am/2 ≡ 1 (mod n), so we replace m by 40990245383080862032888620. Thisis enough, since 2m/2 ≡ 4015382800099 (mod n). Then,gcd(2m/2 − 1, n) = gcd(4015382800098, 32295194023343) = 737531,
and we have found a factor of n. Dividing, we find thatn = 737531 · 43788253.
SAGE Example 3.4.9. We implement Algorithm 3.4.5 in Sage.
sage: def crack_given_decrypt(n, m):...
n = Integer(n); m = Integer(m); # some type checking...
# Step 1: divide out powers of 2...
while True:...
if is_odd(m): break...
divide_out = True...
for i in range(5):...
a = randrange(1,n)...
if gcd(a,n) == 1:...
if Mod(a,n)^(m//2) != 1:...
divide_out = False
...
break

66

3. Public-key Cryptography

...
...
...
...
...
...
...
...
...
...
...

if divide_out:m = m//2
else:
break
# Step 2: Compute GCD
while True:a = randrange(1,n)
g = gcd(lift(Mod(a, n)^(m//2)) - 1, n)
if g != 1 and g != n:return g

We show how to verify Example 3.4.8 using Sage.
sage: n=32295194023343; e=29468811804857; d=11127763319273
sage: crack_given_decrypt(n, e*d - 1)
737531
sage: factor(n)737531 * 43788253
We try a much larger example.
sage: e = 22601762315966221465875845336488389513
sage: d = 31940292321834506197902778067109010093
sage: n = 268494924039590992469444675130990465673
sage: p = crack_given_decrypt(n, e*d - 1)sage: p# random output (could be other prime divisor)
13432418150982799907
sage: n % p
0

3.4.4

Further Remarks

If one were to implement an actual RSA cryptosystem, there are many additional tricks and ideas to keep in mind. For example, one can add someextra random letters to each block of text, so that a given string will encrypt differently each time it is encrypted. This makes it more difficult for
an attacker who knows the encrypted and plaintext versions of one message
to gain information about subsequent encrypted messages. In any particular implementation, there might be attacks that would be devastating inpractice, but which would not require factorization of the RSA modulus.
RSA is in common use, for example, it is used in OpenSSH protocol
version 1 (see http://www.openssh.com/).
We will consider the ElGamal cryptosystem in Sections 6.4.2. It has asimilar flavor to RSA, but is more flexible in some ways.

3.5 Exercises

67

Probably the best general purpose attack on RSA is the number fieldsieve, which is a general algorithm for factoring integers of the form pq. A
description of the sieve is beyond the scope of this book. The elliptic curve
method is another related general algorithm that we will discuss in detailin Section 6.3.
SAGE Example 3.4.10. Here is a simple example of using a variant of thenumber field sieve (called the quadratic sieve) in Sage to factor an RSA
key with about 192 bits:
sage: set_random_seed(0)sage: p = next_prime(randrange(2^96))
sage: q = next_prime(randrange(2^97))
sage: n = p * q
sage: qsieve(n)([6340271405786663791648052309,
46102313108592180286398757159], ’’)

3.5 Exercises
3.1 This problem concerns encoding phrases using numbers using theencoding of Section 3.3.2. What is the longest that an arbitrary sequence of letters (no spaces) can be if it must fit in a number that isless than 1020 ?
3.2 Suppose Michael creates an RSA cryptosystem with a very large modulus n for which the factorization of n cannot be found in a reasonableamount of time. Suppose that Nikita sends messages to Michael byrepresenting each alphabetic character as an integer between 0 and 26(A corresponds to 1, B to 2, etc., and a space to 0), then encrypts
each number separately using Michael’s RSA cryptosystem. Is thismethod secure? Explain your answer.
3.3 For any n ∈ N, let σ(n) be the sum of the divisors of n; for example,
σ(6) = 1 + 2 + 3 + 6 = 12 and σ(10) = 1 + 2 + 5 + 10 = 18. Supposethat n = pqr with p, q, and r distinct primes. Devise an “efficient”
algorithm that given n, ϕ(n) and σ(n), computes the factorizationof n. For example, if n = 105, then p = 3, q = 5, and r = 7, so the
input to the algorithm would be
n = 105,

ϕ(n) = 48,

and σ(n) = 192,

and the output would be 3, 5, and 7.
3.4 You and Nikita wish to agree on a secret key using the Diffie-Hellman
key exchange. Nikita announces that p = 3793 and g = 7. Nikita

68

3. Public-key Cryptography

secretly chooses a number n < p and tells you that g n ≡ 454 (mod p).
You choose the random number m = 1208. What is the secret key?
3.5 You see Michael and Nikita agree on a secret key using the DiffieHellman key exchange. Michael and Nikita choose p = 97 and g = 5.
Nikita chooses a random number n and tells Michael that g n ≡ 3(mod 97), and Michael chooses a random number m and tells Nikita
that g m ≡ 7 (mod 97). Brute force crack their code: What is thesecret key that Nikita and Michael agree upon? What is n? Whatis m?
3.6 In this problem, you will “crack” an RSA cryptosystem. What is thesecret decoding number d for the RSA cryptosystem with public key(n, e) = (5352381469067, 4240501142039)?
3.7 Nikita creates an RSA cryptosystem with public key(n, e) = (1433811615146881, 329222149569169).
In the following two problems, show the steps you take to factor n.
(Don’t simply factor n directly using a computer.)(a) Somehow you discover that d = 116439879930113. Show how to
use the probabilistic algorithm of Section 3.4.3 to factor n.
(b) In part (a) you found that the factors p and q of n are very
close. Show how to use the Fermat Factorization Method of Section 3.4.2 to factor n.

This is page 69
Printer: Opaque this4
Quadratic Reciprocity

A linear equationax ≡ b (mod n)
has a solution if and only if gcd(a, n) divides b (see Proposition 2.1.15).
This chapter is about some amazing mathematics motivated by the search
for a criterion for whether or not a given quadratic equation
ax2 + bx + c ≡ 0(mod n)has a solution. In many cases, the Chinese Remainder Theorem and thequadratic formula reduce this to the key question of whether a given integer
a is a perfect square modulo a prime p.
The Quadratic Reciprocity Law of Gauss provides a precise answer tothe following question: For which primes p is the image of a in (Z/pZ)∗ aperfect square? A deep fact, which we will completely prove in this chapter,is that the answer depends only on the reduction of p modulo 4a. Thus todecide if a is a square modulo p, one only needs to consider the residue of p
modulo 4a, which is extremely surprising. It turns out that this “reciprocitylaw” goes to the heart of modern number theory and touches on advanced
topics such as class field theory and the Langlands program.
There are over a hundred proofs of the Quadratic Reciprocity Law (see[Lem] for a long list). In this chapter, we give two proofs. The first, which
we give in Section 4.3, is completely elementary and involves keeping trackof integer points in intervals. It is satisfying because one can understandevery detail without much abstraction, but it might be unsatisfying if you
find it difficult to conceptualize what is going on. In contrast, our second

70

4. Quadratic Reciprocity

proof, which we give in Section 4.4, is more abstract and uses a conceptual
development of properties of Gauss sums. You should read Sections 4.1 and
4.2, then at least one of Section 4.3 or Section 4.4, depending on your taste
and how much abstract algebra you know.
In Section 4.5, we return to the computational question of actually finding square roots and solving quadratic equations in practice.

4.1 Statement of the Quadratic Reciprocity Law
In this section, we state the Quadratic Reciprocity Law.
Definition 4.1.1 (Quadratic Residue). Fix a prime p. An integer a notdivisible by p is a quadratic residue modulo p if a is a square modulo p;otherwise, a is a quadratic nonresidue.
For example, the squares modulo 5 are
12 = 1,22 = 4,32 = 4,42 = 1,(mod 5)so 1 and 4 are both quadratic residues and 2 and 3 are quadratic nonresidues.
The quadratic reciprocity theorem is the deepest theorem that we will
prove in this book. It connects the question of whether or not a is aquadratic residue modulo p to the question of whether p is a quadratic
residue modulo each of the prime divisors of a. To express it precisely, we
introduce some new notation.
Definition 4.1.2 (Legendre Symbol). Let p be an odd prime and let a bean integer. Set  
if gcd(a, p) 6= 1,
0a= +1 if a is a quadratic residue, andp
−1 if a is a quadratic nonresidue.
We call this symbol the Legendre Symbol.
For example, we have
 
 
1
2
= 1,
= −1,
5
5

 
3
= −1,
5

 
4
= 1,
5

 
5
= 0.
5

This notation is well entrenched in the literature even though it is also
the notation for “a divided by p;” be careful not to confuse the two.
SAGE Example 4.1.3. Use the legendre symbol command to compute theLegendre symbol in Sage.

4.1 Statement of the Quadratic Reciprocity Law

71

sage: legendre_symbol(2,3)-1
sage: legendre_symbol(1,3)
1
sage: legendre_symbol(3,5)-1
sage: legendre_symbol(Mod(3,5), 5)
-1
 
 
Since ap only depends on a (mod p), it makes sense to define ap for a ∈ Z/pZ to be ãp for any lift ã of a to Z.
Recall (see Definition 3.4.6) that a group homomorphism ϕ : G → H isa map such that for every a, b ∈ G we have ϕ(ab) = ϕ(a)ϕ(b). Moreover,
we say that ϕ is surjective if for every c ∈ H there is an a ∈ G with
ϕ(a) = c. The next lemma explains how the quadratic residue symbol
defines a surjective group homomorphism.
 
Lemma 4.1.4. The map ψ : (Z/pZ)∗ → {±1} given by ψ(a) = ap is asurjective group homomorphism.
Proof. By Theorem 2.5.8, primitive roots exist, so there is g ∈ (Z/pZ)∗
such that the elements of (Z/pZ)∗ areg, g 2 , . . . , g (p−1)/2 , g (p+1)/2 , . . . , g p−1 = 1.
Since p − 1 is even, the squares of elements of (Z/pZ)∗ are
g 2 , g 4 , . . . , g (p−1)/2·2 = 1, g p+1 = g 2 , . . . , g 2(p−1) .
Note that the powers of g starting with g p+1 = g 2 all appeared earlier
on the list. Thus, the perfect squares in (Z/pZ)∗ are exactly the powers
g n with n = 2, 4, . . . , p − 1, even, and the nonsquares the powers g n withn = 1, 3, . . . , p − 2, odd. It follows that ψ is a homomorphism since an odd
plus an odd is even, the sum of two evens is even, and odd plus an even isodd. Moreover, since g is not a square, ψ(g) = −1, so ψ is surjective.
Remark 4.1.5. We rephrase the above proof in the language of group theory.
The group G = (Z/pZ)∗ of order p − 1 is a cyclic group. Since p is odd,
p − 1 is even, so the subgroup H of squares of elements of Ghas index 2
in G. (See Exercise 4.2 for why H is a subgroup.) Since ap = 1 if andonly if a ∈ H, we see that ψ is the composition G → G/H ∼= {±1}, where
we identify the nontrivial element of G/H with −1.
Remark 4.1.6. We can alternatively prove that ψ is surjective without usingthat (Z/pZ)∗ is cyclic, as follows. If a ∈ (Z/pZ)∗ is a square, say a ≡ b2

72

4. Quadratic Reciprocity
TABLE 4.1. When is 5 a square modulo p?

p

 
5
p

p mod 5

p

 
5
p

p mod 5

7
11
13
17
19
23

−1
1
−1
−1
1
−1

2
1
3
2
4
3

29
31
37
41
43
47

1
1
−1
1
−1
−1

4
1
2
1
3
2

(mod p), then a(p−1)/2 = bp−1 ≡ 1 (mod p), so a is a root of f = x(p−1)/2 −1. By Proposition 2.5.3, the polynomial f has at most (p−1)/2 roots. Thus,∗
there must be a ∈ (Z/pZ) that is not a root of f , and for that a, we anhave ψ(a) = ap = −1, and trivially ψ(1) = 1, so the map ψ is surjective.
Note that this argument does not prove that ψ is a homomorphism.
 
The symbol ap only depends on the residue class of a modulo p, somaking a table of values a5  for many values of a would be easy. Would it
5
be easy to make a table of p for many p? Perhaps, since there appears 
to be a simple pattern in Table 4.1. It seems that p5 depends only on 
the congruence class of p modulo 5. More precisely, p5 = 1 if and only if p ≡ 1, 4 (mod 5), i.e., p5 = 1 if and only if p is a square modulo 5.
Based on similar observations, in the 18th century various mathematicians found a conjectural explanation for the mystery suggested by Table 4.1. Finally, on April 8, 1796, at the age of 19, Gauss proved the following theorem.
Theorem 4.1.7 (Gauss’s Quadratic Reciprocity Law). Suppose p and q
are distinct odd primes. Then  
p−1 q−1pq= (−1) 2 · 2
.
q
p
Also


−1
p


= (−1)(p−1)/2

and

  (
1
2
=
p
−1

if p ≡ ±1 (mod 8)
if p ≡ ±3 (mod 8).

   We will give two proofs of Gauss’s formula relating pq to pq . The firstelementary proof is in Section 4.3, and the second more algebraic proof isin Section 4.4.

4.2 Euler’s Criterion73

In our example, Gauss’s theorem implies that( p p+1 if p ≡ 1, 4 (mod 5)52· p−1=== (−1) 2p
5
5−1 if p ≡ 2, 3 (mod 5).
As an application, the following example illustrates how to answer questions like “is a a square modulo b” using Theorem 4.1.7.
Example 4.1.8. Is 69 a square modulo the prime 389? We have

 
 
 

69
3 · 23323==·= (−1) · (−1) = 1.
389
389
389
389
Here



3
389




=

389
3


=

 
2
= −1,
3

and


23
389





   

389
21
−2
=
=
=
23
23
23

 
23−1
−1
2
=
= (−1) 2 · 1 = −1.
23
23

Thus 69 is a square modulo 389.
SAGE Example 4.1.9. We could also do this computation in Sage as follows:
sage: legendre_symbol(69,389)
1Though we know that 69 is a square modulo 389, we don’t know an
explicit x such that x2 ≡ 69 (mod 389)! This is reminiscent of how weproved using Theorem 2.1.20 that certain numbers are composite without
knowing a factorization.
Remark 4.1.10. The Jacobi symbol is an extension of the Legendre symbol
to composite moduli. For more details, see Exercise 4.9.

4.2 Euler’s Criterion
Let p be an odd prime and a an integer notby p. Euler used
 divisibleathe existence of primitive roots to show that p is congruent to a(p−1)/2
modulo p. We will use this fact repeatedly below in both proofs of Theorem 4.1.7.
 
Proposition 4.2.1 (Euler’s Criterion). We have ap = 1 if and only ifa(p−1)/2 ≡ 1(mod p).

74

4. Quadratic Reciprocity

Proof. The map ϕ : (Z/pZ)∗ → (Z/pZ)∗ given by ϕ(a) = a(p−1)/2 is agroup homomorphism, since powering is a group homomorphism of any
abelian group (seeExercise4.2). Let ψ : (Z/pZ)∗ → {±1} be the homo
morphism ψ(a) =apof Lemma 4.1.4. If a ∈ ker(ψ), then a = b2 for some∗b ∈ (Z/pZ) , so
ϕ(a) = a(p−1)/2 = (b2 )(p−1)/2 = bp−1 = 1.
Thus ker(ψ) ⊂ ker(ϕ). By Lemma 4.1.4, ker(ψ) has index 2 in (Z/pZ)∗ ,i.e., #(Z/pZ)∗ = 2 · # ker(ψ). Since the kernel of a homomorphism is agroup, and the order of a subgroup divides the order of the group, we have
either ker(ϕ) = ker(ψ) or ϕ = 1. If ϕ = 1, the polynomial x(p−1)/2 − 1 hasp − 1 roots in the field Z/pZ, which contradicts Proposition 2.5.3. Thus
ker(ϕ) = ker(ψ), which proves the proposition.
SAGE Example 4.2.2. From a computational  point of view, Corollary 4.2.3
provides a convenient way to compute ap , which we illustrate in Sage:sage: def kr(a, p):...
if Mod(a,p)^((p-1)//2) == 1:...
return 1
...
else:
...
return -1
sage: for a in range(1,5):...
print a, kr(a,5)
1 1
2 -1
3 -1
4 1
Corollary 4.2.3. The equation x2 ≡
 a (mod p) has no solution if andonly if a(p−1)/2 ≡ −1 (mod p). Thusap

≡ a(p−1)/2 (mod p).

Proof. This follows from Proposition 4.2.1 and the fact that the polynomial x2 − 1 has no roots besides +1 and −1 (which follows from Proposition 2.5.5).
As additional computational
motivation for the value of Corollary 4.2.3,
 anote that to evaluate p using Theorem 4.1.7 would not be practical if aand p are both very large, because it would require factoring a. However,Corollary 4.2.3 provides a method for evaluating ap without factoring a.
Example 4.2.4. Suppose p = 11. By squaring each element of (Z/11Z)∗ , we
see that the squares modulo 11 are {1, 3, 4, 5, 9}. We compute a(p−1)/2 = a5

4.3 First Proof of Quadratic Reciprocity

75

for each a ∈ (Z/11Z)∗ and get
15 = 1, 25 = −1, 35 = 1, 45 = 1, 55 = 1,
65 = −1, 75 = −1, 85 = −1, 95 = 1, 105 = −1.
Thus the a with a5 = 1 are {1, 3, 4, 5, 9}, just as Proposition 4.2.1 predicts.
Example 4.2.5. We determine whether or not 3 is a square modulo theprime p = 726377359.
sage: p = 726377359
sage: Mod(3, p)^((p-1)//2)
726377358so3(p−1)/2 ≡ −1(mod 726377359).

Thus 3 is not a square modulo p. This computation wasn’t difficult, but
it would have been tedious by hand. Since 3 is small, the Quadratic Reciprocity Law provides a way to answer this question, which could easily be
carried out by hand:




3
726377359= (−1)(3−1)/2·(726377359−1)/2
726377359
3
 
1
= −1.
= (−1) ·3

4.3 First Proof of Quadratic Reciprocity
Our first proof of quadratic reciprocity is elementary. The proof involves
keeping track of integer points inProving Gauss’s lemma is the intervals.
afirst step; this lemma computes p in terms of the number of integers ofa certain type that lie in a certain interval. We next prove Lemma 4.3.3,
which controls how the parity of the number of integer points in an interval changes
  when an endpoint of the interval is changed. We then provethat ap depends only on p modulo 4a by applying Gauss’s Lemma andkeeping careful track of intervals as they are rescaled and their endpoints
are changed. Finally, in Section 4.3.2, we use some basic algebra to deduce
the Quadratic Reciprocity Law using the tools we’ve just developed. Our
proof follows the one given in [Dav99] closely.
Lemma 4.3.1 (Gauss’s Lemma). Let p be an odd prime and let a be an
integer 6≡ 0 (mod p). Form the numbersa, 2a, 3a, . . . ,

p−1
a
2

76

4. Quadratic Reciprocity

and reduce them modulo p to lie in the interval (− p2 , p2 ), i.e., for each of theabove products k · a find a number in the interval (− p2 , p2 ) that is congruent
to k · a modulo p. Let ν be the number of negative numbers in the resulting
set. Then a= (−1)ν .
p
Proof. In defining ν, we expressed each number inp−1
S = a, 2a, . . . ,a2
as congruent to a number in the set
p−1 p−1,−
1, −1, 2, −2, . . . ,.
2
2
No number 1, 2, . . . , p−1
2 appears more than once, with either choice ofsign, because if it did then either two elements of S are congruent modulo p
or 0 is the sum of two elements of S, and both events are impossible (theformer case cannot occur because of cancellation modulo p, and in thelatter case we would have ka + ja ≡ 0 (mod p) for 1 ≤ k, j ≤ (p − 1)/2, so
k + j ≡ 0 (mod p), a contradiction). The resulting set must be of the formp−1
T = ε1 · 1, ε2 · 2, . . . , ε(p−1)/2 ·,2where each εi is either +1 or −1. Multiplying together the elements of S
and of T , we see thatp−1(1a) · (2a) · (3a) · · ·
a ≡
2


p−1
(ε1 · 1) · (ε2 · 2) · · · ε(p−1)/2 ·(mod p),2soa(p−1)/2 ≡ ε1 · ε2 · · · ε(p−1)/2(mod p).
 
