Chapter 2

Prime Numbers
The term factoring or factorization refers to the process of expressing an
integer as the product of two or more integers in a nontrivial way, e.g.,
42 = 6 × 7. Prime numbers are those for which this process cannot be
done. We will soon see that prime numbers are the building blocks of theintegers. Together with the theory of divisibility, the properties of primes
are foundational elements of number theory.

2.1

Primes and Composites

Definition. We call an integer p ≥ 2 a prime if it has no divisors strictly
between 1 and p. An integer n ≥ 2 which is not a prime is called composite.
The dichotomy between primes and composites thus takes place:primes: 2, 3,
5,
7,
11,
13,
17,
composites:
4,
6,
8, 9, 10,12,
14, 15, 16,

...
...

The words prime and composite are also used as adjectives, as in a prime
number,1 or a composite integer. Primality and compositeness are the nouns
associated with the two, respectively. Throughout this book, from now on,
we shall designate the notation p to always represent a prime, lest we forget
to remind the reader.
Proposition 2.1. The following statements hold.
1) Other than 2, all primes are odd numbers.2
1 Do not mistake prime numbers (lacking nontrivial factors) for relatively prime numbers (lacking common factors). It is true, two distinct primes are always relatively prime.
2 Being the only even prime, 2 is the odd one out!

12

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132) Every composite has a prime divisor.
3) The number n is composite if and only if n has a prime divisor p ≤√n.

Proof. 1) By definition, even numbers are multiples of 2, hence they are allcomposite, except 2 itself is prime.
2) Suppose, by induction, the statement is true up to n − 1. Either n isprime, and nothing to prove, or else n has a divisor d, with 1 < d < n.
It follows that d has a prime divisor which is also a divisor of n, by
Proposition 1.1(3).
3) A prime has no prime divisor less√than itself.
√ For composite n = ab,where a > 1 and b > 1, either a ≤ n or b ≤ n must√hold. Whicheveris true, by (2) a or b has a prime divisor p, where p ≤ n and p | n. ▽Propositions 2.1(3) implies that in order to test the primality of √a numbern, it suffices to check divisibility by the primes 2, 3, 5, . . . , up
ton. This√testing algorithm is called the trial division. For example, 113 ≈ 10.63,
and the only primes up to 10 are 2, 3, 5, 7—none of which divides 113.
Hence, 113 is a prime.
Exercise 2.1. Determine prime or composite, using trial division.
a) 383b) 447c) 799d) 811e) 1763
Exercise 2.2. Investigate true or false.
a) The number n + 99! is composite for each 2 ≤ n ≤ 100.
b) The number n4 + 4 is composite for each n ≥ 2.
c) The number n2 + n + 41 is prime for each n ≥ 0.
d) The number n2 − 81n + 1681 is prime for each n ≥ 1.
Finding multiples is obviously much easier than finding divisors. Inspired by this fact, the trial division is turned into the sieve of Eratosthenes,
a relatively efficient algorithm to identify the prime numbers up to a predetermined bound N . It works as follows.
We just ignore the even numbers, except the prime 2, then we list the
odd numbers from 3 to N . The smallest in this list, 3, is prime. Now label all
its multiples by this prime 3. Among the unlabeled remnant in the list, 5 isnow smallest. √Label all its multiples by this prime 5. Repeat this procedure
until all up to N have been labeled. The unlabeled numbers which remain
are all primes.

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Theory of Numbers

Example (The Sieve of Eratosthenes). With N = 101, it takes only 3 iterations, with primes 3, 5, 7, and the output is given below. The label following
each number n, separated by a colon (:), also indicates the smallest prime
divisor of n. The odd primes, surviving the sieve, are displayed in bold.
3:3
23
43
63:3
83

5:5
25:5
45:3
65:5
85:5

7:7
27:3
47
67
87:3

9:3
29
49:7
69:3
89

11
31
51:3
71
91:7

13
33:3
53
73
93:3

15:3
35:5
55:5
75:3
95:5

17
37
57:3
77:7
97

19
39:3
59
79
99:3

21:3
41
61
81:3
101

Table 2.1: The sieve of Eratosthenes for odd primes up to 101.
Exercise 2.3. Using the sieve of Eratosthenes, find all primes up to 1,000.
Alternately, implement this algorithm using a programming language of yourchoice, say with N = 10, 000.
Some divisibility properties involving primes will now be presented. The
simplest result is perhaps this next lemma, to be followed by a very usefultheorem, a consequence of Euclid’s lemma.
Lemma 2.2. Let p be a prime. For any integer n, we have gcd(p, n) = p if
p | n, otherwise gcd(p, n) = 1.
Proof. The claim is justified since 1 and p are the only divisors of p.

