aarjaneiro/planetmath huggingface dataset:
proof of Lucas’s theorem

proof of Lucas’s theorem

Let

n

≥

m

∈

ℕ

. Let

a

0

,

b

0

be the least non-negative

residues

of

n

,

m

(

mod

p

)

, respectively. (Additionally, we set

r

=

n

-

m

, and

r

0

is the
least non-negative residue of

r

modulo

p

.) Then the statement follows from

(

n

m

)

≡

(

a

0

b

0

)

⁢

(

⌊

n

p

⌋

⌊

m

p

⌋

)

(

mod

p

)

.

We define the ’carry indicators’

c

i

for all

i

≥

0

as

c

i

=

(

1

if

b

i

+

r

i

≥

p

0

otherwise

,

and additionally

c

-

1

=

0

.

The special case

s

=

1

of Anton’s

congruence

is:

(

n

!

)

p

≡

(

-

1

)

⌊

n

p

⌋

⋅

a

0

!

(

mod

p

)

,

(1)

where

a

0

as defined above, and

(

n

!

)

p

is the product of numbers

≤

n

not

divisible

by

p

.
So we have

n

!

⌊

n

p

⌋

!

⁢

p

⌊

n

p

⌋

=

(

n

!

)

p

≡

(

-

1

)

⌊

n

p

⌋

⋅

a

0

!

(

mod

p

)

When dividing by the left-hand terms of the congruences for

m

and

r

, we see
that the power of

p

is

⌊

n

p

⌋

-

⌊

m

p

⌋

-

⌊

r

p

⌋

=

c

0

So we get the congruence

(

n

m

)

(

⌊

n

p

⌋

⌊

m

p

⌋

)

⋅

p

c

0

≡

(

-

1

)

c

0

⋅

(

a

0

b

0

)

,

or equivalently

(

-

1

p

)

c

0

⋅

(

n

m

)

≡

(

a

0

b

0

)

⁢

(

⌊

n

p

⌋

⌊

m

p

⌋

)

(

mod

p

)

.

(2)

Now we consider

c

0

=

1

. Since

a

0

=

b

0

+

r

0

-

p

⁢

c

0

=

p

⏟

,

b

0

+

r

0

<

b

0

↔

c

0

=

1

↔

b

0

-

(

p

-

r

0

)

=

a

0

<

b

0

↔

(

a

0

b

0

)

=

0

. So both congruences–the one in the
statement and (

2

)– produce the same results.

Title

proof of Lucas’s theorem

Canonical name

ProofOfLucassTheorem

Date of creation

2013-03-22 13:22:56

Last modified on

2013-03-22 13:22:56

Owner

mathcam (2727)

Last modified by

mathcam (2727)

Numerical id

7

Author

mathcam (2727)

Entry type

Proof

Classification

msc 11A07

Related topic

BinomialCoefficient

Generated on Fri Feb 9 15:11:28 2018 by

LaTeXML
Shanks-Tonelli algorithm

Shanks-Tonelli algorithm

The

Shanks-Tonelli algorithm

is a procedure for solving a

congruence

of the form

x

2

≡

n

(

mod

p

)

, where

p

is an

odd prime

and

n

is a

quadratic residue

of

p

. In other words, it can be used to compute modular

square roots

.

First find

positive integers

Q

and

S

such that

p

-

1

=

2

S

⁢

Q

, where

Q

is odd. Then find a

quadratic nonresidue

W

of

p

and compute

V

≡

W

Q

(

mod

p

)

. Then find an

integer

n

′

that is the

multiplicative inverse

of

n

(

mod

p

)

(i.e.,

n

⁢

n

′

≡

1

(

mod

p

)

).

Compute

R

≡

n

Q

+

1

2

(

mod

p

)

and find the smallest integer

i

≥

0

that satisfies

(

R

2

⁢

n

′

)

2

i

≡

1

(

mod

p

)

If

i

=

0

, then

x

=

R

, and the algorithm stops. Otherwise, compute

R

′

≡

R

⁢

V

2

(

S

-

i

-

1

)

(

mod

p

)

and repeat the procedure for

R

=

R

′

.