The lemma then follows from Proposition 4.2.1, since ap = a(p−1)/2 .
SAGE Example 4.3.2. We illustrate Gauss’s Lemma using Sage. The gaussfunction below prints out a list of the normalized numbers appearing inνthe statement  of Gauss’s Lemma, and returns (−1) . In each case below,(−1)ν =

ap.

sage: def gauss(a, p):...
# make the list of numbers reduced modulo p

4.3 First Proof of Quadratic Reciprocity

77...
v = [(n*a)%p for n in range(1, (p-1)//2 + 1)]...
# normalize them to be in the range -p/2 to p/2...
v = [(x if (x < p/2) else x - p) for x in v]...
# sort and print the resulting numbers...
v.sort()...
print v...
# count the number that are negative...
num_neg = len([x for x in v if x < 0])...
return (-1)^num_neg
sage: gauss(2, 13)[-5, -3, -1, 2, 4, 6]-1
sage: legendre_symbol(2,13)-1
sage: gauss(4, 13)[-6, -5, -2, -1, 3, 4]
1sage: legendre_symbol(4,13)
1
sage: gauss(2,31)[-15, -13, -11, -9, -7, -5, -3, -1, 2, 4, 6, 8, 10, 12, 14]1
sage: legendre_symbol(2,31)
1

4.3.1

Euler’s Proposition

For rational numbers a, b ∈ Q, let(a, b) ∩ Z = {x ∈ Z : a < x < b}be the set of integers between a and b. The following lemma will help us tokeep track of how many integers lie in certain open intervals.
Lemma 4.3.3. Let a, b ∈ Q. Then for any integer n,# ((a, b) ∩ Z) ≡ # ((a, b + 2n) ∩ Z)

(mod 2)# ((a, b) ∩ Z) ≡ # ((a − 2n, b) ∩ Z)

(mod 2),

andprovided that each interval involved in the congruence is nonempty.
Note that if one of the intervals is empty, then the statement may be false;for example, if (a, b) = (−1/2, 1/2) and n = −1, then #((a, b) ∩ Z) = 1 but#(a, b − 2) ∩ Z = 0.

78

4. Quadratic Reciprocity

Proof. Let dxe denotes the least integer ≥ x. Since n > 0,(a, b + 2n) = (a, b) ∪ [b, b + 2n),where the union is disjoint. There are 2n integers
dbe, dbe + 1, . . . , dbe + 2n − 1
in the interval [b, b + 2n), so the first congruence of the lemma is true inthis case. We also have(a, b − 2n) = (a, b) minus [b − 2n, b)
and [b−2n, b) contains exactly 2n integers, so the lemma is also true when n
is negative. The statement about # ((a − 2n, b) ∩ Z) is proved in a similarmanner.
Once we have proved the following proposition, it will be easy to deduce
the Quadratic Reciprocity Law.
Proposition 4.3.4 (Euler). Let p be an odd prime and let a 
be a positive 

integer with p - a. If q is a prime with q ≡ ±p (mod 4a), then
Proof. We will apply Lemma 4.3.1 to compute

S=
and


I=

 
a
p

a
p

=

a
q

.

. Let


p−1
a, 2a, 3a, . . . ,
a
2

 




1
3
1
p, p ∪p, 2p ∪ · · · ∪b−p, bp ,
22
2

where b = 12 a or 12 (a − 1), whichever is an integer.
We check that every element of S that is equivalent modulo p to something in the interval (− p2 , 0) lies in I. First suppose that b = 21 a. Then
bp =

1
p
p−1
ap = a >a,2
22

so each element of S that is equivalent modulo p to an element of (− p2 , 0)lies in I. Next suppose that b = 21 (a − 1). Then
bp +

p
a−1
p
p
p−1
=
p+ = a>
a,
2
2
2
2
2

so ((b − 12 )p, bp) is the last interval that could contain an element of S that
reduces to (− p2 , 0). Note that the integer endpoints of I are not in S, since4.3 First Proof of Quadratic Reciprocity

79

those endpoints are divisible by p, but no element of S is divisible by p.
Thus, by Lemma 4.3.1,
 a= (−1)#(S∩I) .
p
To compute #(S ∩ I), first rescale by a to see that




1
1
1
#(S ∩ I) = #S∩ I =# Z∩ I ,
a
a
a
where
1
I=
a



p p
∪
,
2a a



3p 2p
,
2a a




∪ ··· ∪

(2b − 1)p bp
,
2a
a


,

1
1
a S = {1, 2, 3, 4, . . . , (p − 1)/2}, and the second equality is because a I ⊂(0, (p − 1)/2 + 1/2], since

pa
pb
p
p−1 1
≤ 2 = =
+ .
a
a
2
2
2
Write p = 4ac + r, and let





r r
(2b − 1)r br
3r 2r
J=
,
,
∪ ··· ∪
,
.
∪
2a a
2a a
2a
a
The only difference between a1 I and J is that the endpoints of intervals arechanged by addition of an even integer, since
r
p
p
p
−
=
− 2c −= −2c.
2a 2a
2a2a
By Lemma 4.3.3,
1
ν = # Z ∩ I ≡ #(Z ∩ J)a(mod 2).

 ap= (−1)ν depends only on r and a, i.e., only on p modulo 4a.
   Thus if q ≡ p (mod 4a), then ap = aq .

Thus

If q ≡ −p (mod 4a), then the only change in the above computation isthat r is replaced by 4a − r. This changes J into



r
r
3r
2r
K = 2 − ,4 −∪ 6 − ,8 −
∪ ···
2a
a
2a
a


(2b − 1)rbr
∪ 4b − 2 −, 4b −.
2aa

80

4. Quadratic Reciprocity

Thus K is the same as −J, except even integers have been added to theendpoints. By Lemma 4.3.3,

1#(K ∩ Z) ≡ #I ∩Z(mod 2),a   so ap = aq again, which completes the proof.
The following more careful analysis in the special case when a = 2 helps
illustrate the proof of the above lemma, and the result is frequently useful incomputations. For an alternative proof of the proposition, see Exercise 4.6.
Proposition 4.3.5 (Legendre Symbol of 2). Let p be an odd prime. Then  (1 if p ≡ ±1 (mod 8)2=p−1 if p ≡ ±3 (mod 8).
Proof. When a = 2, the set S = {a, 2a, . . . , p−12 a} is{2, 4, 6, . . . , p − 1}.
We must count the parity of the number of elements of S that lie in theinterval I = ( p2 , p). Writing p = 8c + r, we have p p 1# (I ∩ S) = #I ∩Z =#
,
∩Z
2
4 2

 r r 


r
r
,
∩Z ≡#
∩Z
(mod 2),= # 2c + , 4c +42
4 2
where the last equality comes from Lemma 4.3.3. The possibilities for r are
1, 3, 5, 7. When r = 1, the cardinality is 0; when r = 3, 5 it is 1; and whenr = 7 it is 2.

4.3.2

Proof of Quadratic Reciprocity

It is now straightforward to deduce the Quadratic Reciprocity Law.
First Proof of Theorem 4.1.7. First suppose that p ≡ q (mod 4). By swapping p and q if necessary, we may assume that p > q, and write p − q = 4a.
Since p = 4a + q,         p4a + q
4a
4
a
a
=
=
=
=
,
q
q
q
q
q
q
and

  
 
 
  
q
p − 4a
−4a
−1
a
=
=
=
·
.
p
p
p
p
p

4.4 A Proof of Quadratic Reciprocity Using Gauss Sums

Proposition 4.3.4 implies that

 
a
q

=

 
a
p

81

, since p ≡ q (mod 4a). Thus    
p−1
p−1 q−1pq−1·== (−1) 2 = (−1) 2 · 2 ,
q
p
p
q−1
where the last equality is because p−1
2 is even if and only if 2 is even.
Next suppose that p 6≡ q (mod 4), so p ≡ −q (mod 4). Write p + q = 4a.
We have
  
  
  
  
p
4a − qaq
4a − p
a
=
=
, and
=
=
.
q
q
q
p
p
p
   Since p ≡ −q (mod 4a), Proposition 4.3.4 implies that aq = ap . Since
p−1 q−1(−1) 2 · 2 = 1, the proof is complete.

4.4 A Proof of Quadratic Reciprocity Using Gauss
Sums
In this section, we present a beautiful proof of Theorem 4.1.7 using algebraic
identities satisfied by sums of “roots of unity.” The objects we introduce
in the proof are of independent interest, and provide a powerful tool toprove higher-degree analogs of quadratic reciprocity. (For more on higher
reciprocity, see [IR90]. See also Section 6 of [IR90], on which the proof
below is modeled.)
Definition 4.4.1 (Root of Unity). An nth root of unity is a complex
number ζ such that ζ n = 1. A root of unity ζ is a primitive nth root ofunity if n is the smallest positive integer such that ζ n = 1.
√

For example, −1 is a primitive second root of unity, and ζ = −3−1is2
a primitive cube root of unity. More generally, for any n ∈ N the complex
numberζn = cos(2π/n) + i sin(2π/n)
is a primitive nth root of unity (this follows from the identity eiθ = cos(θ)+
i sin(θ)). For the rest of this section, we fix an odd prime p and the primitive
pth root ζ = ζp of unity.
SAGE Example 4.4.2. In Sage, use the CyclotomicField command tocreate an exact pth root of ζ unity. Expressions in ζ are always re-expressedas polynomials in ζ of degree at most p − 1.
sage: K.<zeta> = CyclotomicField(5)sage: zeta^51

82

4. Quadratic Reciprocity

sage: 1/zeta-zeta^3 - zeta^2 - zeta - 1
Definition 4.4.3 (Gauss Sum). Fix an odd prime p. The Gauss sum associated to an integer a isp−1  Xnga =ζ an ,
pn=1
where ζ = ζp = cos(2π/p) + i sin(2π/p) = e2πi/p .
Note that p is implicit in the definition of ga . If we were to change p,then the Gauss sum ga associated to a would be different. The definitionof ga also depends on our choice of ζ; we’ve chosen ζ = ζp , but could havechosen a different ζ and then ga could be different.
SAGE Example 4.4.4. We define a gauss sum function and compute theGauss sum g2 for p = 5:
sage: def gauss_sum(a,p):...
K.<zeta> = CyclotomicField(p)...
return sum(legendre_symbol(n,p) * zeta^(a*n)...
for n in range(1,p))
sage: g2 = gauss_sum(2,5); g2
2*zeta^3 + 2*zeta^2 + 1
sage: g2.complex_embedding()-2.23606797749979 + 3.33066907387547e-16*I
sage: g2^2
5Here, g2 is initially output as a polynomial in ζ5 , so there is no loss of
precision. The complex embedding command shows some embedding of g2
into the complex numbers, which√ is only correct to about the first 15 digits.
Note that g22 = 5, so g2 = − 5.
We compute a graphical representation of the Gauss sum g2 as follows(see Figure 4.1):
zeta = CDF(exp(2*pi*I/5))
v = [legendre_symbol(n,5) * zeta^(2*n) for n in range(1,5)]
S = sum([point(tuple(z), pointsize=100) for z in v])
show(S + point(tuple(sum(v)), pointsize=100, rgbcolor=’red’))Figure 4.1 illustrates the Gauss sum g2 for p = 5. The Gauss sum isobtained by adding the points√ on the unit circle, with signs as indicated,
to obtain the real number − 5. This suggests the following proposition,
whose proof will require some work.
Proposition 4.4.5 (Gauss Sum). For any a not divisible by p,
ga2 = (−1)(p−1)/2 p.

4.4 A Proof of Quadratic Reciprocity Using Gauss Sums

83

1
0.75
0.5
0.25
-1.5

-2

-1

-0.5

-0.25
-0.5
-0.75
-1

FIGURE 4.1. The red dot is the Gauss sum g2 for p = 5

SAGE Example 4.4.6. We illustrate using Sage that the proposition is correct for p = 7 and p = 13:
sage: [gauss_sum(a, 7)^2 for a in range(1,7)][-7, -7, -7, -7, -7, -7]
sage: [gauss_sum(a, 13)^2 for a in range(1,13)][13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13]
In order to prove the proposition, we introduce a few lemmas.
Lemma 4.4.7. For any integer a,(p−1
Xp if a ≡ 0 (mod p),anζ =0 otherwise.
n=0
Proof. If a ≡ 0 (mod p), then ζ a = 1, so the sum equals the number ofsummands, which is p. If a 6≡ 0 (mod p), then we use the identity
xp − 1 = (x − 1)(xp−1 + · · · + x + 1)
with x = ζ a . We have ζ a 6= 1, so ζ a − 1 6= 0 and
p−1
X
n=0

ζ an =

ζ ap − 1
1−1
= a
= 0.
ζa − 1ζ −1

84

4. Quadratic Reciprocity

Lemma 4.4.8. If x and y are arbitrary integers, then(p−1
X
p if x ≡ y (mod p),(x−y)nζ=
0 otherwise.
n=0
Proof. This follows from Lemma 4.4.7 by setting a = x − y.
Lemma 4.4.9. We have g0 = 0.
Proof. By definitiong0 =p−1  
X
n
n=0

p

.

(4.4.1)

By Lemma 4.1.4, the map ·: (Z/pZ)∗ → {±1}pis a surjective homomorphism of groups. Thus, half the elements of (Z/pZ)∗
map to +1
 half map to −1 (the subgroup that maps to +1 has index
 and2). Since0p= 0, the sum (4.4.1) is 0.

Lemma 4.4.10. For any integer a,
 
a
ga =
g1 .
p
Proof. When a ≡ 0 (mod p), the lemma follows from Lemma 4.4.9, so
suppose that a 6≡ 0 (mod p). Then,
 X


 
p−1  
p−1 
p−1 
X
X
a
n
an
m
a
ga =
ζ an =
ζ an =
ζ m = g1 .
p
p n=0 pppn=0m=0Here, we use that multiplication by a is an automorphism of Z/pZ. Finally,  2multiply both sides by ap and use that ap = 1.
We have enough lemmas to prove Proposition 4.4.5.
Pp−1
Proof of Proposition 4.4.5. We evaluate the sum a=0 ga g−a in two different ways. By Lemma 4.4.10, since a 6≡ 0 (mod p) we have  2
  
a
−a
−1
a
ga g−a =g1
g1 =g12 = (−1)(p−1)/2 g12 ,
p
p
p
p
 
where the last step follows from Proposition 4.2.1 and that ap ∈ {±1}.
Thus
p−1
X
ga g−a = (p − 1)(−1)(p−1)/2 g12 .
(4.4.2)a=0

4.4 A Proof of Quadratic Reciprocity Using Gauss Sums

85

On the other hand, by definitionga g−a =p−1  
X
n

p

n=0

=

ζ an ·

p

p


p−1 X
p−1   Xnm
n=0 m=0

p

m=0


p−1 X
p−1   Xnm
n=0 m=0

=


p−1 
X
m

p

p

ζ −am

ζ an ζ −am
ζ an−am .

Let δ(n, m) = 1 if n ≡ m (mod p) and 0 otherwise. By Lemma 4.4.8,
p−1
X
a=0


p−1 X
p−1 X
p−1   Xmn

ζ an−ampp
a=0 n=0 m=0
 p−1
p−1 X
p−1   Xnm X an−am
=
ζ
p
p a=0
n=0 m=0
p−1 X
p−1   Xnm=pδ(n, m)ppn=0 m=0p−1  2
X
n
=
p
p
n=0

ga g−a == p(p − 1).
Equate (4.4.2) and the above equality, then cancel (p − 1) to see that
g12 = (−1)(p−1)/2 p.
Since a 6≡ 0 (mod p), we have

 2

ga2 =

a
p

= 1, so by Lemma 4.4.10, 2ag12 = g12 ,p

and the proposition is proved.

4.4.1

Proof of Quadratic Reciprocity

We are now ready to prove Theorem 4.1.7 using Gauss sums.
Proof. Let q be an odd prime with q 6= p. Set p∗ = (−1)(p−1)/2 p andrecall
Pp−1  n  n∗2that Proposition 4.4.5 asserts that p = g , where g = g1 = n=0 p ζ .

86

4. Quadratic Reciprocity

Proposition 4.2.1 implies that(p∗ )(q−1)/2 ≡

 ∗
p
q

(mod q).

We have g q−1 = (g 2 )(q−1)/2 = (p∗ )(q−1)/2 , so multiplying both sides of thedisplayed equation by g yields a congruence ∗pgq ≡ g(mod q).
(4.4.3)q
q
But wait, what does this congruence mean, ∗  given that g is not an intepqger? It means that the difference g − g q is a multiple of q in the ringZ[ζ] of all polynomials in ζ with coefficients in Z.
The ring Z[ζ]/(q) has characteristic q, so if x, y ∈ Z[ζ], then (x + y)q ≡q
x + y q (mod q). Applying this to (4.4.3), we see that!q p−1  
p−1  p−1  
X
X n q
X
n
n
q
n
nq
g =
ζ
≡
ζ ≡
ζ nq ≡ gq (mod q).
p
p
p
n=0
n=0
n=0
By Lemma 4.4.10,
g q ≡ gq ≡

 
q
g
p

Combining this with (4.4.3) yields
 
 ∗
q
p
g≡
g
p
q

(mod q).

(mod q).

2∗
Sincep and p 6= q, we can cancel g from both sides to find that  g =p∗q(mod q). Since both residue symbols are ±1 and q is odd, itp ≡q   ∗follows that pq = pq . Finally, we note using Corollary 4.2.3 that 
(p−1)/2   ∗  
q−1 p−1(−1)(p−1)/2 p
−1
p
p
p
·
=
=
= (−1) 2 2 ·
.
q
q
q
q
q

4.5 Finding Square Roots
We return in this section to the question of computing square roots. If K
is a field in which 2 6= 0, and a, b, c ∈ K, with a 6= 0, then the two solutions
to the quadratic equation ax2 + bx + c = 0 are√−b ± b2 − 4acx=.
2a4.5 Finding Square Roots87

Now assume K = Z/pZ, with p an odd prime. Using Theorem 4.1.7, we
can decide whether or not b2 − 4ac is a perfect square in Z/pZ, and hence
whether or not ax2 + bx + c = 0 has a solution in Z/pZ. However, Theorem 4.1.7 says nothing about how to actually find a solution when there
is one. Also note that for this problem we do not need the full Quadratic
Reciprocity Law; in practice, deciding whether an element of Z/pZ is aperfect square with Proposition 4.2.1 is quite fast, in view of Section 2.3.
Suppose a ∈ Z/pZ is a nonzero quadratic residue. If p ≡ 3 (mod 4), thenp+1
b = a 4 is a square root of a because
 
p−1
p+1
p−1
a
b2 = a 2 = a 2 +1 = a 2 · a =· a = a.
p
We can compute b in time polynomial in the number of digits of p using
the powering algorithm of Section 2.3.
Suppose next that p ≡ 1 (mod 4). Unfortunately, we do not know adeterministic algorithm that takes a and p as input, outputs a square rootof a modulo p when one exists, and is polynomial-time in log(p).
Remark 4.5.1. There is an algorithm
p due to Schoof [Sch85] that computes
the square root of a in time O(( (|a|)1/2+ε · log(p))9 ). This beautiful algorithm (which makes use of elliptic curves) is not polynomial time in thesense described above, since for large a it takes exponentially longer thanfor small a.
We next describe a probabilistic algorithm to compute a square root of amodulo p, which is very quick in practice. Recall the notion of ring from
Definition 2.1.3. We will also need the notion of ring homomorphism andisomorphism.
Definition 4.5.2 (Homomorphism of Rings). Let R and S be rings. A
homomorphism of rings ϕ : R → S is a map such that for all a, b ∈ R, wehave• ϕ(ab) = ϕ(a)ϕ(b),• ϕ(a + b) = ϕ(a) + ϕ(b), and• ϕ(1) = 1.
An isomorphism ϕ : R → S of rings is a ring homomorphism that isbijective.
Consider the ringR = (Z/pZ)[x]/(x2 − a)defined as follows. We have
R = {u + vα : u, v ∈ Z/pZ}

88

4. Quadratic Reciprocity

with multiplication defined by(u + vα)(z + wα) = (uz + awv) + (uw + vz)α.
Here α corresponds to the class of x in R.
SAGE Example 4.5.3. We define and work with the ring R above in Sage
as follows (for p = 13):sage: S.<x> = PolynomialRing(GF(13))
sage: R.<alpha> = S.quotient(x^2 - 3)sage: (2+3*alpha)*(1+2*alpha)7*alpha + 7
Let b and c be the square roots of a in Z/pZ (though we cannot easily
compute b and c yet, we can consider them in order to deduce an algorithm
to find them). We have ring homomorphisms f : R → Z/pZ and g : R →Z/pZ given by f (u + vα) = u + vb and g(u + vα) = u + vc. Together, thesedefine a ring isomorphism
ϕ : R −→ Z/pZ × Z/pZ
given by ϕ(u + vα) = (u + vb, u + vc). Choose in some way a random
element z of (Z/pZ)∗ , and define u, v ∈ Z/pZ by
p−1

u + vα = (1 + zα) 2 ,p−1

where we compute (1+zα) 2 quickly using an analog of the binary powering algorithm of Section 2.3.2. If v = 0, we try again with another random z.
If v 6= 0, we can quickly find the desired square roots b and c as follows.
The quantity u + vb is a (p − 1)/2 power in Z/pZ, so it equals either 0, 1, or−1, so b = −u/v, (1 − u)/v, or (−1 − u)/v, respectively. Since we know uand v, we can try each of −u/v, (1 − u)/v, and (−1 − u)/v and see which
is a square root of a.
Example 4.5.4. Continuing Example 4.1.8, we find a square root of 69
modulo 389. We apply the algorithm described above in the case p ≡ 1(mod 4). We first choose the random z = 24 and find that (1 + 24α)194 =−1. The coefficient of α in the power is 0, and we try again with z = 51.
This time, we have (1 + 51α)194 = 239α = u + vα. The inverse of 239 in
Z/389Z is 153, so we consider the following three possibilities for a squareroot of 69:
−

u
=0
v

1−u
= 153
v

−1 − u= −153.
v

Thus, 153 and −153 are the square roots of 69 in Z/389Z.
SAGE Example 4.5.5. We implement the above algorithm in Sage andillustrate it with some examples.