▽

Theorem 2.3. If p | mn, then either p | m or p | n. More generally, if aprime p divides the product of integers, then p divides one of them.
Proof. If p ∤ n then by Lemma 2.2, gcd(p, n) = 1, and by Theorem 1.8(1)(Euclid’s lemma) we then have p | m. Repeated use of this argument establishes the general claim.
▽
Exercise 2.4. Show that if p | n2 then p2 | n2 , where p is prime.

2.2

Factoring Composites into Primes

Intuitively, by factoring a given composite n and further factoring its factors
if necessary, we should be able to write n as a product of only prime numbers.
The next theorem, which is of greatest importance in the theory of numbers,
assures not only that factorization into primes can always be done, but alsothat the end collection of prime factors is uniquely determined by n. For
example, in factoring the number 1998, one may obtain 1998 = 2 × 3 × 3 ×

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15

3 × 37, or 1998 = 3 × 37 × 3 × 2 × 3, but it would be impossible to findanother prime factor outside the collection {2, 3, 3, 3, 37}. Here, a collection
is like a set in which repetition of elements is taken into account.
Theorem 2.4 (The Fundamental Theorem of Arithmetic). Every composite
is the product of a unique collection of prime numbers.
Proof. Claim first that every composite is a product of primes. Suppose
this is true up to n − 1. If n is prime, there is nothing to prove. Else by
Proposition 2.1(2), n = pn′ for a prime p and n′ < n. Either n′ is prime or,by the induction hypothesis, a product of primes. Thus the claim is true.
To prove uniqueness, we proceed by contradiction. Suppose we have twodifferent collections of primes, p’s and q’s, whose products both equal n.
Equating these products, after canceling out all common terms, will resultin p1 p2 · · · pj = q1 q2 · · · qk , where none of the p’s equals any of the q’s. By
Theorem 2.3, p1 must divide one of the q’s. But it is impossible for a prime
to divide another prime other than itself.
▽
Exercise 2.5. Factor these numbers into primes.
a) 123b) 400c) 720d) 7575e) 19392

Q
We will become familiar with the notation n = pei i , where n has beenfactored into powers of distinct primes, ei ≥ 1. We call this the factorization
into prime powers—distinct is implicitly assumed,Qalways. Occasionally it
will be more convenient to have the product n = pei i span over all prime
numbers by allowing ei ≥ 0. Of course, ei = 0 for all but finitely many ofthem. With
Q this assumption, if d | n then it is a consequence of Theorem 2.4
that d = pdi i , where each di ≤ ei .
Exercise 2.6. Prove that if d2 | n2 then d | n.

Exercise 2.7. Count how many positive divisors each number has.
a) 323b) 720c) 1024d) 2310e) 19392
CorollaryQ2.5. Suppose that
Q eim and n have been factored into primeQpowers, m =pfi i and n =pi , with ei , fi ≥ 0. Then gcd(m, n) =pdi i ,
where di = min(ei , fi ), the lesser of the two.

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Theory of Numbers

Q
Proof. By Theorem 2.4, a divisor of m must be of the form d = pdi i withdi ≤ fi . Similarly, if d | n then di ≤ ei . So the greatest possible value forsuch d is when di = min(ei , fi ).
▽Example. We evaluate gcd(27720, 61152) using prime factorization,
27720 =
61152 =

23 × 32 × 51 × 71 × 111 × 130
25 × 31 × 50 × 72 × 110 × 131

and get the result, gcd(27720, 61152) = 23 × 3 × 7 = 168.
Exercise 2.8. Evaluate gcd(m, n) by factoring m and n.
a) gcd(400, 720)b) gcd(19392, 29391)
c) gcd(23 × 38 × 54 × 75 , 37 × 52 × 72 )d) gcd(25 × 57 × 113 , 37 × 72 × 139 )e) gcd(24 × 52 × 7 × 113 , 27 × 32 × 52 × 11)
Exercise 2.9. Show that gcd(m2 , n2 ) = gcd(m, n)2 .
Exercise 2.10. The least common multiple of two nonzero integers is thesmallest positive integer which is divisible by both. For example, lcm(4, 6) =12 because it is the smallest positive integer such that 4 | 12 and 6 | 12.
QQa) Suppose m = pfi i and n = pei i , where ei ≥ 0 and fi ≥ 0. Prove thatQ dilcm(m, n) = pi , where di = max(ei , fi ), the greater of the two.
b) Show that if m | k and n | k, then lcm(m, n) | k.
c) Find an equation relating gcd(m, n) to lcm(m, n).
d) Illustrate your answer in (c) using m = 600 and n = 630.
Thus we have now another method for evaluating gcd(m, n), totally independent from the Euclidean algorithm. In contrast, however, factoringis slow and the computation time grows exponentially with the size of theinteger.
More about factorization will be discussed in Chapter 7; in the meantime,
we demonstrate next a factorization technique due to Fermat. Although themethod is old, many modern and powerful factoring algorithms are actuallybased on this principle.
If n = x2 − y 2 , then n factors as n = (x + y)(x − y). This fact is thesimple idea behind the method of Fermat factorization. We seek√a factor ofn by calculating the numbers y 2 = x2 − n for each integer x ≥ n until wefind a perfect square, i.e., the square of an integer.