Title

Shanks-Tonelli algorithm

Canonical name

ShanksTonelliAlgorithm

Date of creation

2013-03-22 11:55:16

Last modified on

2013-03-22 11:55:16

Owner

mathcam (2727)

Last modified by

mathcam (2727)

Numerical id

10

Author

mathcam (2727)

Entry type

Algorithm

Classification

msc 11A07

Classification

msc 11Y99

Generated on Fri Feb 9 15:11:30 2018 by

LaTeXML
short Taylor theorem

short Taylor theorem

If

f

⁢

(

x

)

is a

polynomial

with

integer

coefficients

and

x

0

and

h

integers, then the

congruence

f

⁢

(

x

0

+

h

)

≡

f

⁢

(

x

0

)

+

f

′

⁢

(

x

0

)

⁢

h

(

mod

h

2

)

(1)

is in force.

Proof.

Because of the linear properties of (1) we can confine us to the

monomials

f

⁢

(

x

)

:=

x

n

.  Then

f

′

⁢

(

x

)

=

n

⁢

x

n

-

1

.  By the

binomial theorem

we have

(

x

0

+

h

)

n

=

x

0

n

+

n

⁢

x

0

n

-

1

⁢

h

+

h

2

⁢

P

⁢

(

x

0

)

(2)

where

P

⁢

(

x

0

)

is a polynomial in

x

0

with integer coefficients.  The equality (2) may be written as the asserted congruence (1).

Title

short Taylor theorem

Canonical name

ShortTaylorTheorem

Date of creation

2013-04-01 13:19:12

Last modified on

2013-04-01 13:19:12

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

1

Author

pahio (2872)

Entry type

Definition

Classification

msc 11A07

Generated on Fri Feb 9 15:11:32 2018 by

LaTeXML
Wieferich prime

Wieferich prime

A

Wieferich prime

a is

prime number

p

such that

p

2

divides

2

p

-

1

-

1

; compare this with Fermat’s little

theorem

, which states that every prime

p

divides

2

p

-

1

-

1

. Wieferich primes were first described by Arthur Wieferich in 1909 in works pertaining to Fermat’s last theorem.

The only known Wieferich primes are 1093 and 3511, found by W. Meissner in 1913 and N. G. W. H. Beeger in 1922, respectively; if any others exist, they must be at least

1.25

×

10

15

. The

conjecture

that only finitely many Wieferich primes exist remains unproven, though J. H. Silverman was able to show in 1988 that if the

abc Conjecture

holds, then for any

positive integer

a

>

1

, there exist infinitely many primes

p

such that

p

2

does not divide

a

p

-

1

-

1

. In particular, there are infinitely many primes which are not Wieferich.

Wieferich primes and Fermat’s last theorem

The following theorem connecting Wieferich primes and Fermat’s last theorem was proven by Wieferich in 1909:

Theorem 1

.

Let

p

be prime, and let

x

,

y

,

z

be

integers

such that

x

p

+

y

p

+

z

p

=

0

. Furthermore, assume that

p

does not divide the product

x

⁢

y

⁢

z

. Then

p

is a Wieferich prime.

In 1910, Mirimanoff was able to expand this theorem by showing that, if the preconditions of the theorem hold true for some prime

p

, then

p

2

must also divide

3

p

-

1

. Prime numbers of this kind have been called Mirimanoff primes on occasion, but the name has not entered general mathematical use.

An analysis of Wieferich primes also proved crucial to Preda Mihailescu’s proof of the (formerly-named) Catalan’s conjecture.

References

1

Ireland, Kenneth and Rosen, Michael. A Classical Introduction to Modern

Number Theory

. Springer, 1998.

2

Nathanson, Melvyn B.

Elementary Methods

in Number Theory. Springer, 2000.