4.6 Exercises

89

sage: def find_sqrt(a, p):...
assert (p-1)%4 == 0...
assert legendre_symbol(a,p) == 1...
S.<x> = PolynomialRing(GF(p))...
R.<alpha> = S.quotient(x^2 - a)...
while True:...
z = GF(p).random_element()...
w = (1 + z*alpha)^((p-1)//2)...
(u, v) = (w[0], w[1])...
if v != 0: break...
if (-u/v)^2 == a: return -u/v...
if ((1-u)/v)^2 == a: return (1-u)/v...
if ((-1-u)/v)^2 == a: return (-1-u)/v...
sage: b = find_sqrt(3,13)
sage: b# random: either 9 or 39sage: b^2
3
sage: b = find_sqrt(3,13)
sage: b# see, it’s random
4
sage: find_sqrt(5,389)# random: either 303 or 86
303
sage: find_sqrt(5,389)# see, it’s random86

4.6 Exercises
4.1 Calculate the following by hand:

3
97



,

3
389



,

22
11



, and

5!
7



.

4.2 Let G be an abelian group, and let n be a positive integer.
(a) Prove that the map ϕ : G → G given by ϕ(x) = xn is a grouphomomorphism.
(b) Prove that the subset H of G of squares of elements of G is a
subgroup.
4.3 Use Theorem 4.1.7 to show that for p ≥ 5 prime,  (1 if p ≡ 1, 11 (mod 12),3=p
−1 if p ≡ 5, 7 (mod 12).

90

4. Quadratic Reciprocity

4.4 (*) Use that (Z/pZ)∗ is cyclic to give a direct proof that



−3
p



=1
∗

when p ≡ 1 (mod 3). (Hint: There is an element c ∈ (Z/pZ) oforder 3. Show that (2c + 1)2 = −3.) 
4.5 (*) If p ≡ 1 (mod 5), show directly that p5 = 1 by the method of
Exercise 4.4. (Hint: Let c ∈ (Z/pZ)∗ be an element of order 5. Show
that (c + c4 )2 + (c + c4 ) − 1 = 0, etc.) 
4.6 (*) Let p be an odd prime. In this exercise, you will prove that p2 =1 if and only if p ≡ ±1 (mod 8).
(a) Prove thatx=

1 − t2,
1 + t2y=

2t
1 + t2is a parameterization of the set of solutions to x2 + y 2 ≡ 1(mod p), in the sense that the solutions (x, y) ∈ Z/pZ × Z/pZ
are in bijection with the t ∈ Z/pZ ∪ {∞} such that 1 + t2 6≡ 0(mod p). Here, t = ∞ corresponds to the point (−1, 0). (Hint:
if (x1 , y1 ) is a solution, consider the line y = t(x + 1) through(x1 , y1 ) and (−1, 0), and solve for x1 , y1 in terms of t.)(b) Prove that the number of solutions to x2 + y 2 ≡ 1 (mod p) is
p + 1 if p ≡ 3 (mod 4) and p − 1 if p ≡ 1 (mod 4).
∗∗(c) Consider the set  (a, b) ∈ (Z/pZ) × (Z/pZ) such that Sof pairs

a + b = 1 and

a
p

=

b
p

= 1. Prove that #S = (p + 1 − 4)/4

if p ≡ 3 (mod 4) and #S = (p − 1 − 4)/4 if p ≡ 1 (mod 4).
Conclude that #S is odd if and only if p ≡ ±1 (mod 8).
(d) The map σ(a, b) = (b, a) that swaps coordinates is a bijection of
the set S. It has exactly one fixed point  if and only if there isan a ∈ Z/pZ such that 2a = 1 and ap = 1. Also, prove that 
2a = 1 has a solution a ∈ Z/pZ with ap = 1 if and only if 2
p = 1.
(e) Finish by showing that σ has exactly one fixed point if and only
if #S is odd, i.e., if and only if p ≡ ±1 (mod 8).
Remark: The method of proof of this exercise can be generalized togive a proof of the full Quadratic Reciprocity Law.
4.7 How many natural numbers x < 213 satisfy the equationx2 ≡ 5(mod 213 − 1)?

You may assume that 213 − 1 is prime.

4.6 Exercises

91

4.8 Find the natural number x < 97 such that x ≡ 448 (mod 97). Notethat 97 is prime.
4.9 In this problem, we will formulate an analog of quadratic reciprocityafor a symbol like q , but without the restriction that q be a prime.
Qk
Suppose n is an odd positive integer,
which we factor as i=1 pei i .
We define the Jacobi symbol na as follows:
a
n

=

e
k 
Y
a i
i=1

(a) Give an example to show thata perfect square modulo n.

pi
a
n



.

= 1 need not imply that a is(b) (*) Let n be odd and a and b be integers. Prove that the following
holds: 
ai. na nb = abn . (Thus a 7→ n induces a homomorphismfrom (Z/nZ)∗ to {±1}.)
ii. −1n ≡ n (mod 4).
iii. n2 = 1 if n ≡ ±1 (mod 8) and −1 otherwise.

a−1 n−1
iv. Assume a is positive and odd. Then na = (−1) 2 · 2 na
4.10 (*) Prove that for any n ∈ Z, the integer n2 + n + 1 does not haveany positive divisors of the form 6k − 1.

92

4. Quadratic Reciprocity

This is page 93
Printer: Opaque this5
Continued Fractions√

The golden ratio 1+2 5 is equal to the infinite fraction
1

1+

1

1+
1+

1
1 + ··· ,and the fraction
103993= 3.14159265301190260407 . . .
33102
is an excellent approximation to π. Both of these observations are explainedby continued fractions.
Continued fractions are theoretically beautiful and provide tools thatyield powerful algorithms for solving problems in number theory. For example, continued fractions provide a fast way to write a prime—even ahundred digit prime—as a sum of two squares, when possible.
Continued fractions are thus a beautiful algorithmic and conceptual toolin number theory that has many applications. For example, they provide
a surprisingly efficient way to recognize a rational number given just thefirst few digits of its decimal expansion, and they give a sense in which e
is “less complicated” than π (see Example 5.3.4 and Section 5.4).
In Section 5.2, we study continued fractions of finite length and lay thefoundations for our later investigations. In Section 5.3, we give the continued fraction procedure, which associates to a real number x a continued
fraction that converges to x. In Section 5.5, we characterize (eventually)

94

5. Continued Fractions

periodic continued fractions as the continued fractions of nonrational rootsof quadratic polynomials, then discuss an unsolved mystery concerning
continued fractions of roots of irreducible polynomials of degree greaterthan 2. We conclude the chapter with applications of continued fractions
to recognizing approximations to rational numbers (Section 5.6) and writing integers as sums of two squares (Section 5.7).
The reader is encouraged to read more about continued fractions in[HW79, Ch. X], [Khi63], [Bur89, §13.3], and [NZM91, Ch. 7].

5.1 The Definition
A continued fraction is an expression of the form
1

a0 +

1

a1 +

1
a3 + · · · .

a2 +In this book, we will assume that the ai are real numbers and ai > 0 fori ≥ 1, and the expression may or may not go on indefinitely. More general
notions of continued fractions have been extensively studied, but they arebeyond the scope of this book. We will be most interested in the case when
the ai are all integers.
We denote the continued fraction displayed above by[a0 , a1 , a2 , . . .].
For example,
1
3
= ,
2
2

[1, 2] = 1 +1[3, 7, 15, 1, 292] = 3 +

1

7+
15 +

1
1+

=

1
292

103993= 3.14159265301190260407 . . . ,33102

5.2 Finite Continued Fractions

95

and
1

[2, 1, 2, 1, 1, 4, 1, 1, 6] = 2 +

1

1+

1

2+

1

1+

1

1+

1

4+
1+

1
1+

1
6

1264
465
= 2.7182795698924731182795698 . . .
=The second two examples were chosen to foreshadow that continued fractions can be used to obtain good rational approximations to irrational
numbers. Note that the first approximates π, and the second e.

5.2 Finite Continued Fractions
This section is about continued fractions of the form [a0 , a1 , . . . , am ] forsome m ≥ 0. We give an inductive definition of numbers pn and qn suchthat for all n ≤ m
pn.
(5.2.1)[a0 , a1 , . . . , an ] =qn
We then give related formulas
for the determinants of the 2 × 2 matrices
p pn−2 pn pn−1, which we will repeatedly use to deduce propand qnn qn−2
qn qn−1
erties of the sequence of partial convergents [a0 , . . . , ak ]. We will use Algorithm 1.1.13 to prove that every rational number is represented by acontinued fraction, as in (5.2.1).
Definition 5.2.1 (Finite Continued Fraction). A finite continued fraction
is an expression
1
a0 +
1
a1 +
1
a2 +
· · · + a1nwhere each am is a real number and am > 0 for all m ≥ 1.
Definition 5.2.2 (Simple Continued Fraction). A simple continued fraction is a finite or infinite continued fraction in which the ai are all integers.

96

5. Continued Fractions

To get a feeling for continued fractions, observe that[a0 ] = a0 ,1
a0 a1 + 1=,
a1
a1
a0 a1 a2 + a0 + a2
1=.
[a0 , a1 , a2 ] = a0 +1
a1 a2 + 1
a1 +a2
[a0 , a1 ] = a0 +

Also,


1
[a0 , a1 , . . . , an−1 , an ] = a0 , a1 , . . . , an−2 , an−1 +an1= a0 +[a1 , . . . , an ]= [a0 , [a1 , . . . , an ]].
SAGE Example 5.2.3. The continued fraction command computes continued fractions1 :sage: continued_fraction(17/23)[0, 1, 2, 1, 5]
sage: continued_fraction(e)[2, 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8, 1, 1, 10, 1, 1,
12, 1, 1]
Use the optional second argument bits = n to determine the precision (inbits) of the input number that is used to compute the continued fraction.
sage: continued_fraction(e, bits=21)[2, 1, 2, 1, 1, 4, 1, 1, 6]
sage: continued_fraction(e, bits=30)[2, 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8]
You can obtain the value of a continued fraction and even do arithmetic
with continued fractions:sage: a = continued_fraction(17/23); a[0, 1, 2, 1, 5]
sage: a.value()
17/23
sage: b = continued_fraction(6/23); b[0, 3, 1, 5]
sage: a + b[1]
1 The continued fraction module of Sage has changed a lot. Please refer to the documentation of Sage online for the newest version.

5.2 Finite Continued Fractions

5.2.1

97

Partial Convergents

Fix a finite continued fraction [a0 , . . . , am ]. We do not assume at this pointthat the ai are integers.
Definition 5.2.4 (Partial Convergents). For 0 ≤ n ≤ m, the nth convergent of the continued fraction [a0 , . . . , am ] is [a0 , . . . , an ]. These convergentsfor n < m are also called partial convergents.
For each n with −2 ≤ n ≤ m, define real numbers pn and qn as follows:
p−2 = 0,q−2 = 1,p−1 = 1,q−1 = 0,p 0 = a0 ,q0 = 1,······pn = an pn−1 + pn−2
qn = an qn−1 + qn−2

··· ,
··· .

Proposition 5.2.5 (Partial Convergents). For n ≥ 0 with n ≤ m we havepn
[a0 , . . . , an ] =.
qnProof. We use induction. The assertion is obvious when n = 0, 1. Suppose
the proposition is true for all continued fractions of length n − 1. Then1[a0 , . . . , an ] = [a0 , . . . , an−2 , an−1 +]anan−1 + a1n pn−2 + pn−3= 
an−1 + a1n qn−2 + qn−3
(an−1 an + 1)pn−2 + an pn−3
(an−1 an + 1)qn−2 + an qn−3
an (an−1 pn−2 + pn−3 ) + pn−2=an (an−1 qn−2 + qn−3 ) + qn−2
an pn−1 + pn−2=
an qn−1 + qn−2
pn
=
.
qn
=

SAGE Example 5.2.6. If c is a continued fraction, use c.convergents()
to compute a list of the partial convergents of c.
sage: c = continued_fraction(pi,bits=35); c[3, 7, 15, 1, 292, 1]
sage: c.convergents()[3, 22/7, 333/106, 355/113, 103993/33102, 104348/33215]
As we will see, the convergents of a continued fraction are the best rational approximations to the value of the continued fraction. In the example
above, the listed convergents are the best rational approximations of π withgiven denominator size.

98

5. Continued Fractions

Proposition 5.2.7. For n ≥ 0 with n ≤ m we havepn qn−1 − qn pn−1 = (−1)n−1(5.2.2)pn qn−2 − qn pn−2 = (−1)n an .

(5.2.3)andEquivalently,
pn−1
pn
1
−
= (−1)n−1 ·qn
qn−1
qn qn−1
and

pn
pn−2
an
−
= (−1)n ·.
qn
qn−2
qn qn−2

Proof. The case for n = 0 is obvious from the definitions. Now supposen > 0 and the statement is true for n − 1. Thenpn qn−1 − qn pn−1 = (an pn−1 + pn−2 )qn−1 − (an qn−1 + qn−2 )pn−1= pn−2 qn−1 − qn−2 pn−1
= −(pn−1 qn−2 − pn−2 qn−1 )= −(−1)n−2 = (−1)n−1 .
This completes the proof of (5.2.2). For (5.2.3), we havepn qn−2 − pn−2 qn = (an pn−1 + pn−2 )qn−2 − pn−2 (an qn−1 + qn−2 )= an (pn−1 qn−2 − pn−2 qn−1 )= (−1)n an .

Remark 5.2.8. Expressed in terms of matrices, the proposition
asserts thatp pn−2 
p pn−1is (−1)n an .
is (−1)n−1 , and of qnn qn−2
the determinant of qnn qn−1
SAGE Example 5.2.9. We use Sage to verify Proposition 5.2.7 for the first
few terms of the continued fraction of π.
sage: c = continued_fraction(pi); c[3, 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, 1, 14]
sage: for n in range(-1, len(c)):...
print c.pn(n)*c.qn(n-1) - c.qn(n)*c.pn(n-1),
1 -1 1 -1 1 -1 1 -1 1 -1 1 -1 1 -1
sage: for n in range(len(c)):...
print c.pn(n)*c.qn(n-2) - c.qn(n)*c.pn(n-2),
3 -7 15 -1 292 -1 1 -1 2 -1 3 -1 14
Corollary 5.2.10 (Convergents in lowest terms). If [a0 , a1 , . . . , am ] is a
simple continued fraction, so each ai is an integer, then the pn and qn are
integers and the fraction pn /qn is in lowest terms.

5.2 Finite Continued Fractions

99

Proof. It is clear that the pn and qn are integers, from the formula thatdefines them. If d is a positive divisor of both pn and qn , then d | (−1)n−1 ,so d = 1.
SAGE Example 5.2.11. We illustrate Corollary 5.2.10 using Sage.
sage: c = continued_fraction([1,2,3,4,5])
sage: c.convergents()[1, 3/2, 10/7, 43/30, 225/157]
sage: [c.pn(n) for n in range(len(c))][1, 3, 10, 43, 225]
sage: [c.qn(n) for n in range(len(c))][1, 2, 7, 30, 157]

5.2.2

The Sequence of Partial Convergents

Let [a0 , . . . , am ] be a continued fraction and for n ≤ m let
cn = [a0 , . . . , an ] =pn
qn

denote the nth convergent. Recall that by definition of continued fraction, an > 0 for n > 0, which gives the partial convergents of a continued fraction additional structure. For example, the partial convergents of[2, 1, 2, 1, 1, 4, 1, 1, 6] are
2, 3, 8/3, 11/4, 19/7, 87/32, 106/39, 193/71, 1264/465.
To make the size of these numbers clearer, we approximate them using
decimals. We also underline every other number, to illustrate some extra
structure.
2, 3, 2.66667, 2.75000, 2.71429, 2.71875, 2.71795, 2.71831, 2.71828
The underlined numbers are smaller than all of the nonunderlined numbers,
and the sequence of underlined numbers is strictly increasing, whereas thenonunderlined numbers strictly decrease.
SAGE Example 5.2.12. Figure 5.1 illustrates the above pattern on anothercontinued fraction using Sage.
sage: c = continued_fraction([1,1,1,1,1,1,1,1])
sage: v = [(i, c.pn(i)/c.qn(i)) for i in range(len(c))]
sage: P = point(v, rgbcolor=(0,0,1), pointsize=40)
sage: L = line(v, rgbcolor=(0.5,0.5,0.5))
sage: L2 = line([(0,c.value()),(len(c)-1,c.value())], \...
thickness=0.5, rgbcolor=(0.7,0,0))
sage: (L+L2+P).show(xmin=0,ymin=1)

100

5. Continued Fractions

2

1.5

2

4

6

FIGURE 5.1. Graph of a Continued Fraction

We next prove that this extra structure is a general phenomenon.
Proposition 5.2.13 (How Convergents Converge). The even indexed convergents c2n increase strictly with n, and the odd indexed convergents c2n+1
decrease strictly with n. Also, the odd indexed convergents c2n+1 are greaterthan all of the even indexed convergents c2m .
Proof. The an are positive for n ≥ 1, so the qn are positive. By Proposition 5.2.7, for n ≥ 2,cn − cn−2 = (−1)n ·an,
qn qn−2

which proves the first claim.
Suppose for the sake of contradiction that there exist integers r and msuch that c2m+1 < c2r . Proposition 5.2.7 implies that for n ≥ 1,cn − cn−1 = (−1)n−1 ·1
qn qn−1

has sign (−1)n−1 , so for all s ≥ 0 we have c2s+1 > c2s . Thus it is impossiblethat r = m. If r < m, then by what we proved in the first paragraph,c2m+1 < c2r < c2m , a contradiction (with s = m). If r > m, then c2r+1 <c2m+1 < c2r , which is also a contradiction (with s = r).

5.2.3

Every Rational Number is Represented

Proposition 5.2.14 (Rational Continued Fractions). Every nonzero rational number can be represented by a simple continued fraction.

5.3 Infinite Continued Fractions

101

Proof. Without loss of generality, we may assume that the rational number
is a/b, with b ≥ 1 and gcd(a, b) = 1. Algorithm 1.1.13 gives:a = b · a0 + r1 ,0 < r1 < bb = r1 · a1 + r2 ,0 < r2 < r1···
rn−2 = rn−1 · an−1 + rn ,0 < rn < rn−1

rn−1 = rn · an + 0.
Note that ai > 0 for i > 0 (also rn = 1, since gcd(a, b) = 1). Rewrite theequations as follows:
a/b = a0 + r1 /b = a0 + 1/(b/r1 ),b/r1 = a1 + r2 /r1 = a1 + 1/(r1 /r2 ),r1 /r2 = a2 + r3 /r2 = a2 + 1/(r2 /r3 ),···rn−1 /rn = an .
It follows thata= [a0 , a1 , . . . , an ].
b

The proof of Proposition 5.2.14 leads to an algorithm for computing thecontinued fraction of a rational number.
A nonzero rational number can be represented in exactly two ways; forexample, 2 = [1, 1] = [2] (see Exercise 5.2).

5.3 Infinite Continued Fractions
This section begins with the continued fraction procedure, which associates
a sequence a0 , a1 , . . . of integers to a real number x. After giving several
examples, we prove that x = limn→∞ [a0 , a1 , . . . , an ] by proving that the
odd and even partial convergents become arbitrarily close to each other.
We also show that if a0 , a1 , . . . is any infinite sequence of positive integers,then the sequence of cn = [a0 , a1 , . . . , an ] converges.PMore generally, if an∞
is an arbitrary sequence of positive reals such that n=0 an diverges then(cn ) converges.