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Example (Fermat Factorization). Let n = 4277. We haveso let us start with x = 66.

√

4277 ≈ 65.39,

662 − 4277 = 79672 − 4277 = 212

682 − 4277 = 347

692 − 4277 = 484 = 222

The result is, 4277 = 692 − 222 = (69 + 22)(69 − 22) = 91 × 47.
Exercise 2.11. Illustrate Fermat factorization using the following numbers.
a) 2117b) 16781c) 17933d) 70027
Fermat factorization works for all odd numbers, for if n = ab with botha and b odd, then n = x2 − y 2 , where x = (a + b)/2 and y = (a − b)/2.
Moreover, this shows that we should terminate the algorithm when we reachx = (n + 1)/2, in which case the result is trivial,√i.e., n = n × 1, and n isprime. For large n, however, the iterations from n to (n + 1)/2 make thisalgorithm too slow to be practical.
Exercise 2.12. Fermat √factorization is efficient when n has at least one
factor relatively close to n. Why is this statement true?

2.3

The Infinitude of Primes

One relevant question concerning primes is whether or not there exist infinitely many primes of a special form, e.g., 4n + 3 or n2 + 1. This willturn to generate very difficult problems, many of which are still unsolved.
But first, of course, we need to be convinced that prime numbers are indeed
infinitely many—and this fact is not hard to demonstrate.
Theorem 2.6. There are infinitely many prime numbers.
Proof. If there were only finitely many primes, let n be the product of themall. Too large to be prime, the number n + 1 would be composite anddivisible by p, which is one of the primes dividing n. Then p would divide(n + 1) − n = 1, by Proposition 1.1(4). This is absurd since p ≥ 2.
▽
Furthermore, we actually have a way to estimate the distribution ofprimes among the positive integers in a given interval. To make precise thestatement, we need the next definition.

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Theory of Numbers

Definition. The prime counting function π(x) denotes the number of primes
up to x, where x can be any real number.3 For example, π(13) = 6 because
there are exactly six primes in this range, i.e., 2, 3, 5, 7, 11, 13. Similarly,
π(100) = 25, according to Table 2.1.
For large values of x, the function π(x) behaves much like x/ log x, where
log x denotes the natural logarithm function. We state this result as the next
theorem, the proof of which requires advanced techniques from complex
analysis and, unfortunately, will not be provided here.
Theorem 2.7 (The Prime Number Theorem). We have
lim

x→∞

π(x)
= 1
x/ log x

Moreover, it has been found that x/(log x − 1) is a slightly better function
than x/ log x in approximating π(x) for large values of x.
Example. Up to 25 billion, the number of primes is estimated by
25, 000, 000, 000≈ 1, 089, 697, 743
log 25, 000, 000, 000 − 1which is comparable to the actual count, π(25 × 109 ) = 1, 091, 987, 405.
Exercise 2.13. Estimate how many prime numbers there are,a) up to one million.
b) up to ten million.
c) between 9 million and 10 million.
d) among the ten-digit integers.
Now back to primes of special forms. The first case we shall consider
involves primes that come in the sequence {an + b}, where a and b are fixed.
According to Proposition 1.6(1), every number of this form is a multiple ofgcd(a, b). Hence if gcd(a, b) > 1, then the sequence {an+b} can contain only
composites, except perhaps gcd(a, b) itself, if prime. So to avoid triviality,
we assume that gcd(a, b) = 1—a condition which, claimed below, is sufficient
to ensure the infinitude of such primes.
Theorem 2.8 (Dirichlet’s Theorem on Primes in Arithmetic Progressions).
Primes of the form an + b are infinitely many if, and only if, gcd(a, b) = 1.
This theorem is a very advanced general result whose proof lies in thedomain of analytic number theory and, unfortunately again, cannot be given
within the scope of this book. Instead we will supply, by way of illustration,a simple proof for the specific case a = 4 and b = 3.
3 For example, π(π) = 2. Ha!