3

Wikipedia, the free encyclopedia, entry on Wieferich primes. All text is available under the terms of the GNU Free Documentation License

Title

Wieferich prime

Canonical

name

WieferichPrime

Date of creation

2013-03-22 13:50:21

Last modified on

2013-03-22 13:50:21

Owner

mathcam (2727)

Last modified by

mathcam (2727)

Numerical id

7

Author

mathcam (2727)

Entry type

Definition

Classification

msc 11A07

Generated on Fri Feb 9 15:11:34 2018 by

LaTeXML
Wilson’s theorem for prime powers

Wilson’s theorem for prime powers

For every

natural number

n

, let

(

n

⁢

!

¯

)

p

denote the

product

of numbers

1

≤

m

≤

n

with

g

⁢

c

⁢

d

⁢

(

m

,

p

)

=

1

.

For prime

p

and

s

∈

ℕ

(

p

s

!

¯

)

p

≡

(

1

for

⁢

p

=

2

,

s

≥

3

-

1

otherwise

(

mod

p

s

)

.

Proof:

We pair up all factors of the product

(

p

s

⁢

!

¯

)

p

into those
numbers

m

where

m

≢

m

-

1

(

mod

p

s

)

and those where this is not
the case. So

(

p

s

⁢

!

¯

)

p

is

congruent

(modulo

p

s

) to the product of those
numbers

m

where

m

≡

m

-

1

(

mod

p

s

)

↔

m

2

≡

1

(

mod

p

s

)

.

Let

p

be an

odd prime

and

s

∈

ℕ

. Since

2

|̸

p

s

,

p

s

|

(

m

2

-

1

)

implies

p

s

|

(

m

+

1

)

either or

p

s

|

(

m

-

1

)

. This
leads to

(

p

s

⁢

!

¯

)

p

≡

-

1

(

mod

p

s

)

for odd prime

p

and any

s

∈

ℕ

.

Now let

p

=

2

and

s

≥

2

. Then

(

1

+

t

.2

s

-

1

)

2

≡

1

(

mod

2

s

)

,

t

=

+

-

1

.

Since

(

2

s

-

1

+

1

)

⁢

(

2

s

-

1

-

1

)

≡

-

1

(

mod

2

s

)

,

we have

(

p

s

⁢

!

¯

)

p

≡

(

-

1

)

.

(

-

1

)

=

1

(

mod

p

s

)

For

p

=

2

,

s

≥

3

, but

-

1

for

s

=

1

,

2

.

□

Title

Wilson’s

theorem

for prime powers

Canonical name

WilsonsTheoremForPrimePowers

Date of creation

2013-03-22 13:22:14

Last modified on

2013-03-22 13:22:14

Owner

Thomas Heye (1234)

Last modified by

Thomas Heye (1234)

Numerical id

8

Author

Thomas Heye (1234)

Entry type

Theorem

Classification

msc 11A07

Classification

msc 11A41

Generated on Fri Feb 9 15:11:36 2018 by

LaTeXML
any nonzero integer is quadratic residue

any nonzero integer is quadratic residue

Theorem.

For every nonzero

integer

a

there exists an

odd prime

number

p

such that

a

is a

quadratic residue

modulo

p

.

Proof.

1

∘

.

a

=

2

.  We see that

3

2

≡

2

(

mod

7

)

and

7

∤

2

,  whence 2 is a quadratic residue modulo

7

.

2

∘

.

2

∣

a

but

a

≠

2

.  The number

1

2

-

a

=

1

-

a

(which is odd and

≠

±

1

) has an odd prime factor

p

which does not divide

a

.  Thus

a

is a quadratic residue modulo

p

.

3

∘

.

a

=

3

.  We state that

4

2

-

3

=

13

≡

0

(

mod

13

)

and

13

∤

3

.  Therefore 3 is a quadratic residue modulo 13.