5.3.1

The Continued Fraction Procedure

Let x ∈ R and write
x = a0 + t0

102

5. Continued Fractions

with a0 ∈ Z and 0 ≤ t0 < 1. We call the number a0 the floor of x, and we
also sometimes write a0 = bxc. If t0 6= 0, write1= a1 + t1
t01= [0, a1 + t1 ], which is awith a1 ∈ N and 0 ≤ t1 < 1. Thus t0 = a1 +t
1
continued fraction expansion of t0 , which need not be simple. Continue inthis manner so long as tn 6= 0 writing

1= an+1 + tn+1
tnwith an+1 ∈ N and 0 ≤ tn+1 < 1. We call this procedure, which associates
to a real number x the sequence of integers a0 , a1 , a2 , . . ., the continued
fraction process.
Example 5.3.1. Let x = 38 . Then x = 2 + 23 , so a0 = 2 and t0 = 23 . Then
3
1
1
1
1
t0 = 2 = 1 + 2 , so a1 = 1 and t1 = 2 . Then t1 = 2, so a2 = 2, t2 = 0, andthe sequence terminates. Notice that8
= [2, 1, 2],
3
so the continued fraction procedure produces the continued fraction of 83 .
√Example 5.3.2. Let x = 1+2 5 . Then
−1 +
x=1+
2

√

5

,

√

so a0 = 1 and t0 = −1+2 5 . We have
√
√
1
2
1+ 5
−2 − 2 5
√ =
=
=
,
t0
−4
2
−1 + 5√so a1 = 1 and t1 = −1+2 5 . Likewise, an = 1 for all n. As we will see below,
the following exciting equality makes sense.
√
1
1+ 5
=1+
1
2
1+
1
1+
1
1+
1
1+
1 + ···SAGE Example 5.3.3. The equality of Example 5.3.2 is consistent with thefollowing Sage calculation:

5.3 Infinite Continued Fractions

103

sage: def cf(bits):...
x = (1 + sqrt(RealField(bits)(5))) / 2...
return continued_fraction(x)sage: cf(10)[1, 1, 1, 1, 1, 1, 1]
sage: cf(30)[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1]
sage: cf(50)[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]
Example 5.3.4. Suppose x = e = 2.71828182 . . .. Using the continued fraction procedure, we find thata0 , a1 , a2 , . . . = 2, 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8, 1, 1, 10, . . .
For example, a0 = 2 is the floor of 2. Subtracting 2 and inverting, we
obtain 1/0.718 . . . = 1.3922 . . ., so a1 = 1. Subtracting 1 and invertingyields 1/0.3922 . . . = 2.5496 . . ., so a2 = 2. We will prove in Section 5.4
that the continued fraction of e obeys a simple pattern.
The 5th partial convergent of the continued fraction of e is[a0 , a1 , a2 , a3 , a4 , a5 ] =87= 2.71875,
32

which is a good rational approximation to e, in the sense that87
− e = 0.000468 . . . .
32
Note that 0.000468 . . . < 1/322 = 0.000976 . . ., which illustrates the boundin Corollary 5.3.11.
Let’s do the same thing with π = 3.14159265358979 . . .. Applying thecontinued fraction procedure, we find that the continued fraction of π is
a0 , a1 , a2 , . . . = 3, 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, 1, 14, . . .
The first few partial convergents are3,

22 333 355 103993
,
,
,
,···
7 106 113 33102

These are good rational approximations to π; for example,
103993= 3.14159265301 . . . .
33102
Notice that the continued fraction of e exhibits a nice pattern (see Section 5.4 for a proof), whereas the continued fraction of π exhibits no pattern

104

5. Continued Fractions

that is obvious to the author. The continued fraction of π has been extensively studied, and over 20 million terms have been computed. The datasuggests that every integer appears infinitely often as a partial convergent.
For much more about the continued fraction of π, or of any other sequencein this book, type the first few terms of the sequence into [Slo].

5.3.2

Convergence of Infinite Continued Fractions

Lemma 5.3.5. For every n such that an is defined, we havex = [a0 , a1 , . . . , an + tn ],and if tn 6= 0, then x = [a0 , a1 , . . . , an , t1n ].
Proof. We use induction. The statements are both true when n = 0. If thesecond statement is true for n − 1, thenx = a0 , a1 , . . . , an−1 ,

1



tn−1

= [a0 , a1 , . . . , an−1 , an + tn ]1.
= a0 , a1 , . . . , an−1 , an ,tn
Similarly, the first statement is true for n if it is true for n − 1.
Theorem 5.3.6 (Continued Fraction Limit). Let a0 , a1 , . . . be a sequenceof integers such that an > 0 for all n ≥ 1, and for each n ≥ 0, set cn =[a0 , a1 , . . . an ]. Then lim cn exists.
n→∞

Proof. For any m ≥ n, the number cn is a partial convergent of [a0 , . . . , am ].
By Proposition 5.2.13, the even convergents c2n form a strictly increasing
sequence and the odd convergents c2n+1 form a strictly decreasing sequence.
Moreover, the even convergents are all ≤ c1 and the odd convergents areall ≥ c0 . Hence α0 = limn→∞ c2n and α1 = limn→∞ c2n+1 both exist, andα0 ≤ α1 . Finally, by Proposition 5.2.7|c2n − c2n−1 | =

1
1
≤
→ 0,
q2n · q2n−12n(2n − 1)so α0 = α1 .
We define[a0 , a1 , . . .] = lim cn .
n→∞5.3 Infinite Continued Fractions

105

Example 5.3.7. We illustrate the theorem with x = π. As in the proof of
Theorem 5.3.6, let cn be the nth partial convergent to π. The cn with n
odd converge down to π
c1 = 3.1428571 . . . , c3 = 3.1415929 . . . , c5 = 3.1415926 . . .
whereas the cn with n even converge up to πc2 = 3.1415094 . . . , c4 = 3.1415926 . . . , c6 = 3.1415926 . . . .
Theorem 5.3.8. Let a0 , a1 , a2 , . . . be a sequence of real numbers such thatan > 0 for all n ≥ 1, and for each n
0, set cn = [a0 , a1 , . . . an ]. ThenP≥∞
lim cn exists if and only if the sum n=0 an diverges.
n→∞P
Proof. We only prove that if an diverges, then limn→∞ cn exists. A proofof the converse can be found in [Wal48, Ch. 2, Thm. 6.1].
Let qn be the sequence of “denominators” of the partial convergents, as
defined in Section 5.2.1, so q−2 = 1, q−1 = 0, and for n ≥ 0, we have
qn = an qn−1 + qn−2 .
As we saw in the proof of Theorem 5.3.6, the limit limn→∞ cn exists provided that the sequence {qn qn−1 } diverges to positive infinity.
For n even,
qn = an qn−1 + qn−2
= an qn−1 + an−2 qn−3 + qn−4
= an qn−1 + an−2 qn−3 + an−4 qn−5 + qn−6
= an qn−1 + an−2 qn−3 + · · · + a2 q1 + q0

and for n odd,
qn = an qn−1 + an−2 qn−3 + · · · + a1 q0 + q−1 .
Since an > 0 for n > 0, the sequence {qn } is increasing, so qi ≥ 1 for alli ≥ 0. Applying this fact to the above expressions for qn , we see that for neven
qn ≥ an + an−2 + · · · + a2 ,and for n odd
qn ≥ an + an−2 + · · · + a1 .
PPIfan diverges, then at least one of
a2n ora2n+1 must diverge.
The above inequalities then imply that at least one of the sequences {q2n }or {q2n+1 } diverge to infinity. Since {qn } is an increasing sequence, it follows
that {qn qn−1 } diverges to infinity.
P106

5. Continued Fractions

1
Example 5.3.9. Let an = n log(n)for n ≥ 2 and a0 = a1 = 0. By theP
integral test,
an diverges, so by Theorem 5.3.8, the continued fraction[a0 , a1 , a2 , . . .] converges. This convergence is very slow, since, e.g.

[a0 , a1 , . . . , a9999 ] = 0.5750039671012225425930 . . .
yet[a0 , a1 , . . . , a10000 ] = 0.7169153932917378550424 . . . .
Theorem 5.3.10. Let x ∈ R be a real number. Then x is the value of the(possibly infinite) simple continued fraction [a0 , a1 , a2 , . . .] produced by thecontinued fraction procedure.
Proof. If the sequence is finite, then some tn = 0 and the result follows by
Lemma 5.3.5. Suppose the sequence is infinite. By Lemma 5.3.5,
x = [a0 , a1 , . . . , an ,

1
].
tn

By Proposition 5.2.5 (which we apply in a case when the partial quotientsof the continued fraction are not integers), we have1
· pn + pn−1
tn
x=
.
1
· qn + qn−1
tn
Thus, if cn = [a0 , a1 , . . . , an ], then
pn
qn
1
1
p
n qn + pn−1 qn − tn pn qn − pn qn−1


= tn
.
qn t1n qn + qn−1

x − cn = x −=pn−1 qn − pn qn−1

qn t1n qn + qn−1
(−1)n

=
qn



1
tn qn + qn−1

.

Thus
1

|x − cn | =qn

1
tn qn + qn−1

1
qn (an+1 qn + qn−1 )
1
1
=
≤
→ 0.
qn · qn+1
n(n + 1)<

5.4 The Continued Fraction of e107

In the inequality, we use that an+1 is the integer part of t1n , and is hence≤ t1n .
This corollary follows from the proof of Theorem 5.3.10.
Corollary 5.3.11 (Convergence of continued fraction). Let a0 , a1 , . . . define a simple continued fraction, and let x = [a0 , a1 , . . .] ∈ R be its value.
Then for all m,
pm
1
x−
<
.
qm
qm · qm+1
Proposition 5.3.12. If x is a rational number, then the sequence a0 , a1 , . . .
produced by the continued fraction procedure terminates.
Proof. Let [b0 , b1 , . . . , bm ] be the continued fraction representation of x that
we obtain using Algorithm 1.1.13, so the bi are the partial quotients at eachstep. If m = 0, then x is an integer, so we may assume m > 0. Thenx = b0 + 1/[b1 , . . . , bm ].
If [b1 , . . . , bm ] = 1, then m = 1 and b1 = 1, which will not happen using
Algorithm 1.1.13, since it would give [b0 +1] for the continued fraction of theinteger b0 + 1. Thus [b1 , . . . , bm ] > 1, so in the continued fraction algorithm
we choose a0 = b0 and t0 = 1/[b1 , . . . , bm ]. Repeating this argument enoughtimes proves the claim.

5.4 The Continued Fraction of e
The continued fraction expansion of e begins [2, 1, 2, 1, 1, 4, 1, 1, 6, . . .]. The
obvious pattern in fact does continue, as Euler proved in 1737 (see [Eul85]),
and we will prove in this section. As an application, Euler gave a proof
that e is irrational by noting that its continued fraction is infinite.
The proof we give below draws heavily on the proof in [Coh], which
describes a slight variant of a proof of Hermite (see [Old70]). The continued
fraction representation of e is also treated in the German book [Per57], but
the proof requires substantial background from elsewhere in that text.

5.4.1

Preliminaries

First, we write the continued fraction of e in a slightly different form.
Instead of [2, 1, 2, 1, 1, 4, . . .], we can start the sequence of coefficients[1, 0, 1, 1, 2, 1, 1, 4, . . .]to make the pattern the same throughout. (Everywhere else in this chapter we assume that the partial quotients an for n ≥ 1 are positive, but

108

5. Continued Fractions

temporarily relax that condition here and allow a1 = 0.) The numeratorsand denominators of the convergents given by this new sequence satisfy a
simple recurrence. Using ri as a stand-in for pi or qi , we have
r3n = r3n−1 + r3n−2
r3n−1 = r3n−2 + r3n−3
r3n−2 = 2(n − 1)r3n−3 + r3n−4 .
Our first goal is to collapse these three recurrences into one recurrence
that only makes mention of r3n , r3n−3 , and r3n−6 . We have
r3n = r3n−1 + r3n−2= (r3n−2 + r3n−3 ) + (2(n − 1)r3n−3 + r3n−4 )= (4n − 3)r3n−3 + 2r3n−4 .
This same method of simplification also shows us thatr3n−3 = 2r3n−7 + (4n − 7)r3n−6 .
To get rid of 2r3n−4 in the first equation, we make the substitutions
2r3n−4 = 2(r3n−5 + r3n−6 )= 2((2(n − 2)r3n−6 + r3n−7 ) + r3n−6 )= (4n − 6)r3n−6 + 2r3n−7 .
Substituting for 2r3n−4 and then 2r3n−7 , we finally have the needed collapsed recurrence,
r3n = 2(2n − 1)r3n−3 + r3n−6 .

5.4.2

Two Integral Sequences

We define the sequences xn = p3n , yn = q3n . Since the 3n-convergents will
converge to the same real number that the n convergents do, xn /yn also
converges to the limit of the continued fraction. Each sequence {xn }, {yn }will obey the recurrence relation derived in the previous section (where zn
is a stand-in for xn or yn ):zn = 2(2n − 1)zn−1 + zn−2 , for all n ≥ 2.

(5.4.1)

The two sequences can be found in Table 5.1. (The initial conditions
x0 = 1, x1 = 3, y0 = y1 = 1 are taken straight from the first few convergentsof the original continued fraction.) Notice that since we are skipping several
convergents at each step, the ratio xn /yn converges to e very quickly.

5.4 The Continued Fraction of e

109

TABLE 5.1. Convergentsnxn
yn
xn /yn

5.4.3

0
1
1
1

1
3
1
3

2
19
7
2.714 . . .

3
193
71
2.71830 . . .

4
2721
1001
2.7182817 . . .

···
···
···
···

A Related Sequence of Integrals

Now, we define a sequence of real numbers T0 , T1 , T2 , . . . by the following
integrals:

Tn =

Z 1 n
t (t − 1)n t
e dt.
n!
0

Below, we compute the first two terms of this sequence explicitly. (When
we compute T1 , we are doing the integration by parts u = t(t−1), dv = et dt.
Since the integral runs from 0 to 1, the boundary condition is 0 whenevaluated at each of the endpoints. This vanishing will be helpful when wedo the integral in the general case.)
Z 1
T0 =

et dt = e − 1,

0

Z 1
T1 =

t(t − 1)et dt

0

Z 1
=−

((t − 1) + t)et dt

01= −(t − 1)e

t

1

− te
0

t

Z 1
+2

0

et dt

0

= −1 − e + 2(e − 1) = e − 3.
The reason that we defined this series now becomes apparent: T0 =y0 e − x0 and T1 = y1 e − x1 . In general, it will be true that Tn = yn e − xn .
We will now prove this fact.
It is clear that if Tn were to satisfy the same recurrence that the xi andyi do in (5.4.1), then the above statement holds by induction. (The initialconditions are correct, as needed.) So, we simplify Tn by integrating by

110

5. Continued Fractions

parts twice in succession:
Z 1 nt (t − 1)n t
Tn =
e dt
n!
0
Z 1 n−1t(t − 1)n + tn (t − 1)n−1 t=−
e dt(n − 1)!
0
Z 1  n−2t(t − 1)ntn−1 (t − 1)n−1=+n(n − 2)!
(n − 1)!
0n−1t(t − 1)n−1
tn (t − 1)n−2  t
+n
e dt
+
(n − 1)!
(n − 2)!
Z 1 n−2t(t − 1)n−2 2= 2nTn−1 +(2t − 2t + 1) et dt(n − 2)!
0
Z 1 n−2Z 1 n−1t(t − 1)n−2 tt(t − 1)n−1 t
e dt +
e dt= 2nTn−1 + 2(n − 2)!
(n − 2)!
00= 2nTn−1 + 2(n − 1)Tn−1 + Tn−2
= 2(2n − 1)Tn−1 + Tn−2 ,which is the desired recurrence.
Therefore, Tn = yn e − xn . To conclude the proof, we consider the limit
as n approaches infinity:
Z 1 nt (t − 1)n t
lim
e dt = 0,n→∞ 0n!
by inspection, and thereforelimxn

n→∞ yn= lim (e −n→∞Tn
) = e.
yn

Therefore, the ratio xn /yn approaches e, and the continued fraction expansion [2, 1, 2, 1, 1, 4, 1, 1, . . .] does in fact converge to e.

5.4.4

Extensions of the Argument

The method of proof of this section generalizes to show that the continued
fraction expansion of e1/n is[1, (n − 1), 1, 1, (3n − 1), 1, 1, (5n − 1), 1, 1, (7n − 1), . . .]for all n ∈ N (see Exercise 5.6).

5.5 Quadratic Irrationals
The main result of this section is that the continued fraction expansion of
a number is eventually repeating if and only if the number is a quadratic

5.5 Quadratic Irrationals

111

irrational. This can be viewed as an analog for continued fractions of thefamiliar fact that the decimal expansion of x is eventually repeating if andonly if x is rational. The proof that continued fractions of quadratic irrationals eventually repeats is surprisingly difficult and involves an interesting
finiteness argument. Section 5.5.2 emphasizes our striking ignorance about
continued fractions of real roots of irreducible polynomials over Q of degree
bigger than 2.
Definition 5.5.1 (Quadratic Irrational). A quadratic irrational is a real
number α ∈ R that is irrational and satisfies a quadratic polynomial withcoefficients in Q.
√
Thus, for example, (1 + 5)/2 is a quadratic irrational. Recall that√
1+ 5
= [1, 1, 1, . . .].
2√
The continued
fraction of 2 is [1, 2, 2, 2, 2, 2, . . .], and the continued frac√tion of 389 is[19, 1, 2, 1, 1, 1, 1, 2, 1, 38, 1, 2, 1, 1, 1, 1, 2, 1, 38, . . .].
Does the [1, 2, 1, 1, 1, 1, 2, 1, 38] pattern repeat over and over again?
SAGE Example√ 5.5.2. We compute more terms of the continued fraction
expansion of 389 using Sage:sage: def cf_sqrt_d(d, bits):...
x = sqrt(RealField(bits)(d))...
return continued_fraction(x)sage: cf_sqrt_d(389,50)[19, 1, 2, 1, 1, 1, 1, 2, 1, 38, 1, 2, 1, 1, 1, 1, 2, 1, 38, 2]
sage: cf_sqrt_d(389,100)[19, 1, 2, 1, 1, 1, 1, 2, 1, 38, 1, 2, 1, 1, 1, 1, 2, 1, 38,
1, 2, 1, 1, 1, 1, 2, 1, 38, 1, 2, 1, 1, 1, 1, 2, 1, 38, 1,
2, 1, 1]

5.5.1

Periodic Continued Fractions

Definition 5.5.3 (Periodic Continued Fraction). A periodic continued
fraction is a continued fraction [a0 , a1 , . . . , an , . . .] such thatan = an+hfor some fixed positive integer h and all sufficiently large n. We call theminimal such h the period of the continued fraction.

112

5. Continued Fractions

Example 5.5.4. Consider the periodic continued fraction [1, 2, 1, 2, . . .] =[1, 2]. What does it converge to? We have1

[1, 2] = 1 +

,

1

2+

1

1+
2+

1
1 + ···

so if α = [1, 2] then
α=1+

1
1
2+
α

=1+

1
α
3α + 1=1+=2α + 12α + 12α + 1αThus 2α2 − 2α − 1 = 0, so
√
1+ 3
α=
.
2
Theorem 5.5.5 (Periodic Characterization). An infinite simple continued
fraction is periodic if and only if it represents a quadratic irrational.
Proof. (=⇒) First suppose that[a0 , a1 , . . . , an , an+1 , . . . , an+h ]is a periodic continued fraction. Set α = [an+1 , an+2 , . . .]. Thenα = [an+1 , . . . , an+h , α],so by Proposition 5.2.5α=αpn+h + pn+h−1.
αqn+h + qn+h−1

Here we use that α is the last partial quotient. Thus, α satisfies a quadratic
equation with coefficients in Q. Computing as in Example 5.5.4 and rationalizing the denominators, and using that the ai are all integers, showsthat[a0 , a1 , . . .] = [a0 , a1 , . . . , an , α]1
= a0 +
1a1 +
1a2 + · · · +αis of the form c + dα, with c, d ∈ Q, so [a0 , a1 , . . .] also satisfies a quadratic
polynomial over Q.

5.5 Quadratic Irrationals

113

The continued fraction procedure applied to the value of an infinite simple continued fraction yields that continued fraction back, so by Proposition 5.3.12, α 6∈ Q because it is the value of an infinite continued fraction.
(⇐=) Suppose α ∈ R is an irrational number that satisfies a quadratic
equationaα2 + bα + c = 0(5.5.1)
with a, b, c ∈ Z and a 6= 0. Let [a0 , a1 , . . .] be the continued fraction expansion of α. For each n, let
rn = [an , an+1 , . . .],so
α = [a0 , a1 , . . . , an−1 , rn ].
We will prove periodicity by showing that the set of rn ’s is finite. If wehave shown finiteness, then there exists n, h > 0 such that rn = rn+h , so
[a0 , . . . , an−1 , rn ] = [a0 , . . . , an−1 , an , . . . , an+h−1 , rn+h ]= [a0 , . . . , an−1 , an , . . . , an+h−1 , rn ]= [a0 , . . . , an−1 , an , . . . , an+h−1 , an , . . . , an+h−1 , rn+h ]= [a0 , . . . , an−1 , an , . . . , an+h−1 ].
It remains to show there are only finitely many distinct rn . We haveα=pn
rn pn−1 + pn−2=.
qn
rn qn−1 + qn−2

Substituting this expression for α into the quadratic equation (5.5.1), we
see that
An rn2 + Bn rn + Cn = 0,
where2
An = ap2n−1 + bpn−1 qn−1 + cqn−1,

Bn = 2apn−1 pn−2 + b(pn−1 qn−2 + pn−2 qn−1 ) + 2cqn−1 qn−2 , and2
Cn = ap2n−2 + bpn−2 qn−2 + cqn−2.

Note that An , Bn , Cn ∈ Z, that Cn = An−1 , and that
Bn2 − 4An Cn = (b2 − 4ac)(pn−1 qn−2 − qn−1 pn−2 )2 = b2 − 4ac.
Recall from the proof of Theorem 5.3.10 that
α−

1
pn−1
<
.
qn−1
qn qn−1

114

5. Continued Fractions

Thus,|αqn−1 − pn−1 | <
so

1
1
<
,
qn
qn−1

pn−1 = αqn−1 +δ
qn−1with |δ| < 1.

2



δ

Hence,
An = a αqn−1 +δ
qn−1+ b αqn−1 +



qn−1

2
qn−1 + cqn−1δ2
2= (aα2 + bα + c)qn−1+ 2aαδ + a 2 + bδ
qn−1
δ2
= 2aαδ + a 2 + bδ.
qn−1
Thus,
δ2
|An | = 2aαδ + a 2 + bδ < 2|aα| + |a| + |b|.
qn−1
We conclude that there are only finitely many possibilities for the integer
An . Also,p|Cn | = |An−1 | and |Bn | = b2 − 4(ac − An Cn ),so there are only finitely many triples (An , Bn , Cn ), and hence only finitely
many possibilities for rn as n varies, which completes the proof. (The proof
above closely follows [HW79, Thm. 177, pg.144–145].)