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There are infinitely many primes of the form 4n + 3. To see this, first notethat every odd prime has the form either 4n+1 or 4n+3. Second, the productof two numbers of the form 4n + 1 is again of the same form. Therefore, anumber of the form 4n + 3 must have a prime divisor of the form 4n + 3.
If primes of the form 4n+3 were finite in number, let n be the product ofthem all. As noted, one of these prime divisors of n must divide 4(n − 1) + 3,
hence it would also divide 4(n − 1) + 3 − 4n = −1, a contradiction.
▽
Exercise 2.14. Prove the infinitude of primes of the form 6n + 5.
There remain many open problems today concerning the infinitude ofprimes of a curious type, such as n2 + 1 or n! + 1. Among the most popular
is the so-called Mersenne primes—of the form 2p − 1, e.g., the prime 31 =25 − 1. The exponent p must be a prime as a necessary, but not sufficient,condition for a Mersenne prime. This claim is in the next exercise.
Exercise 2.15. Let m and n be two positive integers. Prove these facts.
a) (2m − 1) % (2n − 1) = 2m % n − 1.
b) 2m − 1 is composite if m is.
c) gcd(2m − 1, 2n − 1) = 2gcd(m,n) − 1.
d) 2m − 1 and 2n − 1 are relatively prime if and only if m and n are.
It is unsettled whether or not there are infinitely many Mersenne primes.
Computational evidence suggests that there probably are. The largest finding, announced August 2008 at the Prime Pages [eCal], was the 12,978,189digit Mersenne prime 2p − 1 corresponding to p = 43112609.
This is quite common where, along with the search for a settling proof,
computational efforts are being done to find the largest prime of the specifictype—here, largest means record breaking.
Another famous prime search is that after twin primes, i.e., a pair ofprimes differing by 2. At the writing of this revision, the record of twin
primes at the Prime Pages shows the pair 65516468355 × 2333333 ± 1, of
100,355 decimal digits each.
The search for numerical support, however, has not always been thatfruitful—at least not with the next class of primes, to be found among the
Fermat numbers, i.e.,nFn = 22 + 1
for n ≥ 0. The first five Fermat numbers happen to be primes—such primes
are named Fermat primes. But no other Fermat primes have been discovered
until now, if any more exists.
The big challenge, since Fn grows quite rapidly, is not merely in evaluating a Fermat number or proving its primality, but even more in finding oneof its factors, if composite. We will see more about this topic in Section 9.3.

20

Theory of NumbersnExercise 2.16. Let Fn = 22 + 1 denote a Fermat number, n ≥ 0.
a) Find the first five Fermat primes.
b) Verify the recurrence relation Fn = F0 F1 F2 · · · Fn−1 + 2 for n ≥ 1.
c) Prove that Fermat numbers are relatively prime one to another.
d) Use (c) to prove again that there are infinitely many prime numbers.
Exercise 2.17. Show that 2k + 1 is composite if k has an odd prime factor.

2.4

Euler’s Factorization Method [Project 2]

While Fermat factorization is based on the difference of two squares, there
is another factoring technique, due to Euler, which involves the sum of two
squares.
Suppose that an odd number n can be expressed as a sum of two squares
in two different ways, n = x2 + y 2 = z 2 + w2 . Without loss of generality,
assume that x and z are even, while y and w odd. Let d = gcd(x − z, w − y)and c = gcd(x + z, w + y). This will lead us to finding a factor of n, givenby (c/2)2 + (d/2)2 .
Example (Euler’s Factorization Method). Given that 493 = 222 +32 = 182 +132 . We have d = gcd(22 − 18, 13 − 3) = 2 and c = gcd(22 + 18, 13 + 3) = 8.
Then (c/2)2 + (d/2)2 = 17 divides 493. Indeed, 493 = 17 × 29.
Project 2.4.1. Factor the following numbers by Euler’s method.
a) 10049 (1002 + 72 = 322 + 952 )b) 10081 (1002 + 92 = 842 + 552 )c) 10121d) 1000049
Project 2.4.2. Justify the claim of Euler’s method by showing thatn=(x − z)2 + (w − y)2c2 + d2×4
d2This method of factorization obviously has its limitations; for one thing,
there is no reason to expect that every number can be represented as a sumof two squares. The aim of the next project (Section 3.4) is to identify the
positive integers which are representable in this way. As a preview, theanswer will be given by the following theorem, first discovered by Fermat.
Q ei
Theorem 2.9. Let n =pi be the usual factorization of n into prime
powers. The diophantine equation n = x2 + y 2 has a solution if and only if
ei is even whenever pi % 4 = 3.
In particular, a consequence of Euler’s method, Theorem 2.9 implies thatprimes of the form 4k + 1 can be uniquely written as a sum of two squares!