4

∘

.

a

=

5

.  We see that

4

2

-

5

=

11

≡

0

(

mod

11

)

and

11

∤

5

, i.e. 5 is a quadratic residue modulo 11.

5

∘

.

2

∤

a

but

a

≠

3

,

a

≠

5

.  Now the number

2

2

-

a

=

4

-

a

(which is odd and

≠

±

1

) has an odd prime factor

p

.  Moreover,

p

∤

a

since

p

∤

4

.  Accordingly,

a

is a quadratic residue modulo

p

.

Title

any nonzero integer is quadratic residue

Canonical name

AnyNonzeroIntegerIsQuadraticResidue

Date of creation

2013-03-22 18:01:03

Last modified on

2013-03-22 18:01:03

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

6

Author

pahio (2872)

Entry type

Theorem

Classification

msc 11A15

Related topic

FundamentalTheoremOfArithmetic

Generated on Fri Feb 9 15:11:39 2018 by

LaTeXML
cases when minus one is a quadratic residue

cases when minus one is a quadratic residue

Theorem

.

Let

p

be an

odd prime

. Then

-

1

is a

quadratic residue

modulo

p

if and only if

p

≡

1

mod

4

.

Proof.

Let

p

be an odd prime. Notice that

p

is

congruent

to either

1

or

3

modulo

4

. By the definition of the

Legendre symbol

, we need to verify that

(

-

1

p

)

=

1

if and only if

p

≡

1

mod

4

. By Euler’s criterion

(

-

1

p

)

≡

(

-

1

)

(

p

-

1

)

/

2

mod

p

.

Finally, note that the

integer

p

-

1

2

is even if

p

≡

1

mod

4

and odd if

p

≡

3

mod

4

.
∎

Title

cases when minus one is a quadratic residue

Canonical name

CasesWhenMinusOneIsAQuadraticResidue

Date of creation

2013-03-22 16:18:10

Last modified on

2013-03-22 16:18:10

Owner

alozano (2414)

Last modified by

alozano (2414)

Numerical id

6

Author

alozano (2414)

Entry type

Theorem

Classification

msc 11A15

Related topic

EulersCriterion

Related topic

ValuesOfTheLegendreSymbol

Generated on Fri Feb 9 15:11:40 2018 by

LaTeXML
discrete logarithm

discrete logarithm

Let

p

be a prime. We know that the group

G

:=

(

ℤ

/

p

⁢

ℤ

)

*

is cyclic. Let

g

be a

primitive root

of

G

, i.e.

G

=

{

1

,

g

,

g

2

,

…

,

g

p

-

2

}

.
For a number

x

∈

G

we want to know the unique number

0

≤

n

≤

p

-

2

with

x

=

g

n

.

This number

n

is called the

discrete logarithm

or

index

of

x

to the basis

g

and is denoted as

ind

g

⁡

(

x

)

. For

x

,

y

∈

G

it satisfies the following

properties

:

ind

g

⁡

(

x

⁢

y

)

=

ind

g

⁡

(

x

)

+

ind

g

⁡

(

y

)

;

ind

g

⁡

(

x

-

1

)

=

-

ind

g

⁡

(

x

)

;

ind

g

⁡

(

x

k

)

=

k

⋅

ind

g

⁡

(

x

)

.

Furthermore, for a pair

g

,

h

of distinct primitive roots, we also have, for any

x

∈

G

:

ind

h

⁡

(

x

)

=

ind

h

⁡

(

g

)

⋅

ind

g

⁡

(

x

)

;

1

=

ind

g

⁡

(

h

)

⋅

ind

h

⁡

(

g

)

;

ind

g

⁡

(

-

1

)

=

p

-

1

2

.

It is a difficult problem to compute the discrete logarithm, while powering is very easy. Therefore this is of some

interest

to cryptography.