5.5.2

Continued Fractions of Algebraic Numbers of Higher
DegreeDefinition 5.5.6 (Algebraic Number). An algebraic number is a root of apolynomial f ∈ Q[x].
Open Problem 5.5.7. Give a simple description√ of the complete continued fractions expansion of the algebraic number 3 2. It begins[1, 3, 1, 5, 1, 1, 4, 1, 1, 8, 1, 14, 1, 10, 2, 1, 4, 12, 2, 3, 2, 1, 3, 4, 1, 1, 2, 14,
3, 12, 1, 15, 3, 1, 4, 534, 1, 1, 5, 1, 1, . . .]The author does not see a pattern, and the 534 reduces his confidence
that he will. Lang and√ Trotter (see [LT72]) analyzed many terms√of thecontinued fraction of 3 2 statistically, and their work suggests that 3 2 has
an “unusual” continued fraction; later work in [LT74] suggests that maybe
it does not.

5.6 Recognizing Rational Numbers

115

Khintchine (see [Khi63, pg. 59])
No properties of the representing continued fractions, analogous
to those which have just been proved, are known for algebraic
numbers of higher degree [as of 1963]. [...] It is of interest to
point out that up till the present time no continued fraction
development of an algebraic number of higher degree than thesecond is known [emphasis added]. It is not even known if sucha development has bounded elements. Generally speaking the
problems associated with the continued fraction expansion of algebraic numbers of degree higher than the second are extremely
difficult and virtually unstudied.
Richard Guy (see [Guy94, pg. 260])
Is there an algebraic number of degree greater than two whose
simple continued fraction has unbounded partial quotients? Does
every such number have unbounded partial quotients?
Baum and Sweet [BS76] answered the analog of Richard Guy’s question,
but with algebraic numbers replaced by elements of a field K other than Q.
(The field K is F2 ((1/x)), the field of Laurent series in the variable 1/x overthe finite field with two elements. An element of K is a polynomial in x plus
a formal power series in 1/x.) They found an α of degree 3 over K whose
continued fraction has all terms of bounded degree, and other elements ofvarious degrees greater than 2 over K whose continued fractions have termsof unbounded degree.

5.6 Recognizing Rational Numbers
Suppose that somehow you can compute approximations to some rational
number, and want to figure what the rational number probably is. Computing the approximation to high enough precision to find a period in thedecimal expansion is not a good approach, because the period can be huge(see below). A much better approach is to compute the simple continued
fraction of the approximation, and truncate it before a large partial quotient an , then compute the value of the truncated continued fraction. This
results in a rational number that has a relatively small numerator and denominator, and is close to the approximation of the rational number, since
the tail end of the continued fraction is at most 1/an .
We begin with a contrived example, which illustrates how to recognize arational number. Let
x = 9495/3847 = 2.46815700545879906420587470756433584611385 . . . .

116

5. Continued Fractions

The continued fraction of the truncation 2.468157005458799064 is[2, 2, 7, 2, 1, 5, 1, 1, 1, 1, 1, 1, 328210621945, 2, 1, 1, 1, . . .]We have

9495
.
3847
Notice that no repetition is evident in the digits of x given above, though
we know that the decimal expansion of x must be eventually periodic, since
all decimal expansions of rational numbers are eventually periodic. In fact,
the length of the period of the decimal expansion of 1/3847 is 3846, whichis the order of 10 modulo 3847 (see Exercise 5.7).
For a slightly less contrived application of this idea, suppose f (x) ∈ Z[x]is a polynomial with integer coefficients, and we know for some reason that
one root of f is a rational number. We can find that rational number, byusing Newton’s method to approximate each root, and continued fractions
to decide whether each root is a rational number (we can substitute thevalue of the continued fraction approximation into f to see if it is actuallya root). One could also use the well-known Rational Root Theorem, whichasserts that any rational root n/d of f , with n, d ∈ Z coprime, has theproperty that n divides the constant term of f and d the leading coefficientof f . However, using that theorem to find n/d would require factoring theconstant and leading terms of f , which could be completely impractical
if they have a few hundred digits (see Section 1.1.3). In contrast, Newton’s method and continued fractions should quickly find n/d, assuming
the degree of f isn’t too large.
For example, suppose f = 3847x2 − 14808904x + 36527265. To apply
Newton’s method, let x0 be a guess for a root of f . Iterate using the recurrencef (xn ).
xn+1 = xn − 0
f (xn )
Choosing x0 = 0, approximations of the first two iterates are[2, 2, 7, 2, 1, 5, 1, 1, 1, 1, 1, 1] =x1 = 2.466574501394566404103909378,
andx2 = 2.468157004807401923043166846.
The continued fraction of the approximations x1 and x2 are[2, 2, 6, 1, 47, 2, 1, 4, 3, 1, 5, 8, 2, 3]
and[2, 2, 7, 2, 1, 5, 1, 1, 1, 1, 1, 1, 103, 8, 1, 2, 3, . . .].
Truncating the continued fraction of x2 before 103 gives[2, 2, 7, 2, 1, 5, 1, 1, 1, 1, 1, 1],
which evaluates to 9495/3847, which is a rational root of f .

5.7 Sums of Two Squares

117

SAGE Example 5.6.1. We do the above calculation using SAGE. First we
implement the Newton iteration:
sage: def newton_root(f, iterates=2, x0=0, prec=53):...
x = RealField(prec)(x0)...
R = PolynomialRing(ZZ,’x’)...
f = R(f)...
g = f.derivative()...
for i in range(iterates):...
x = x - f(x)/g(x)...
return x
Next we run the Newton iteration, and compute the continued fraction of
the result:
sage: a = newton_root(3847*x^2 - 14808904*x + 36527265); a2.46815700480740
sage: cf = continued_fraction(a); cf[2, 2, 7, 2, 1, 5, 1, 1, 1, 1, 1, 1, 103, 8, 1, 2, 3]
We truncate the continued fraction and compute its value.
sage: c = cf[:12]; c[2, 2, 7, 2, 1, 5, 1, 1, 1, 1, 1, 1]
sage: c.value()
9495/3847
Another computational application of continued fractions, which we canonly hint at, is that there are functions in certain parts of advanced number
theory (that are beyond the scope of this book) that take rational values
at certain points, and which can only be computed efficiently via approximations; using continued fractions as illustrated above to evaluate suchfunctions is crucial.

5.7 Sums of Two Squares
In this section, we apply continued fractions to prove the following theorem.
Theorem 5.7.1. A positive integer n is a sum of two squares if and only
if all prime factors of p | n such that p ≡ 3 (mod 4) have even exponent in
the prime factorization of n.
We first consider some examples. Notice that 5 = 12 + 22 is a sum of
two squares, but 7 is not a sum of two squares. Since 2001 is divisibleby 3 (because 2 + 1 is divisible by 3), but not by 9 (since 2 + 1 is not),
Theorem 5.7.1 implies that 2001 is not a sum of two squares. The theorem
also implies that 2 · 34 · 5 · 72 · 13 is a sum of two squares.

118

5. Continued Fractions

SAGE Example 5.7.2. We use Sage to write a short program that naively
determines whether or not an integer n is a sum of two squares, and if so
returns a, b such that a2 + b2 = n.
sage: def sum_of_two_squares_naive(n):...
for i in range(int(sqrt(n))):...
if is_square(n - i^2):...
return i, (Integer(n-i^2)).sqrt()...
return "%s is not a sum of two squares"%n
We next use our function in a couple of cases.
sage: sum_of_two_squares_naive(23)’23 is not a sum of two squares’
sage: sum_of_two_squares_naive(389)(10, 17)
sage: sum_of_two_squares_naive(2007)’2007 is not a sum of two squares’
sage: sum_of_two_squares_naive(2008)’2008 is not a sum of two squares’
sage: sum_of_two_squares_naive(2009)(28, 35)
sage: 28^2 + 35^22009
sage: sum_of_two_squares_naive(2*3^4*5*7^2*13)(189, 693)
Definition 5.7.3 (Primitive). A representation n = x2 + y 2 is primitive
if x and y are coprime.
Lemma 5.7.4. If n is divisible by a prime p ≡ 3 (mod 4), then n has no
primitive representations.
Proof. Suppose n has a primitive representation, n = x2 + y 2 , and let p be
any prime factor of n. Thenp | x2 + y 2

andgcd(x, y) = 1,

so p - x and p - y. Since Z/pZ is a field, we may divide by y 2 in the equation2x2 + y 2 ≡ 0 (mod p) to see that (x/y) ≡ −1 (mod p). Thus the Legendre

symbol−1pequals +1. However, by Proposition 4.2.1,


so



−1
p



−1
p



= (−1)(p−1)/2= 1 if and only if (p − 1)/2 is even, which is to say p ≡ 1 (mod 4).

5.7 Sums of Two Squares

119

Proof of Theorem 5.7.1 (=⇒). Suppose that p ≡ 3 (mod 4) is a prime,that pr | n but pr+1 - n with r odd, and that n = x2 + y 2 . Letting d =gcd(x, y), we havex = dx0 ,y = dy 0 ,andn = d2 n0

with gcd(x0 , y 0 ) = 1 and(x0 )2 + (y 0 )2 = n0 .
Because r is odd, p | n0 , so Lemma 5.7.4 implies that gcd(x0 , y 0 ) > 1, which
is a contradiction.
To prepare for our proof of the implication (⇐=) of Theorem 5.7.1, wereduce the problem to the case when n is prime. Write n = n21 n2 , where n2
has no prime factors p ≡ 3 (mod 4). It suffices to show that n2 is a sum of
two squares, since(x21 + y12 )(x22 + y22 ) = (x1 x2 − y1 y2 )2 + (x1 y2 + x2 y1 )2 ,(5.7.1)

so a product of two numbers that are sums of two squares is also a sum of
two squares. Since 2 = 12 + 12 is a sum of two squares, it suffices to showthat any prime p ≡ 1 (mod 4) is a sum of two squares.
aLemma 5.7.5. If x ∈ R and n ∈ N, then there is a fraction
in lowestbterms such that 0 < b ≤ n and
x−

a
1
≤
.
b
b(n + 1)

Proof. Consider the continued fraction [a0 , a1 , . . .] of x. By Corollary 5.3.11,
for each m
pm
1
x−
<
.
qm
qm · qm+1
Since qm+1 ≥ qm + 1 and q0 = 1, either there exists an m such thatqm ≤ n < qm+1 , or the continued fraction expansion of x is finite and nis larger than the denominator of the rational number x, in which case we
take ab = x and are done. In the first case,
x−
so

1
pm
1
<
≤
,
qm
qm · qm+1
qm · (n + 1)apm=
satisfies the conclusion of the lemma.
bqm

Proof of Theorem 5.7.1 (⇐=). As discussed above, it suffices to prove thatany prime p ≡ 1 (mod 4) is a sum of two squares. Since p ≡ 1 (mod 4),(−1)(p−1)/2 = 1,

120

5. Continued Fractions

Proposition 4.2.1 implies that −1 is a square modulo p; i.e., there exists r ∈√Z such that r2 ≡ −1 (mod p). Lemma 5.7.5, with n = b pc and x = − pr ,√implies that there are integers a, b such that 0 < b < p and
r
1
a
1
− −
≤
< √ .
p
b
b(n + 1)b p
Letting c = rb + pa, we have thatpbp√|c| < √ = √ = pb ppso
0 < b2 + c2 < 2p.
But c ≡ rb (mod p), so
b2 + c2 ≡ b2 + r2 b2 ≡ b2 (1 + r2 ) ≡ 0(mod p).

Thus b2 + c2 = p.
Remark 5.7.6. Our proof of Theorem 5.7.1 leads to an efficient algorithm
to compute a representation of any p ≡ 1 (mod 4) as a sum of two squares.
SAGE Example 5.7.7. We next use Sage and Theorem 5.7.1 to give an
efficient algorithm for writing a prime p ≡ 1 (mod 4) as a sum of two
squares. First we implement the algorithm that comes out of the proof of
the theorem.
sage: def sum_of_two_squares(p):...
p = Integer(p)...
assert p%4 == 1, "p must be 1 modulo 4"...
r = Mod(-1,p).sqrt().lift()...
v = continued_fraction(-r/p)...
n = floor(sqrt(p))...
for x in v.convergents():...
c = r*x.denominator() + p*x.numerator()...
if -n <= c and c <= n:...
return (abs(x.denominator()),abs(c))
Next we use the algorithm to write the first 10-digit prime ≡ 1 (mod 4) asa sum of two squares:
sage: p = next_prime(next_prime(10^10))
sage: sum_of_two_squares(p)(55913, 82908)
The above calculation was essentially instantanoues. If instead we use thenaive algorithm from before, it takes several seconds to write p as a sum of
two squares.
sage: sum_of_two_squares_naive(p)(55913, 82908)

5.8 Exercises

121

5.8 Exercises
5.1 If cn = pn /qn is the nth convergent of [a0 , a1 , . . . , an ] and a0 > 0,show thatpn[an , an−1 , . . . , a1 , a0 ] =pn−1and[an , an−1 , . . . , a2 , a1 ] =(Hint: In the first case, notice that

qn
qn−1

.

pn
pn−2
1
= an += an + pn−1 .)
pn−1
pn−1
pn−2

5.2 Show that every nonzero rational number can be represented in exactly two ways by a finite simple continued fraction. (For example, 2
can be represented by [1, 1] and [2], and 1/3 by [0, 3] and [0, 2, 1].)
5.3 Evaluate the infinite continued fraction [2, 1, 2, 1].
√

5.4 Determine the infinite continued fraction of 1+2 13 .
5.5 Let a0 ∈ R and a1 , . . . , an and b be positive real numbers. Prove that[a0 , a1 , . . . , an + b] < [a0 , a1 , . . . , an ]if and only if n is odd.
5.6 (*) Extend the method presented in the text to show that the continued fraction expansion of e1/k is[1, (k − 1), 1, 1, (3k − 1), 1, 1, (5k − 1), 1, 1, (7k − 1), . . .]for all k ∈ N.
(a) Compute p0 , p3 , q0 , and q3 for the above continued fraction.
Your answers should be in terms of k.
(b) Condense three steps of the recurrence for the numerators anddenominators of the above continued fraction. That is, produce
a simple recurrence for r3n in terms of r3n−3 and r3n−6 whose
coefficients are polynomials in n and k.
(c) Define a sequence of real numbers by
1
Tn (k) = n
k

Z 1/k
0

(kt)n (kt − 1)n t
e dt.
n!

i. Compute T0 (k), and verify that it equals q0 e1/k − p0 .
ii. Compute T1 (k), and verify that it equals q3 e1/k − p3 .

122

5. Continued Fractions

iii. Integrate Tn (k) by parts twice in succession, as in Section 5.4, and verify that Tn (k), Tn−1 (k), and Tn−2 (k) satisfy
the recurrence produced in part 6b, for n ≥ 2.
(d) Conclude that the continued fraction[1, (k − 1), 1, 1, (3k − 1), 1, 1, (5k − 1), 1, 1, (7k − 1), . . .]represents e1/k .
5.7 Let d be an integer that is coprime to 10. Prove that the decimal
expansion of d1 has a period equal to the order ofP10 modulo d. (Hint:1−rn
For every positive integer r, we have 1−10.)−r =n≥1 105.8 Find a positive integer that has at least three different representations
as the sum of two squares, disregarding signs and the order of thesummands.
5.9 Show that if a natural number n is the sum of two rational squares
it is also the sum of two integer squares.
5.10 (*) Let p be an odd prime. Show that p ≡ 1, 3 (mod 8) if and only
if p can be written as p = x2 +2y 2 for some choice of integers x and y.
5.11 Prove that of any four consecutive integers, at least one is not representable as a sum of two squares.

This is page 123
Printer: Opaque this6
Elliptic Curves

Elliptic curves are number theoretic objects that are central to both pureand applied number theory. Deep problems in number theory such as thecongruent number problem—which integers are the area of a right triangle
with rational side lengths?—translate naturally into questions about elliptic curves. Other questions, such as the famous Birch and Swinnerton-Dyer
conjecture, describe mysterious structure that mathematicians expect elliptic curves to have. One can also associate finite abelian groups to elliptic
curves, and in many cases these groups are well suited to the construction of cryptosystems. In particular, elliptic curves are widely believed toprovide good security with smaller key sizes, something that is useful inmany applications, for example, if we are going to print an encryption key
on a postage stamp, it is helpful if the key is short! Morover, there is a
way to use elliptic curves to factor integers, which plays a crucial role insophisticated attacks on the RSA public-key cryptosystem of Section 3.3.
This chapter is a brief introduction to elliptic curves that builds on theideas of Chapters 1–3 and introduces several deep theorems and ideas that
we will not prove. In Section 6.1, we define elliptic curves and draw somepictures of them, and then in Section 6.2 we describe how to put a groupstructure on the set of points on an elliptic curve. Sections 6.3 and 6.4
are about how to apply elliptic curves to two cryptographic problems—
constructing public-key cryptosystems and factoring integers. Finally, in
Section 6.5, we consider elliptic curves over the rational numbers, and explain a deep connection between elliptic curves and a 1,000-year old unsolved problem.

124

6. Elliptic Curves

6
5
4
3
2
1
-3

-2

-1

-1

1

2

3

4

-2
-3
-4
-5
-6

FIGURE 6.1. The elliptic curve y 2 = x3 − 5x + 4 over R

6.1 The Definition
Definition 6.1.1 (Elliptic Curve). An elliptic curve over a field K is a
curve defined by an equation of the form
y 2 = x3 + ax + b,
where a, b ∈ K and −16(4a3 + 27b2 ) 6= 0.
The condition that −16(4a3 + 27b2 ) 6= 0 implies that the curve has no“singular points,” which will be essential for the applications we have inmind (see Exercise 6.1).
SAGE Example 6.1.2. We use the EllipticCurve command to create anelliptic curve over the rational field Q and draw the plot in Figure 6.1.
sage: E = EllipticCurve([-5, 4])
sage: E
Elliptic Curve defined by y^2 = x^3 - 5*x + 4
over Rational Field
sage: P = E.plot(thickness=4,rgbcolor=(0.1,0.7,0.1))
sage: P.show(figsize=[4,6])
We will use elliptic curves over finite fields to factor integers in Section 6.3
and to construct cryptosystems in Section 6.4. The following Sage code
creates an elliptic curve over the finite field of order 37 and plots it, as
illustrated in Figure 6.2.

6.2 The Group Structure on an Elliptic Curve125

sage: E = EllipticCurve(GF(37), [1,0])
sage: E
Elliptic Curve defined by y^2 = x^3 + x over
Finite Field of size 37
sage: E.plot(pointsize=45)

35
30
25
20
15
10
5
5

10

15

20

25

30

35

FIGURE 6.2. The elliptic curve y 2 = x3 + x over Z/37Z

In Section 6.2, we will put a natural abelian group structure on the set
E(K) = {(x, y) ∈ K × K : y 2 = x3 + ax + b} ∪ {O}of K-rational points on an elliptic curve E over K. Here, O may be thoughtof as a point on E “at infinity.” Figure 6.2 contains a plot of the pointsof y 2 = x3 + x over the finite field Z/37Z, though note that we do not
explicitly draw the point at O at infinity.
Remark 6.1.3. If K has characteristic 2 (i.e., we have 1 + 1 = 0 in K), thenfor any choice of a, b, the quantity −16(4a3 + 27b2 ) ∈ K is 0, so according
to Definition 6.1.1 there are no elliptic curves over K. There is a similar
problem in characteristic 3. If we instead consider equations of the form
y 2 + a1 xy + a3 y = x3 + a2 x2 + a4 x + a6 ,we obtain a more general definition of elliptic curves, which correctly allows
for elliptic curves in characteristics 2 and 3; these elliptic curves are popularin cryptography because arithmetic on them is often easier to efficiently
implement on a computer.

6.2 The Group Structure on an Elliptic CurveLet E be an elliptic curve over a field K, given by an equation y 2 =x3 + ax + b. We begin by defining a binary operation + on E(K).

126

6. Elliptic Curves

Algorithm 6.2.1 (Elliptic Curve Group Law). Given P1 , P2 ∈ E(K), thisalgorithm computes a third point R = P1 + P2 ∈ E(K).
1. [Is Pi = O?] If P1 = O set R = P2 or if P2 = O set R = P1 andterminate. Otherwise write (xi , yi ) = Pi .
2. [Negatives] If x1 = x2 and y1 = −y2 , set R = O and terminate.
((3x21 + a)/(2y1 )if P1 = P2 ,3. [Compute λ] Set λ =(y1 − y2 )/(x1 − x2 ) otherwise.

4. [Compute Sum] Then R = λ2 − x1 − x2 , −λx3 − ν , where ν = y1 −λx1 and x3 = λ2 − x1 − x2 is the x-coordinate of R.
Note that in Step 3, if P1 = P2 , then y1 6= 0; otherwise, we would haveterminated in the previous step.
Theorem 6.2.2. The binary operation + defined in Algorithm 6.2.1 endows the set E(K) with an abelian group structure, with identity O.
Before discussing why the theorem is true, we reinterpret + geometrically, so that it will be easier for us to visualize. We obtain the sum P1 + P2
by finding the third point P3 of intersection between E and the line L determined by P1 and P2 , then reflecting P3 about the x-axis. (This description
requires suitable interpretation in cases 1 and 2, and when P1 = P2 .) This is
illustrated in Figure 6.3, in which (0, 2) + (1, 0) = (3, 4) on y 2 = x3 − 5x + 4.
SAGE Example 6.2.3. We create the elliptic curve y 2 = x3 −5x+4 in Sage,
then add together P = (1, 0) and Q = (0, 2). We also compute P +P , whichis the point O at infinity, which is represented in Sage by (0 : 1 : 0), andcompute the sum P + Q + Q + Q + Q, which is surprisingly large.
sage: E = EllipticCurve([-5,4])
sage: P = E([1,0]); Q = E([0,2])
sage: P + Q(3 : 4 : 1)sage: P + P(0 : 1 : 0)sage: P + Q + Q + Q + Q(350497/351649 : 16920528/208527857 : 1)
To further clarify the above geometric interpretation of the group law,
we prove the following proposition.
Proposition 6.2.4 (Geometric Group Law). Suppose Pi = (xi , yi ), i =1, 2 are distinct points on an elliptic curve y 2 = x3 + ax + b, and thatx1 6= x2 . Let L be the unique line through P1 and P2 . Then L intersects thegraph of E at exactly one other point
Q = λ2 − x1 − x2 , λx3 + ν ,where λ = (y1 − y2 )/(x1 − x2 ) and ν = y1 − λx1 .

6.2 The Group Structure on an Elliptic Curve

127

8
6
4
2

-3

-2

-1

1

2

3

4

-2
-4
-6

FIGURE 6.3. The Group Law: (1, 0) + (0, 2) = (3, 4) on y 2 = x3 − 5x + 4

Proof. The line L through P1 , P2 is y = y1 + (x − x1 )λ. Substituting this
into y 2 = x3 + ax + b, we get(y1 + (x − x1 )λ)2 = x3 + ax + b.
Simplifying, we get f (x) = x3 − λ2 x2 + · · · = 0, where we omit the coefficients of x and the constant term since they will not be needed. Since
P1 and P2 are in L ∩ E, the polynomial f has x1 and x2 as roots. By
Proposition
2.5.3, the polynomial f can have at most three roots. WritingQ
f = (x − xi ) and equating terms, we see that x1 + x2 + x3 = λ2 . Thus,
x3 = λ2 − x1 − x2 , as claimed. Also, from the equation for L we see thaty3 = y1 + (x3 − x1 )λ = λx3 + ν, which completes the proof.
To prove Theorem 6.2.2 means to show that + satisfies the three axiomsof an abelian group with O as identity element: existence of inverses, commutativity, and associativity. The existence of inverses follows immediatelyfrom the definition, since (x, y) + (x, −y) = O. Commutativity is also clear
from the definition of group law, since in Parts 1–3, the recipe is unchanged
if we swap P1 and P2 ; in Part 4 swapping P1 and P2 does not change theline determined by P1 and P2 , so by Proposition 6.2.4 it does not change
the sum P1 + P2 .
It is more difficult to prove that + satisfies the associative axiom, i.e.,that (P1 + P2 ) + P3 = P1 + (P2 + P3 ). This fact can be understood from atleast three points of view. One is to reinterpret the group law geometrically(extending Proposition 6.2.4 to all cases), and thus transfer the problem
to a question in plane geometry. This approach is beautifully explained

128

6. Elliptic Curves

with exactly the right level of detail in [ST92, §I.2]. Another approach is to
use the formulas that define + to reduce associativity to checking specificalgebraic identities; this is something that would be extremely tedious to
do by hand, but can be done using a computer (also tedious). A third
approach (see [Sil86] or [Har77]) is to develop a general theory of “divisorson algebraic curves,” from which associativity of the group law falls out
as a natural corollary. The third approach is the best, because it opens
up many new vistas; however, we will not pursue it further because it isbeyond the scope of this book.
SAGE Example 6.2.5. In the following Sage session, we use the formulafrom Algorithm 6.2.1 to verify that the group law holds for any choiceof points P1 , P2 , P3 on any elliptic curve over Q such that the points
P1 , P2 , P3 , P1 + P2 , P2 + P3 are all distinct and nonzero. We define a polynomial ring R in 8 variables.
sage: R.<x1,y1,x2,y2,x3,y3,a,b> = QQ[]
We define the relations the xi will satisfy, and a quotient ring Q in whichthose relations are satisfied. (Quotients of polynomial rings are a generalization of the construction Z/nZ that may be viewed as the quotient of thering Z of integers by the relation that sets n to equal 0.)
sage: rels = [y1^2 - (x1^3 + a*x1 + b),...
y2^2 - (x2^3 + a*x2 + b),...
y3^2 - (x3^3 + a*x3 + b)]...
sage: Q = R.quotient(rels)
We define the group operation, which assumes the points are distinct.
sage: def op(P1,P2):...
x1,y1 = P1; x2,y2 = P2...
lam = (y1 - y2)/(x1 - x2); nu = y1 - lam*x1...
x3 = lam^2 - x1 - x2; y3 = -lam*x3 - nu...
return (x3, y3)
We define three points, add them together via P1 + (P2 + P3 ) and (P1 +(P2 + P3 )), and observe that the results are the same modulo the relations.
sage: P1 = (x1,y1); P2 = (x2,y2); P3 = (x3,y3)
sage: Z = op(P1, op(P2,P3)); W = op(op(P1,P2),P3)
sage: (Q(Z[0].numerator()*W[0].denominator() ...
Z[0].denominator()*W[0].numerator())) == 0
True
sage: (Q(Z[1].numerator()*W[1].denominator() ...
Z[1].denominator()*W[1].numerator())) == 0
True

6.3 Integer Factorization Using Elliptic Curves

129

6.3 Integer Factorization Using Elliptic Curves
In 1987, Hendrik Lenstra published the landmark paper [Len87] that introduces and analyzes the Elliptic Curve Method (ECM), which is a powerfulalgorithm for factoring integers using elliptic curves. Lenstra’s method is
also described in [ST92, §IV.4], [Dav99, §VIII.5], and [Coh93, §10.3].
Lenstra’s algorithm is well suited for finding“medium-sized” factors of an integer N , which today means between 10 to 40 decimal digits. The
ECM method is not directly used for factoring RSA
challenge numbers (see Section 1.1.3), but it isused on auxiliary numbers as a crucial step in the“number field sieve,” which is the best known algorithm for hunting for such factorizations. Also,
implementation of ECM typically requires little
memory.

6.3.1

H. Lenstra

Pollard’s (p − 1)-Method

Lenstra’s discovery of ECM was inspired by Pollard’s (p−1)-method, which
we describe in this section.
Definition 6.3.1 (Power Smooth). Let B be aQpositive integer. If n isa positive integer with prime factorization n = pei i , then n is B-powersmooth if pei i ≤ B for all i.
For example, 30 = 2·3·5 is B power smooth for B = 5, 7, but 150 = 2·3·52
is not 5-power smooth (it is B = 25-power smooth).
We will use the following algorithm in both the Pollard p − 1 and elliptic
curve factorization methods.
Algorithm 6.3.2 (Least Common Multiple of First B Integers). Given a
positive integer B, this algorithm computes the least common multiple ofthe positive integers up to B.
1. [Sieve] Using, for example, the prime sieve (Algorithm 1.2.3), computea list P of all primes p ≤ B.
Q
2. [Multiply] Compute and output the product p∈P pblogp (B)c .
Proof. Set m = lcm(1, 2, . . . , B). Then,
ordp (m) = max({ordp (n) : 1 ≤ n ≤ B}) = ordp (pr ),where pr is the largest power of p that satisfies pr ≤ B. Since pr ≤ B <pr+1 , we have r = blogp (B)c.

130

6. Elliptic Curves

SAGE Example 6.3.3. We implement Algorithm 6.3.2 in Sage and computethe least common multiple for B = 100 using both the above algorithm anda naive algorithm. We use math.log below so that logp (B) is computed
quickly using double precision numbers.
sage: def lcm_upto(B):...
return prod([p^int(math.log(B)/math.log(p))...
for p in prime_range(B+1)])
sage: lcm_upto(10^2)69720375229712477164533808935312303556800
sage: LCM([1..10^2])69720375229712477164533808935312303556800
Algorithm 6.3.2 as implemented above in Sage takes about a second for
B = 106 .
Let N be a positive integer that we wish to factor. We use the Pollard(p − 1)-method to look for a nontrivial factor of N as follows. First, wechoose a positive integer B, usually with at most six digits. Suppose that
there is a prime divisor p of N such that p − 1 is B-power smooth. We try
to find p using the following strategy. If a > 1 is an integer not divisible
by p, then by Theorem 2.1.20,
ap−1 ≡ 1(mod p).

Let m = lcm(1, 2, 3, . . . , B), and observe that our assumption that p − 1 is
B-power smooth implies that p − 1 | m, so
am ≡ 1(mod p).

Thus
p | gcd(am − 1, N ) > 1.
If gcd(am −1, N ) < N also then gcd(am −1, N ) is a nontrivial factor of N . If
gcd(am − 1, N ) = N , then am ≡ 1 (mod q r ) for every prime power divisorq r of N . In this case, repeat the above steps but with a smaller choice of B
or possibly a different choice of a. Also, it is a good idea to check from thestart whether or not N is not a perfect power M r and, if so, replace N
by M . We formalize the algorithm as follows:
Algorithm 6.3.4 (Pollard p − 1 Method). Given a positive integer N anda bound B, this algorithm attempts to find a nontrivial factor g of N . (Eachprime p | g is likely to have the property that p − 1 is B-power smooth.)
1. [Compute lcm] Use Algorithm 6.3.2 to compute m = lcm(1, 2, . . . , B).
2. [Initialize] Set a = 2.
3. [Power and gcd] Compute x = am − 1 (mod N ) and g = gcd(x, N ).
4. [Finished?] If g 6= 1 or N , output g and terminate.

6.3 Integer Factorization Using Elliptic Curves

131

5. [Try Again?] If a < 10 (say), replace a by a + 1 and go to step 3.
Otherwise, terminate.
For fixed B, Algorithm 6.3.4 often splits N when N is divisible by aprime p such that p − 1 is B-power smooth. Approximately 15 percent ofprimes p in the interval from 1015 and 1015 + 10000 are such that p − 1
is 106 power smooth, so the Pollard method with B = 106 already fails
nearly 85 percent of the time at finding 15-digit primes in this range (seealso Exercise 6.10). We will not analyze Pollard’s method further, since it
was mentioned here only to set the stage for the elliptic curve factorization
method.
The following examples illustrate the Pollard (p − 1)-method.
Example 6.3.5. In this example, Pollard works perfectly. Let N = 5917.
We try to use the Pollard p − 1 method with B = 5 to split N . We havem = lcm(1, 2, 3, 4, 5) = 60; taking a = 2, we have
260 − 1 ≡ 3416

(mod 5917)

andgcd(260 − 1, 5917) = gcd(3416, 5917) = 61,
so 61 is a factor of 5917.
Example 6.3.6. In this example, we replace B with a larger integer. Let
N = 779167. With B = 5 and a = 2, we have
260 − 1 ≡ 710980(mod 779167),and gcd(260 − 1, 779167) = 1. With B = 15, we havem = lcm(1, 2, . . . , 15) = 360360,
2360360 − 1 ≡ 584876(mod 779167),

andgcd(2360360 − 1, N ) = 2003,
so 2003 is a nontrivial factor of 779167.
Example 6.3.7. In this example, we replace B by a smaller integer. Let
N = 4331. Suppose B = 7, so m = lcm(1, 2, . . . , 7) = 420,
2420 − 1 ≡ 0(mod 4331),

and gcd(2420 − 1, 4331) = 4331, so we do not obtain a factor of 4331. If wereplace B by 5, Pollard’s method works:
260 − 1 ≡ 1464

(mod 4331),

and gcd(260 − 1, 4331) = 61, so we split 4331.

132

6. Elliptic Curves

Example 6.3.8. In this example, a = 2 does not work, but a = 3 does. Let
N = 187. Suppose B = 15, so m = lcm(1, 2, . . . , 15) = 360360,
2360360 − 1 ≡ 0(mod 187),

and gcd(2360360 − 1, 187) = 187, so we do not obtain a factor of 187. If wereplace a = 2 by a = 3, then Pollard’s method works:3360360 − 1 ≡ 66

(mod 187),

and gcd(3360360 − 1, 187) = 11. Thus 187 = 11 · 17.
SAGE Example 6.3.9. We implement the Pollard (p − 1)-method in Sage
and use our implementation to do all of the above examples.
sage: def pollard(N, B=10^5, stop=10):...
m = prod([p^int(math.log(B)/math.log(p))...
for p in prime_range(B+1)])...
for a in [2..stop]:...
x = (Mod(a,N)^m - 1).lift()...
if x == 0: continue...
g = gcd(x, N)...
if g != 1 or g != N: return g...
return 1
sage: pollard(5917,5)61
sage: pollard(779167,5)
1
sage: pollard(779167,15)
2003
sage: pollard(4331,7)
1
sage: pollard(4331,5)61
sage: pollard(187, 15, 2)
1
sage: pollard(187, 15)
11

6.3.2

Motivation for the Elliptic Curve Method

Fix a positive integer B. If N = pq with p and q prime, and we assume
that p − 1 and q − 1 are not B-power smooth, then the Pollard (p − 1)method is unlikely to work. For example, let B = 20 and suppose thatN = 59 · 101 = 5959. Note that neither 59 − 1 = 2 · 29 nor 101 − 1 = 4 · 25
is B-power smooth. With m = lcm(1, 2, 3, . . . , 20) = 232792560, we have
2m − 1 ≡ 5944

(mod N ),6.3 Integer Factorization Using Elliptic Curves

133

and gcd(2m − 1, N ) = 1, so we do not find a factor of N .
As remarked above, the problem is that p − 1 is not 20-power smooth foreither p = 59 or p = 101. However, notice that p − 2 = 3 · 19 is 20-powersmooth. Lenstra’s ECM replaces (Z/pZ)∗ , which has order p − 1, by thegroup of points on an elliptic curve E over Z/pZ. It is a theorem that
#E(Z/pZ) = p + 1 ± s√
for some nonnegative integer s < 2 p (see [Sil86, §V.1] for a proof). Also,
every value of s subject to this bound occurs, as one can see using “complex
multiplication theory.” For example, if E is the elliptic curvey 2 = x3 + x + 54
over Z/59Z, then by enumerating points one sees that E(Z/59Z) is cyclicof order 57. The set of numbers 59 + 1 ± s for s ≤ 15 contains 14 numbersthat are B-power smooth for B = 20, which illustrates that working with
an elliptic curve gives us more flexibility. For example, 60 = 59 + 1 + 0 is
5-power smooth and 70 = 59 + 1 + 10 is 7-power smooth.

6.3.3

Lenstra’s Elliptic Curve Factorization Method

Algorithm 6.3.10 (Elliptic Curve Factorization Method). Given a positive integer N and a bound B, this algorithm attempts to find a nontrivial
factor g of N or outputs “Fail.”
1. [Compute lcm] Use Algorithm 6.3.2 to compute m = lcm(1, 2, . . . , B).
2. [Choose Random Elliptic Curve] Choose a random a ∈ Z/N Z suchthat 4a3 + 27 ∈ (Z/N Z)∗ . Then P = (0, 1) is a point on the elliptic
curve y 2 = x3 + ax + 1 over Z/N Z.
3. [Compute Multiple] Attempt to compute mP using an elliptic curveanalog of Algorithm 2.3.13. If at some point we cannot compute a sumof points because some denominator in Step 3 of Algorithm 6.2.1 is
not coprime to N , we compute the greatest common divisor g of thisdenominator with N . If g is a nontrivial divisor, output it. If every
denominator is coprime to N , output “Fail.”If Algorithm 6.3.10 fails for one random elliptic curve, there is an optionthat is unavailable with Pollard’s (p−1)-method—we may repeat the above
algorithm with a different elliptic curve. With Pollard’s method we always
work with the group (Z/N Z)∗ , but here we can try many groups E(Z/N Z)
for many curves E. As mentioned above, the number of points on E over√
Z/pZ is of the form p + 1 − t for some t with |t| < 2 p; Algorithm 6.3.10√
thus has a chance if p + 1 − t is B-power smooth for some t with |t| < 2 p.

134

6. Elliptic Curves

6.3.4

ExamplesFor simplicity, we use an elliptic curve of the form
y 2 = x3 + ax + 1,
which has the point P = (0, 1) already on it.
We factor N = 5959 using the elliptic curve method. Letm = lcm(1, 2, . . . , 20) = 232792560 = 11011110000000100001111100002 ,where x2 means x is written in binary. First, we choose a = 1201 at random
and consider y 2 = x3 + 1201x + 1 over Z/5959Z. Using the formula fori
P + P from Algorithm 6.2.1 we compute 2P· P = 2i · (0, 1) for i ∈ B ={4, 5, 6, 7, 8, 13, 21, 22, 23, 24, 26, 27}. Then i∈B 2i P = mP . It turns outthat during no step of this computation does a number not coprime to 5959
appear in any denominator, so we do not split N using a = 1201. Next, we
try a = 389 and at some stage in the computation we add P = (2051, 5273)
and Q = (637, 1292). When computing the group law explicitly, we try tocompute λ = (y1 − y2 )/(x1 − x2 ) in (Z/5959Z)∗ , but we fail since x1 − x2 =1414 and gcd(1414, 5959) = 101. We thus find a nontrivial factor 101 of
5959.
SAGE Example 6.3.11. We implement elliptic curve factorization in Sage,
then use it to do the above example and some other examples.
sage: def ecm(N, B=10^3, trials=10):...
m = prod([p^int(math.log(B)/math.log(p))...
for p in prime_range(B+1)])...
R = Integers(N)...
# Make Sage think that R is a field:...
R.is_field = lambda : True...
for _ in range(trials):...
while True:...
a = R.random_element()...
if gcd(4*a.lift()^3 + 27, N) == 1: break
...
try:
...
m * EllipticCurve([a, 1])([0,1])...
except ZeroDivisionError, msg:...
# msg: "Inverse of <int> does not exist"...
return gcd(Integer(str(msg).split()[2]), N)...
return 1
sage: set_random_seed(2)sage: ecm(5959, B=20)
101
sage: ecm(next_prime(10^20)*next_prime(10^7), B=10^3)
10000019

6.4 Elliptic Curve Cryptography

6.3.5

135

A Heuristic Explanation

Let N be a positive integer and, for simplicity of exposition, assume thatN = p1 · · · pr with the pi distinct primes. It follows from Lemma 2.2.5 that
there is a natural isomorphismf : (Z/N Z)∗ −→ (Z/p1 Z)∗ × · · · × (Z/pr Z)∗ .
When using Pollard’s method, we choose an a ∈ (Z/N Z)∗ , compute am ,then compute gcd(am − 1, N ). This gcd is divisible exactly by the primes pisuch that am ≡ 1 (mod pi ). To reinterpret Pollard’s method using themm
above isomorphism, let (a1 , . . . , ar ) = f (a). Then (am1 , . . . , ar ) = f (a ),mmand the pi that divide gcd(a − 1, N ) are exactly the pi such that ai = 1.
By Theorem 2.1.20, these pi include the primes pj such that pj − 1 is
B-power smooth, where m = lcm(1, . . . , B).
We will not define E(Z/N Z) when N is composite, since this is notneeded for the algorithm (where we assume that N is prime and hope fora contradiction). However, for the remainder of this paragraph, we pretendthat E(Z/N Z) is meaningful and describe a heuristic connection between
Lenstra and Pollard’s methods. The significant difference between Pollard’s
method and the elliptic curve method is that the isomorphism f is replacedby an isomorphism (in quotes)“g : E(Z/N Z) → E(Z/p1 Z) × · · · × E(Z/pr Z)”
where E is y 2 = x3 + ax + 1, and the a of Pollard’s method is replaced by
P = (0, 1). We put the isomorphism in quotes to emphasize that we havenot defined E(Z/N Z). When carrying out the elliptic curve factorization
algorithm, we attempt to compute mP , and if some components of f (Q)are O, for some point Q that appears during the computation, but others
are nonzero, we find a nontrivial factor of N .

6.4 Elliptic Curve Cryptography
The idea to use elliptic curves in cryptography was independently proposed
by Neil Koblitz and Victor Miller in the mid 1980s. In this section, we
discuss an analog of Diffie-Hellman that uses an elliptic curve instead of(Z/pZ)∗ . We then discuss the ElGamal elliptic curve cryptosystem.

6.4.1

Elliptic Curve Analogs of Diffie-Hellman

The Diffie-Hellman key exchange from Section 3.2 works well on an elliptic
curve with no serious modification. Michael and Nikita agree on a secretkey as follows:

136

6. Elliptic Curves

1. Michael and Nikita agree on a prime p, an elliptic curve E over Z/pZ,
and a point P ∈ E(Z/pZ).
2. Michael secretly chooses a random m and sends mP .
3. Nikita secretly chooses a random n and sends nP .
4. The secret key is nmP , which both Michael and Nikita can compute.
Presumably, an adversary can not compute nmP without solving the discrete logarithm problem (see Problem 3.2.2 and Section 6.4.3 below) in
E(Z/pZ). For well-chosen E, P , and p, experience suggests that the discrete logarithm problem in E(Z/pZ) is much more difficult than the discrete
logarithm problem in (Z/pZ)∗ (see Section 6.4.3 for more on the elliptic
curve discrete log problem).

6.4.2

The ElGamal Cryptosystem and Digital Rights
Management

This section is about the ElGamal cryptosystem, which works well on an
elliptic curve. This section draws on a paper by a computer hacker named
Beale Screamer who cracked a “Digital Rights Management” (DRM) system.
The elliptic curve used in the DRM is an elliptic curve over the finite
field k = Z/pZ, wherep = 785963102379428822376694789446897396207498568951.
The number p in base 16 is
89ABCDEF012345672718281831415926141424F7,
which includes counting in hexadecimal, and digits of e, π, andelliptic curve E is√

2. They 2 = x3 + 317689081251325503476317476413827693272746955927x+ 79052896607878758718120572025718535432100651934.
We have#E(k) = 785963102379428822376693024881714957612686157429,
and the group E(k) is cyclic with generator
B = (771507216262649826170648268565579889907769254176,
390157510246556628525279459266514995562533196655).

6.4 Elliptic Curve Cryptography

137

Our heroes Nikita and Michael share digital music when they are notout fighting terrorists. When Nikita installed the DRM software on her
computer, it generated a private key
n = 670805031139910513517527207693060456300217054473,
which it hides in bits and pieces of files. In order for Nikita to play Juno
Reactor’s latest hit juno.wma, her web browser contacts a website thatsells music. After Nikita sends her credit card number, that website allows
Nikita to download a license file that allows her audio player to unlock andplay juno.wma.
As we will see below, the license file was created using the ElGamal
public-key cryptosystem in the group E(k). Nikita can now use her license
file to unlock juno.wma. However, when she shares both juno.wma and thelicense file with Michael, he is frustrated because even with the license, his
computer still does not play juno.wma. This is because Michael’s computer
does not know Nikita’s computer’s private key (the integer n above), so
Michael’s computer can not decrypt the license file.
We now describe the ElGamal cryptosystem, which lends itself well to
implementation in the group E(Z/pZ). To illustrate ElGamal, we describe
how Nikita would set up an ElGamal cryptosystem that anyone could use
to encrypt messages for her. Nikita chooses a prime p, an elliptic curve E
over Z/pZ, and a point B ∈ E(Z/pZ), and publishes p, E, and B. She alsochooses a random integer n, which she keeps secret, and publishes nB. Her
public key is the four-tuple (p, E, B, nB).
Suppose Michael wishes to encrypt a message for Nikita. If the message is
encoded as an element P ∈ E(Z/pZ), Michael computes a random integer r
and the points rB and P + r(nB) on E(Z/pZ). Then P is encrypted as thepair (rB, P + r(nB)). To decrypt the encrypted message, Nikita multiplies
rB by her secret key n to find n(rB) = r(nB), then subtracts this fromP + r(nB) to obtain
P = P + r(nB) − r(nB).
Remark 6.4.1. It also make sense to construct an ElGamal cryptosystemin the group (Z/pZ)∗ .
Returning to our story, Nikita’s license file is an encrypted message toher. It contains the pair of points (rB, P + r(nB)), whererB = (179671003218315746385026655733086044982194424660,
697834385359686368249301282675141830935176314718)
andP + r(nB) = (137851038548264467372645158093004000343639118915,
110848589228676224057229230223580815024224875699).

138

6. Elliptic Curves

When Nikita’s computer plays juno.wma, it loads the secret key
n = 670805031139910513517527207693060456300217054473
into memory and computesn(rB) = (328901393518732637577115650601768681044040715701,
586947838087815993601350565488788846203887988162).
It then subtracts this from P + r(nB) to obtain
P = (14489646124220757767,
669337780373284096274895136618194604469696830074).
The x-coordinate 14489646124220757767 is the key that unlocks juno.wma.
If Nikita knew the private key n that her computer generated, she could
compute P herself and unlock juno.wma and share her music with Michael.
Beale Screamer found a weakness in the implementation of this system thatallows Nikita to detetermine n, which is not a huge surprise since n is stored
on her computer after all.
SAGE Example 6.4.2. We do the above examples in Sage:sage: p = 785963102379428822376694789446897396207498568951
sage: E = EllipticCurve(GF(p), \...
[317689081251325503476317476413827693272746955927,...
79052896607878758718120572025718535432100651934])
sage: E.cardinality()
785963102379428822376693024881714957612686157429
sage: E.cardinality().is_prime()True
sage: B = E([...
771507216262649826170648268565579889907769254176,...
390157510246556628525279459266514995562533196655])
sage: n=670805031139910513517527207693060456300217054473
sage: r=70674630913457179596452846564371866229568459543
sage: P = E([14489646124220757767,...
669337780373284096274895136618194604469696830074])
sage: encrypt = (r*B, P + r*(n*B))
sage: encrypt[1] - n*encrypt[0] == P
# decrypting works
True

6.4.3

The Elliptic Curve Discrete Logarithm Problem

Problem 6.4.3 (Elliptic Curve Discrete Log Problem). Suppose E is anelliptic curve over Z/pZ and P ∈ E(Z/pZ). Given a multiple Q of P , theelliptic curve discrete log problem is to find n ∈ Z such that nP = Q.

6.4 Elliptic Curve Cryptography

139

For example, let E be the elliptic curve given by y 2 = x3 + x + 1 overthe field Z/7Z. We haveE(Z/7Z) = {O, (2, 2), (0, 1), (0, 6), (2, 5)}.
If P = (2, 2) and Q = (0, 6), then 3P = Q, so n = 3 is a solution to thediscrete logarithm problem.
If E(Z/pZ) has order p or p ± 1, or is a product of reasonably small
primes, then there are some methods for attacking the discrete log problem
on E, which are beyond the scope of this book. It is therefore important
to be able to compute #E(Z/pZ) efficiently, in order to verify that theelliptic curve one wishes to use for a cryptosystem doesn’t have any obvious
vulnerabilities. The naive algorithm to compute #E(Z/pZ) is to try eachvalue of x ∈ Z/pZ and count how often x3 +ax+b is a perfect square mod p,
but this is of no use when p is large enough to be useful for cryptography.
Fortunately, there is an algorithm due to Schoof, Elkies, and Atkin forcomputing #E(Z/pZ) efficiently (polynomial time in the number of digitsof p), but this algorithm is beyond the scope of this book.
In Section 3.2.1, we discussed the discrete log problem in (Z/pZ)∗ . There
are general attacks called “index calculus attacks” on the discrete log problem in (Z/pZ)∗ that are slow, but still faster than the known algorithms
for solving the discrete log in a “general” group (one with no extra structure). For most elliptic curves, there is no known analog of index calculus
attacks on the discrete log problem. At present, it appears that given p,
the discrete log problem in E(Z/pZ) is much harder than the discrete log
problem in the multiplicative group (Z/pZ)∗ . This suggests that by using an elliptic curve-based cryptosystem instead of one based on (Z/pZ)∗ ,one gets equivalent security with much smaller numbers, which is one reason why building cryptosystems using elliptic curves is attractive to somecryptographers. For example, Certicom, a company that strongly supports
elliptic curve cryptography, claims:“[Elliptic curve crypto] devices require less storage, less power,
less memory, and less bandwidth than other systems. This allows you to implement cryptography in platforms that are constrained, such as wireless devices, handheld computers, smart
cards, and thin-clients. It also provides a big win in situations
where efficiency is important.”
For an up-to-date list of elliptic curve discrete log challenge problems
that Certicom sponsors, see [Cer]. For example, in April 2004, a specificcryptosystem was cracked that was based on an elliptic curve over Z/pZ,
where p has 109 bits. The first unsolved challenge problem involves an
elliptic curve over Z/pZ, where p has 131 bits, and the next challenge afterthat is one in which p has 163 bits. Certicom claims at [Cer] that the 163-bit
challenge problem is computationally infeasible.

140

6. Elliptic Curves

FIGURE 6.4. Louis J. Mordell

6.5 Elliptic Curves Over the Rational Numbers
Let E be an elliptic curve defined over Q. The following is a deep theorem
about the group E(Q).
Theorem 6.5.1 (Mordell). The group E(Q) is finitely generated. That is,there are points P1 , . . . , Ps ∈ E(Q) such that every element of E(Q) is of
the form n1 P1 + · · · + ns Ps for integers n1 , . . . ns ∈ Z.
Mordell’s theorem implies that it makes sense to ask whether or not
we can compute E(Q), where by “compute” we mean find a finite set
P1 , . . . , Ps of points on E that generate E(Q) as an abelian group. Thereis a systematic approach to computing E(Q) called “descent” (see, forexample, [Cre97, Cre, Sil86]). It is widely believed that the method ofdescent will always succeed, but nobody has yet proved that it will. Provingthat descent works for all curves is one of the central open problems in
number theory, and is closely related to the Birch and Swinnerton-Dyer
conjecture (one of the Clay Math Institute’s million dollar prize problems).
The crucial difficulty amounts to deciding whether or not certain explicitly
given curves have any rational points on them or not (these are curves thathave points over R and modulo n for all n).
The details of using descent to compute E(Q) are beyond the scope of
this book. In several places below, we will simply assert that E(Q) has
a certain structure or is generated by certain elements. In each case, we
computed E(Q) using a computer implementation of this method.

6.5.1

The Torsion Subgroup of E(Q)For any abelian group G, let Gtor be the subgroup of elements of finite
order. If E is an elliptic curve over Q, then E(Q)tor is a subgroup ofE(Q), which must be finite because of Theorem 6.5.1 (see Exercise 6.6).

6.5 Elliptic Curves Over the Rational Numbers

141

One can also prove that E(Q)tor is finite by showing that there is a prime
p and an injective reduction homomorphism E(Q)tor ,→ E(Z/pZ), thennoting that E(Z/pZ) is finite. For example, if E is y 2 = x3 − 5x + 4, thenE(Q)tor = {O, (1, 0)} ∼= Z/2Z.
The possibilities for E(Q)tor are known.
Theorem 6.5.2 (Mazur, 1976). Let E be an elliptic curve over Q. Then
E(Q)tor is isomorphic to one of the following 15 groups:
Z/nZ
Z/2Z × Z/2n

for n ≤ 10 or n = 12,
for n ≤ 4.

SAGE Example 6.5.3. We compute the structure of the torsion subgroupsof some elliptic curves. In each case, the output of the function T (a, b)
below is a pair c, d ∈ Z (or integer c) such that the torsion subgroup of
y 3 = x3 + ax + b is Z/cZ × Z/dZ.
sage: T = lambda v: EllipticCurve(v...
).torsion_subgroup().invariants()
sage: T([-5,4])(2,)
sage: T([-43,166])(7,)
sage: T([-4,0])(2, 2)
sage: T([-1386747, 368636886])(2, 8)

6.5.2

The Rank of E(Q)The quotient E(Q)/E(Q)tor is a finitely generated free abelian group, so
it is isomorphism to Zr for some integer r, called the rank of E(Q). For
example, one can prove that if E is y 2 = x3 − 5x + 4, then E(Q)/E(Q)tor
is generated by the point (0, 2).
SAGE Example 6.5.4. We use Sage to compute the ranks of some elliptic
curves y 2 = x3 + ax + b. The function r(a, b) below returns the rank of thiscurve over Q.
sage: r = lambda v: EllipticCurve(v).rank()
sage: r([-5,4])
1
sage: r([0,1])
0
sage: r([-3024, 46224])2
sage: r([-112, 400])

142

6. Elliptic Curves

3
sage: r([-102627, 12560670])
4
The following is a folklore conjecture, not associated with any particularmathematician:Conjecture 6.5.5. There are elliptic curves over Q of arbitrarily large
rank.
The world record is the following curve, whose rank is at least 28:
y 2 +xy + y = x3 − x2 −20067762415575526585033208209338542750930230312178956502x+
344816117950305564670329856903907203748559443593191803612 . . .
. . . 66008296291939448732243429
It was discovered in May 2006 by Noam Elkies of Harvard University.

6.5.3

The Congruent Number Problem

Definition 6.5.6 (Congruent Number). We call a nonzero rational number n a congruent number if ±n is the area of a right triangle with rational
side lengths. Equivalently, n is a congruent number if the system of two
equations
a2 + b2
1
ab
2

= c2
= n

has a solution with a, b, c ∈ Q.
For example, 6 is the area of the right triangle with side lengths 3, 4,
and 5, so 6 is a congruent number. Less obvious is that 5 is also a congruent
number; it is the area of the right triangle with side lengths 3/2, 20/3, and41/6. It is nontrivial to prove that 1, 2, 3, and 4 are not congruent numbers.
Here is a list of the integer congruent numbers up to 50:
5, 6, 7, 13, 14, 15, 20, 21, 22, 23, 24, 28, 29, 30, 31, 34, 37, 38, 39, 41, 45, 46, 47.

Every congruence class modulo 8 except 3 is represented in this list,
which incorrectly suggests that if n ≡ 3 (mod 8) then n is not a congruent
number. Though no n ≤ 218 with n ≡ 3 (mod 8) is a congruent number,n = 219 is a congruent number and 219 ≡ 3 (mod 8).
Deciding whether an integer n is a congruent number can be subtle, since
the simplest triangle with area n can be very complicated. For example,

6.5 Elliptic Curves Over the Rational Numbers

143

as Zagier pointed out, the number 157 is a congruent number, and the“simplest” rational right triangle with area 157 has side lengthsa=6803298487826435051217540
411340519227716149383203
and b =.
411340519227716149383203
21666555693714761309610

This solution would be difficult to find by a brute force search.
We call congruent numbers “congruent” because of the following proposition, which asserts that any congruent number is the common “congruence”
between three perfect squares.
Proposition 6.5.7. Suppose n is the area of a right triangle with rational
side lengths a, b, c, with a ≤ b < c. Let A = (c/2)2 . Then
A − n,A, and A + n

are all perfect squares of rational numbers.
Proof. We havea2 + b2
1
ab
2

= c2
= n

Add or subtract 4 times the second equation to the first to get
a2 ± 2ab + b2
(a ± b)2
2
a±b2= c2 ± 4n= c2 ± 4n c 2
=
±n
2
= A±n

The main motivating open problem related to congruent numbers is to
give a systematic way to recognize them.
Open Problem 6.5.8. Give an algorithm which, given n, outputs whether
or not n is a congruent number.
Fortunately, the vast theory developed about elliptic curves has something to say about the above problem. In order to understand this connection, we begin with an elementary algebraic proposition that establishes alink between elliptic curves and the congruent number problem.
Proposition 6.5.9 (Congruent Numbers and Elliptic Curves). Let n be arational number. There is a bijection betweenab
A = (a, b, c) ∈ Q3 := n, a2 + b2 = c22

144

6. Elliptic Curvesand
B = (x, y) ∈ Q2 : y 2 = x3 − n2 x, with y 6= 0
given explicitly by the mapsf (a, b, c) =

nb
2n2
−
,
a+c a+cand
g(x, y) =2xn n2 + x2
n2 − x2
, −
,
y
y
y


.

The proof of this proposition is not deep, but involves substantial (elementary) algebra and we will not prove it in this book.
For n 6= 0, let En be the elliptic curve y 2 = x3 − n2 x.
Proposition 6.5.10 (Congruent Number Criterion). The rational number n is a congruent number if and only if there is a point P = (x, y) ∈En (Q) with y 6= 0.
Proof. The number n is a congruent number if and only if the set A from
Proposition 6.5.9 is nonempty. By the proposition A is nonempty if andonly if B is nonempty.
Example 6.5.11. Let n = 5. Then En is y 2 = x3 − 25x, and we notice that(−4, −6) ∈ En (Q). We next use the bijection of Proposition 6.5.9 to findthe corresponding right triangle: 
3 20 41
25 − 16 −40 25 + 16,−,
= − ,− ,−.
g(−4, −6) =
−6
−6
−6
2
3
6
Multiplying through by −1 yields the side lengths of a rational right triangle
with area 5. Are there any others?
Observe that we can apply g to any point in En (Q) with y 6= 0. Using
the group law, we find that 2(−4, −6) = (1681/144, 62279/1728) andg(2(−4, −6)) =1519 4920 3344161
−
,−
,
492
1519 747348

.

This example foreshadows Theorem 6.5.14.
Example 6.5.12. Let n = 1, so E1 is defined by y 2 = x3 − x. Since 1 is nota congruent number, the elliptic curve E1 has no point with y 6= 0. See
Exercise 6.11.
SAGE Example 6.5.13. We implement the cong function in Sage, which
returns a triple (a, b, c) whose entries are the sides of a rational right triangleof area n if one exists, and returns False if there are no such triangles.

6.5 Elliptic Curves Over the Rational Numbers

145

sage: def cong(n):...
G = EllipticCurve([-n^2,0]).gens()...
if len(G) == 0: return False...
x,y,_ = G[0]...
return ((n^2-x^2)/y,-2*x*n/y,(n^2+x^2)/y)sage: cong(6)(3, 4, 5)
sage: cong(5)(3/2, 20/3, 41/6)
sage: cong(1)False
sage: cong(13)(323/30, 780/323, 106921/9690)
sage: (323/30 * 780/323)/213
sage: (323/30)^2 + (780/323)^2 == (106921/9690)^2
True
Theorem 6.5.14 (Infinitely Many Triangles). If n is a congruent number,then there are infinitely many distinct right triangles with rational side
lengths and area n.
We will not prove this theorem, except to note that one proves it byshowing that En (Q)tor = {O, (0, 0), (n, 0), (−n, 0)}, so the elements of theset B in Proposition 6.5.9 all have infinite order. Hence, B is infinite so A
is infinite.
Tunnell has proved that the Birch and Swinnerton-Dyer conjecture (alluded to above), implies the existence of an elementary way to decide
whether or not an integer n is a congruent number. We state Tunnell’s
elementary way in the form of a conjecture.
Conjecture 6.5.15. Let a, b, c denote integers. If n is an even square-free
integer, then n is a congruent number if and only ifonn# (a, b, c) ∈ Z3 : 4a2 + b2 + 8c2 = : c is even2non
= # (a, b, c) : 4a2 + b2 + 8c2 = : c is odd .
2
If n is odd and square free then n is a congruent number if and only if# (a, b, c) : 2a2 + b2 + 8c2 = n : c is even
= # (a, b, c) : 2a2 + b2 + 8c2 = n : c is odd .
Enough of the Birch and Swinnerton-Dyer conjecture is known to prove
one direction of Conjecture 6.5.15. In particular, it is a very deep theorem
that if we do not have equality of the displayed cardinalities, then n is nota congruent number.

146

6. Elliptic Curves

The even more difficult (and still open!) part of Conjecture 6.5.15 is theconverse: If one has equality of the displayed cardinalities, prove that n isa congruent number. The difficulty in this direction, which appears to bevery deep, is that we must somehow construct (or prove the existence of)elements of En (Q). This has been accomplished in some cases due to thegroundbreaking work of Gross and Zagier ([GZ86]) but much work remains
to be done.
The excellent book [Kob84] is about congruent numbers and Conjecture 6.5.15, and we encourage the reader to consult it. The Birch and
Swinnerton-Dyer conjecture is a Clay Math Institute million dollar millennium prize problem (see [Cla, Wil00]).

6.6 Exercises
6.1 Write down an equation y 2 = x3 + ax + b over a field K such that−16(4a3 +27b2 ) = 0. Precisely what goes wrong when trying to endow
the set E(K) = {(x, y) ∈ K × K : y 2 = x3 + ax + b} ∪ {O} with agroup structure?
6.2 One rational solution to the equation y 2 = x3 − 2 is (3, 5). Find arational solution with x 6= 3 by drawing the tangent line to (3, 5) andcomputing the second point of intersection.
6.3 Let E be the elliptic curve over the finite field K = Z/5Z defined by
the equationy 2 = x3 + x + 1.
(a) List all 9 elements of E(K).
(b) What is the structure of E(K), as a product of cyclic groups?
6.4 Let E be the elliptic curve defined by the equation y 2 = x3 + 1. For
each prime p ≥ 5, let Np be the cardinality of the group E(Z/pZ)
of points on this curve having coordinates in Z/pZ. For example, wehave that N5 = 6, N7 = 12, N11 = 12, N13 = 12, N17 = 18, N19 =12, , N23 = 24, and N29 = 30 (you do not have to prove this).
(a) For the set of primes satisfying p ≡ 2 (mod 3), can you see apattern for the values of Np ? Make a general conjecture for thevalue of Np when p ≡ 2 (mod 3).
(b) (*) Prove your conjecture.
6.5 Let E be an elliptic curve over the real numbers R. Prove that E(R)is not a finitely generated abelian group.
6.6 (*) Suppose G is a finitely generated abelian group. Prove that the
subgroup Gtor of elements of finite order in G is finite.

6.6 Exercises

147

6.7 Suppose y 2 = x3 + ax + b with a, b ∈ Q defines an elliptic curve. Show
that there is another equation Y 2 = X 3 + AX + B with A, B ∈ Z
whose solutions are in bijection with the solutions to y 2 = x3 +ax+b.
6.8 Suppose a, b, c are relatively prime integers with a2 + b2 = c2 . Then
there exist integers x and y with x > y such that c = x2 + y 2 andeither a = x2 − y 2 , b = 2xy or a = 2xy, b = x2 − y 2 .
6.9 (*) Fermat’s Last Theorem for exponent 4 asserts that any solution
to the equation x4 + y 4 = z 4 with x, y, z ∈ Z satisfies xyz = 0. Prove
Fermat’s Last Theorem for exponent 4, as follows.
(a) Show that if the equation x2 + y 4 = z 4 has no integer solutions
with xyz 6= 0, then Fermat’s Last Theorem for exponent 4 istrue.
(b) Prove that x2 + y 4 = z 4 has no integer solutions with xyz 6= 0 as
follows. Suppose n2 + k 4 = m4 is a solution with m > 0 minimal
among all solutions. Show that there exists a solution with msmaller using Exercise 6.8 (consider two cases).
6.10 This problem requires a computer.
(a) Show that the set of numbers 59 + 1 ± s for s ≤ 15 contains 14
numbers that are B-power smooth for B = 20.
(b) Find the proportion of primes p in the interval from 1012 and1012 + 1000 such that p − 1 is B = 105 power smooth.
6.11 (*) Prove that 1 is not a congruent number by showing that theelliptic curve y 2 = x3 − x has no rational solutions except (0, ±1) and(0, 0), as follows:(a) Write y = pq and x = rs , where p, q, r, s are all positive integers
and gcd(p, q) = gcd(r, s) = 1. Prove that s | q, so q = sk forsome k ∈ Z.
(b) Prove that s = k 2 , and substitute to see that p2 = r3 − rk 4 .
(c) Prove that r is a perfect square by supposing that there is aprime ` such that ord` (r) is odd, and analyzing ord` of bothsides of p2 = r3 − rk 4 .
(d) Write r = m2 , and substitute to see that p2 = m6 −m2 k 4 . Divide
through by m2 and deduce a contradiction to Exercise 6.9.

148

6. Elliptic Curves

This is page 149
Printer: Opaque thisAnswers and Hints• Chapter 1. Prime Numbers
2. They are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59,
61, 67, 71, 73, 79, 83, 89, 97.
3. Emulate the proof of Proposition 1.2.5.
• Chapter 2. The Ring of Integers Modulo n
2. They are 5, 13, 3, and 8.
3. For example, x = 22, y = −39.
4. Hint: Use the binomial theorem and prove that if r ≥ 1, then p
divides pr .
7. For example, S1 = {0, 1, 2, 3, 4, 5, 6}, S2 = {1, 3, 5, 7, 9, 11, 13},
S3 = {0, 2, 4, 6, 8, 10, 12}, and S4 = {2, 3, 5, 7, 11, 13, 29}. In each
we find Si by listing the first seven numbers satisfying the ith
condition, then adjust the last number if necessary so that the
reductions will be distinct modulo 7.
8. An integer is divisible by 5 if and only if the last digits is 0 or 5.
An integer is divisible by 9 if and only if the sum of the digits
is divisible by 9. An integer is divisible by 11 if and only if thealternating sum of the digits is divisible by 11.
9. Hint for part (a): Use the divisibility rule you found in Exercise 1.8.

150

6. Elliptic Curves

10. 71
11. 8
12. As explained on page 23, we know that Z/nZ is a ring for any n.
Thus to show that Z/pZ is a field it suffices to show that every
nonzero element a ∈ Z/pZ has an inverse. Lift a to an elementa ∈ Z, and set b = p in Proposition 2.3.1. Because p is prime,
gcd(a, p) = 1, so there exists x, y such that ax+py = 1. Reducing
this equality modulo p proves that a has an inverse x (mod p).
Alternatively, one could argue just like after Definition 2.1.16
that am = 1 for some m, so some power of a is the inverse of a.
13. 302
15. Only for n = 1, 2. If n > 2, then n is either divisible by an
odd prime p or 4. If 4 | n, then 2e − 2e−1 divides ϕ(n) for somee ≥ 2, so ϕ(n) is even. If an odd p divides n, then the evennumber pe − pe−1 divides ϕ(n) for some e ≥ 1.
16. The map ψ is a homomorphism since both reduction maps
Z/mnZ → Z/mZ and Z/mnZ → Z/nZ
are homomorphisms. It is injective because if a ∈ Z is such that
ψ(a) = 0, then m | a and n | a, so mn | a (since m and n are
coprime), so a ≡ 0 (mod mn). The cardinality of Z/mnZ is mn
and the cardinality of the product Z/mZ × Z/nZ is also mn,so ψ must be an isomorphism. The units (Z/mnZ)∗ are thus inbijection with the units (Z/mZ)∗ × (Z/nZ)∗ .
For the second part of the exercise, let g = gcd(m, n) and set
a = mn/g. Then a 6≡ 0 (mod mn), but m | a and n | a, so
a ker(ψ).
17. We express the question as a system of linear equations modulo
various numbers, and use the Chinese remainder theorem. Let
x be the number of books. The problem asserts thatx≡6(mod 7)x≡2(mod 6)x≡1(mod 5)x≡0(mod 4)

Applying CRT to the first pair of equations, we find that x ≡ 20
(mod 42). Applying CRT to this equation and the third, we findthat x ≡ 146 (mod 210). Since 146 is not divisible by 4, we add
multiples of 210 to 146 until we find the first x that is divisible
by 4. The first multiple works, and we find that the aspiring
mathematicians have 356 math books.

6.6 Exercises151

18. Note that p = 3 works, since 11 = 32 + 2 is prime. Now suppose
p 6= 3 is any prime such that p and p2 + 2 are both prime.
We must have p ≡ 1 (mod 3) or p ≡ 2 (mod 3). Then p2 ≡ 1(mod 3), so p2 + 2 ≡ 0 (mod 3). Since p2 + 2 is prime, we musthave p2 + 2 = 3, so p = 1, a contradiction as p is assumed prime.
19. For (a) n = 1, 2, see solution to Exercise 2.15. For (b), yes there
are many such examples. For example, m = 2, n = 4.
20. By repeated application of multiplicativity
and Equation (2.2.2)
Q
on page 31, we see that if n = i pei i is the prime factorizationof n, thenYYYpiei −1 · (pi − 1).
ϕ(n) =(pei i − piei −1 ) =iii

23. 1, 6, 29, 34
24. Let g = gcd(12n+1, 30n+2). Then g | 30n+2−2·(12n+1) = 6n.
For the same reason, g also divides 12n + 1 − 2 · (6n) = 1, so
g = 1, as claimed.
27. There is no primitive root modulo 8, since (Z/8Z)∗ has order4, but every element of (Z/8Z)∗ has order 2. Prove that if ζ isa primitive root modulo 2n , for n ≥ 3, then the reduction of ζ
mod 8 is a primitive root, a contradiction.
28. 2 is a primitive root modulo 125.
Qm
29. Let i=1 pei i be the prime factorization of n. Slightly generalizing Exercise 16, we see that
Y
(Z/nZ)∗ ∼(Z/pei i Z)∗ .
=Thus (Z/nZ)∗ is cyclic if and only if the product (Z/pei i Z)∗ iscyclic. If 8 | n, then there is no chance (Z/nZ)∗ is cyclic, so
assume 8 - n. Then by Exercise 2.28, each group (Z/pei i Z)∗ isitself cyclic. A product of cyclic groups is cyclic if and only theorders of the factors in the product are coprime (this follows from
Exercise 2.16). Thus (Z/nZ)∗ is cyclic if and only if the numberspi (pi − 1), for i = 1, . . . , m are pairwise coprime. Since pi − 1 iseven, there can be at most one odd prime in the factorization ofn, and we see that (Z/nZ)∗ is cyclic if and only if n is an odd
prime power, twice an odd prime power, or n = 4.
• Chapter 3. Public-Key Cryptography
1. The best case is that each letter is A. Then the question is to findthe largest n such that 1 + 27 + · · · + 27n ≤ 1020 . By computing

152

6. Elliptic Curves

log27 (1020 ), we see that 2713 < 1020 and 2714 > 1020 . Thusn ≤ 13, and since 1 + 27 + · · · + 27n−1 < 27n , and 2 · 2713 < 1020 ,it follows that n = 13.
2. This is not secure, since it is just equivalent to a “Ceaser Cipher,” that is a permutation of the letters of the alphabet, which
is well-known to be easily broken using a frequency analysis.
3. If we can compute the polynomialf = (x−p)(x−q)(x−r) = x3 −(p+q+r)x2 +(pq+pr+qr)x−pqr,then we can factor n by finding the roots of f , for example,
using Newton’s method (or Cardona’s formula for the roots of a
cubic). Because p, q, r, are distinct odd primes, we haveϕ(n) = (p − 1)(q − 1)(r − 1) = pqr − (pq + pr + qr) + p + q + r,
andσ(n) = 1 + (p + q + r) + (pq + pr + qr) + pqr.
Since we know n, ϕ(n), and σ(n), we know
σ(n) − 1 − n = (p + q + r) + (pq + pr + qr),and

ϕ(n) − n = (p + q + r) − (pq + pr + qr).
We can thus compute both p + q + r and pq + pr + qr, hence
deduce f and find p, q, r.
• Chapter 4. Quadratic Reciprocity
1. They are all 1, −1, 0, and 1.
3. By Proposition 4.3.4, the value of

 
3
p

depends only on the re-duction ±p (mod 12). List enough primes p such that ±p reduce
to 1, 5, 7, 11 modulo 12 and verify that the asserted formula holds
for each of them.
7. Since p = 213 − 1 is prime, there are either two solutions or nosolutions to x2 ≡ 5 (mod p), and we can decide which using
quadratic reciprocity. We have p p5= (−1)(p−1)/2·(5−1)/2
=
,
p
5
5
so there are two solutions if and only if p = 213 − 1 is ±1 mod 5.
In fact, p ≡ 1 (mod 5), so there are two solutions.
8. We have 448 = 296 . By Euler’s Theorem, 296 = 1, so x = 1.

6.6 Exercises

153

9. For (a), take a = 19 and n = 20. We found this example using the Chinese remainder theorem
applied to 4 (mod 5) and 3
19
19(mod 4), and used that 19=·= (−1)(−1) = 1, yet
20
54
19 is not a square modulo either 5 or 4, so is certainly not asquare modulo 20.
10. Hint: First reduce to the case that 6k − 1 is prime, by usingthat if p and q are primes not of the form 6k − 1, then neither
is their product. If p = 6k − 1 divides n2 + n + 1, it divides
4n2 + 4n + 4 = (2n + 1)2 + 3, so −3 is a quadratic residue
modulo p. Now use quadratic reciprocity to show that −3 is nota quadratic residue modulo p.
• Chapter 5. Continued Fractions
9. Suppose n = x2 + y 2 , with x, y ∈ Q. Let d be such that dx, dy ∈Z. Then d2 n = (dx)2 + (dy)2 is a sum of two integer squares, so
by Theorem 5.7.1, if p | d2 n and p ≡ 3 (mod 4), then ordp (d2 n)
is even. We have ordp (d2 n) is even if and only if ordp (n) is even,so Theorem 5.7.1 implies that n is also a sum of two squares.
11. The squares modulo 8 are 0, 1, 4, so a sum of two squares reduces
modulo 8 to one of 0, 1, 2, 4, or 5. Four consecutive integers thatare sums of squares would reduce to four consecutive integers inthe set {0, 1, 2, 4, 5}, which is impossible.
• Chapter 6. Elliptic Curves
2. The second point of intersection is (129/100, 383/1000).
3. The group is cyclic of order 9, generated by (4, 2). The elementsof E(K) are{O, (4, 2), (3, 4), (2, 4), (0, 4), (0, 1), (2, 1), (3, 1), (4, 3)}.
4. In part (a), the pattern is that Np = p + 1. For part (b), a hint
is that when p ≡ 2 (mod 3), the map x 7→ x3 on (Z/pZ)∗ is an
automorphism, so x 7→ x3 + 1 is a bijection. Now use what you
learned about squares in Z/pZ from Chapter 4.
5. For all sufficiently large real x, the equation y 2 = x3 + ax + b
has a real solution y. Thus, the group E(R) is not countable,
since R is not countable. But any finitely generated group iscountable.
6. In a course on abstract algebra, one often proves the nontrivial
fact that every subgroup of a finitely generated abelian groupis finitely generated. In particular, the torsion subgroup Gtor is

154

6. Elliptic Curves

finitely generated. However, a finitely generated abelian torsion
group is finite.
7. Hint: Multiply both sides of y 2 = x3 + ax + b by a power of acommon denominator, and “absorb” powers into x and y.
8. Hint: see Exercise 4.6.

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This is page 160
Printer: Opaque thisIndex

B-power smooth, 129dxe, 62
Z/nZ, 21 ap , 70abelian group, 22
algebraic number, 114
algorithm, 4Chinese Remainder Theorem,
29
Compute Power, 35
Division Algorithm, 5Elliptic Curve Factorization
Method, 133
Elliptic Curve Group Law, 126
Extended Euclidean Algorithm,
33
Greatest Common Division,
5
Inverse Modulo n, 33
Least Common Multiple of First
B Integers, 129
Miller-Rabin Primality Test,
38
Pollard p − 1 Method, 130

Prime Sieve, 12
Primitive Root, 44
Probabilistic Algorithm to Factor n, 64
Write a Number in Binary,34
Artin, 43
Artin’s conjecture, 43binary, writing number in, 34
cancellation proposition, 23
Carmichael numbers, 37
Certicom challenges, 139
Chinese remainder theorem, 29
commutative ring, 22complete set of residues, 24, 24
composite, 2
computecontinued fraction, 101
gcd, 5greatest common divisor, 4
inverse modulo n, 31
powers modulo n, 31, 34
square roots mod p, 86–89

Indexcongruences, 22
congruent number, 142
157 is, 143
all ≤ 50 are, 142
and arithmetic progression, 143
and elliptic curves, 143
problem, 142
why called congruent, 143congruent number criterion proposition, 144congruent numbers and elliptic curves
proposition, 143conjecture
Artin, 43
continued fraction, 94, 94–122
algorithm, 101convergents, 99
every
rational number has, 100√
32, 114of √of 2, 111of e, 103, 107of algebraic number, 115
of finite length, 95
of higher degree number, 114
of quadratic irrational, 110partial convergents of, 97
periodic, 111
recognizing rational numbers,
115
continued fraction convergence theorem, 105
continued fraction existence theorem, 106
continued fraction limit theorem,
104
continued fraction procedure, 107
continued fraction process, 102convergence of continued fraction
proposition, 107convergent, 97
convergents
partial, 99
convergents in lowest terms corollary, 98
corollary161

convergents in lowest terms,
98
cryptography, 13
using elliptic curves, 135
cryptosystem
Diffie-Hellman, 50, 51
ElGamal, 136, 137
RSA, 56–66
decryption key proposition, 57
density of primes, 14
deterministic primality test, 38
Diffie-Hellman cryptosystem, 50,
51
on elliptic curve, 135
digital signatures, 56
Dirichlet theorem, 14
discrete log problem, 52, 53difficulty of, 53
on elliptic curve, 136
on elliptic curve, 138
divides, 2, 2
divisibility by 3 proposition, 23
divisibility tests, 23
division algorithm, 5divisor, 2does not divide, 2ECM, 129ElGamal cryptosystem, 136, 137
elliptic curve, 124and congruent numbers, 143
cryptography, 135
Diffie-Hellman, 135
discrete log problem, 136, 138
factorization, 129, 133
group structure, 125
rank, 142
rational points on, 140
torsion subgroup, 140
elliptic curve discrete log problem,
138
elliptic curve group law theorem,
126
equivalence relation

162

Index

congruence modulo n, 22
Euclid, 2
Euclid theorem, 7
Euclid’s theorem
on divisibility, 7
Euler, 73, 107
phi function, 22, 26, 30
is multiplicative, 31
Euler ϕ-function, 30
Euler proposition, 78
Euler’s criterion proposition, 73
Euler’s proposition, 77
Euler’s theorem, 25, 26
group-theoretic interpretation,
26
extended Euclidean algorithm, 33
extended Euclidean proposition,
32
factorization
and breaking RSA, 61, 63difficulty of, 8
Pollard’s (p−1)-method, 129–
132
quantum, 8
using elliptic curves, 129
Fermat Factorization Method, 62
field, 23of integers modulo p, 23, 46
finite continued fraction, 95
finite field, 23
floor, 102
fundamental theorem of arithmetic,
3, 7, 10Gauss, 15, 69, 72, 73, 75
Gauss sum, 82
Gauss sum proposition, 82
Gauss’s lemma, 75
gcd, 3
gcd algorithm, 5Generalized Riemann Hypothesis,
44
geometric group law proposition,
126

graph
of group law, 127
greatest common divisor, 3group, 22(Z/mZ)∗ , 26of units, 22
structure of elliptic curve, 125
group homomorphism, 64
group lawillustrated, 127
Hadamard, 16
homomorphism of rings, 87
Hooley, 44
how convergents converge proposition, 100
infinitely many primes proposition,
13
infinitely many primes theorem,
11
infinitely many triangles theorem,
145
injective, 64
integers, 2factor, 7
factor uniquely, 3, 10
modulo n, 22
integers modulo n, 22
isomorphism, 87
joke, 11kernel, 64
Lagrange, 27Lang, 114
largest known
elliptic curve rank, 142
prime, 12value of π(x), 16
Legendre Symbol, 70
Legendre symbol of 2 proposition,
80
Lenstra, 11, 129–133
lift, 23

Indexlinear equations modulo n, 23
long division proposition, 4
man in the middle attack, 56
Mazur theorem, 141
Mersenne prime, 13
Michael, 56, 135, 137
modular arithmetic
and linear equations, 23
order of element, 25
Mordell, 140
Mordell theorem, 140
multiplicative, 31
functions, 30
order, 22
multiplicative of Euler’s function
proposition, 31
natural numbers, 2
Nikita, 61, 135, 137
normal, 46
notation, x
number of primitive roots proposition, 43
one-way function, 56
open problem
congruent numbers, 142
decide if congruent number,143
fast integer factorization, 8order, 25, 42of element, 25
partial convergents, 97
partial convergents proposition, 97
period continued fraction theorem,
112
period of the continued fraction,
111
periodic continued fraction, 111
ϕ function, 22
phi function
is multiplicative, 31
Pieter, 44

163

Pollard’s (p−1)-method, 129–132
polynomial time, 8
polynomials
over Z/pZ, 40
power smooth, 129
powering algorithm, 34
primality test
deterministic, 38
Miller-Rabin, 37
probabilistic, 31
pseudoprime, 36
prime, 2prime factorization proposition, 7
prime number theorem, 11, 16
primes, 2density of, 14
infinitely many, 11largest known, 12
Mersenne, 13
of form 4x − 1, 13
of form ax + b, 13of the form 6x − 1, 19
sequence of, 10
testing for, 36
primitive, 81, 118
representation, 118
primitive root, 39
existence, 42
mod power of two, 40
primitive root mod prime powers
theorem, 43
primitive root of unity, 81
primitive root theorem, 42
propositioncancellation, 23
congruent number criterion,
144
congruent numbers and elliptic curves, 143convergence of continued fraction, 107
decryption key, 57
divisibility by 3, 23
Euler, 78
Euler’s criterion, 73

164

Index

extended Euclidean, 32
Gauss sum, 82
geometric group law, 126
how convergents converge, 100
infinitely many primes, 13
Legendre symbol of 2, 80
long division, 4multiplicative of Euler’s function, 31
number of primitive roots, 43partial convergents, 97
prime factorization, 7
rational continued fractions,100root bound, 40
solvability, 25
units, 24
Wilson, 27
Pseudoprimality theorem, 36
pseudoprime, 36
public key, 57
quadratic irrational, 111
continued fraction of, 110
quadratic nonresidue, 70
quadratic reciprocity, 69
elementary proof, 75–81
Gauss sums proof, 81
quadratic reciprocity theorem, 72
quadratic residue, 70
quantum computer, 8, 53
rank, 141, 142
rational continued fractions proposition, 100
rational point, 140
recognizing rational numbers, 115
reduction modulo n, 23
Riemann Hypothesis, 18
Riemann Hypothesis, 11, 15
bound on π(x), 18
ring, 22root bound proposition, 40
root of unity, 81
primitive, 81

RSA cryptosystem, 56–66
RSA-155, 9
RSA-576, 8
Shor, 8, 53
simple continued fraction, 95
smooth, 129
solvability proposition, 25
square roots
how to find mod p, 86–89
squaressum of two, 117
subgroup, 64
sum of two squares theorem, 117
sums of two squares, 117
surjective, 64
tablecomparing π(x) to x/(log(x)−1), 17
values of π(x), 16
when 5 a square mod p, 72
The Man, 56
theorem
Chinese remainder, 29
continued fraction convergence,105
continued fraction existence,106
continued fraction limit, 104
Dirichlet, 14
elliptic curve group law, 126
Euclid, 7
Euler’s, 25, 26
infinitely many primes, 11
infinitely many triangles, 145
Mazur, 141
Mordell, 140
of Dirichlet, 11
of Wilson, 27
period continued fraction, 112
prime number, 16
primitive root, 42
primitive root mod prime powers, 43Index

Pseudoprimality, 36
quadratic reciprocity, 72
sum of two squares, 117
unique factorization, 3
torsion subgroup, 140
Trotter, 114
unique factorization, 3
unique factorization theorem, 3unit group, 22
unitsof Z/pZ are cyclic, 39
roots of unity, 81
units proposition, 24
Vallée Poussin, 16
Wilson proposition, 27
Wilson’s theorem, 27
Zagier, 143165