Title

discrete logarithm

Canonical

name

DiscreteLogarithm

Date of creation

2013-03-22 14:54:27

Last modified on

2013-03-22 14:54:27

Owner

mathwizard (128)

Last modified by

mathwizard (128)

Numerical id

7

Author

mathwizard (128)

Entry type

Definition

Classification

msc 11A15

Synonym

index

Generated on Fri Feb 9 15:11:46 2018 by

LaTeXML
Euler’s criterion

Euler’s criterion

Let

p

be an

odd prime

and

n

an

integer

such that

(

n

,

p

)

=

1

(that is,

n

and

p

are relatively prime).

Then

(

n

|

p

)

≡

n

(

p

-

1

)

/

2

(

mod

p

)

where

(

n

|

p

)

is the

Legendre symbol

.

Title

Euler’s criterion

Canonical name

EulersCriterion

Date of creation

2013-03-22 12:20:00

Last modified on

2013-03-22 12:20:00

Owner

drini (3)

Last modified by

drini (3)

Numerical id

5

Author

drini (3)

Entry type

Theorem

Classification

msc 11A15

Related topic

GaussLemma

Related topic

QuadraticReciprocityRule

Related topic

LegendreSymbol

Related topic

QuadraticResidue

Related topic

ProofOfGaussLemma

Related topic

PropertiesOfTheLegendreSymbol

Related topic

1IsQuadraticResidueIfAndOnlyIfPequiv1Mod4

Generated on Fri Feb 9 15:11:47 2018 by

LaTeXML
Gauss’ lemma

Gauss’ lemma

Gauss’ lemma on

quadratic residues

is:

Proposition

1:

Let

p

be an

odd prime

and let

n

be an

integer

which is not a

multiple

of

p

. Let

u

be the
number of

elements of

the set

{

n

,

2

⁢

n

,

3

⁢

n

,

…

,

p

-

1

2

⁢

n

}

whose least

positive

residues

, modulo

p

, are

greater

than

p

/

2

.
Then

(

n

p

)

=

(

-

1

)

u

where

(

n

p

)

is the

Legendre symbol

.

That is,

n

is a quadratic residue modulo

p

when

u

is even and it is a

quadratic nonresidue

when

u

is odd.

Gauss’ Lemma is the special case

S

=

{

1

,

2

,

…

,

p

-

1

2

}

of the slightly more general statement below.
Write

𝔽

p

for the field of

p

elements, and
identify

𝔽

p

with the set

{

0

,

1

,

…

,

p

-

1

}

,
with its

addition

and

multiplication

mod

p

.

Proposition 2:

Let

S

be a subset of

𝔽

p

*

such that

x

∈

S

or

-

x

∈

S

, but not both, for any

x

∈

𝔽

p

*

.
For

n

∈

𝔽

p

*

let

u

⁢

(

n

)

be the number of elements

k

∈

S

such that

k

⁢

n

∉

S

. Then

(

n

p

)

=

(

-

1

)

u

⁢

(

n

)

.

Proof:

If

a

and

b

are distinct elements of

S

, we cannot have

a

⁢

n

=

±

b

⁢

n

, in view of the

hypothesis

on

S

. Therefore

∏

a

∈

S

a

⁢

n

=

(

-

1

)

u

⁢

(

n

)

⁢

∏

a

∈

S

a

.

On the left we have

n

(

p

-

1

)

/

2

⁢

∏

a

∈

S

a

=

(

n

p

)

⁢

∏

a

∈

S

a

by Euler’s criterion. So

(

n

p

)

⁢

∏

a

∈

S

a

=

(

-

1

)

u

⁢

(

n

)

⁢

∏

a

∈

S

a

The product is nonzero, hence can be cancelled, yielding the proposition.

Remarks:

Using Gauss’ Lemma, it is straightforward to prove that
for any odd prime

p

:

(

-

1

p

)

=

{

1

if

p

≡

1

(

mod

4

)

-

1

if

p

≡

-

1

(

mod

4

)

(

2

p

)

=

{

1

if

p

≡

±

1

(

mod

8

)

-

1

if

p

≡

±

3

(

mod

8

)

The condition on

S

can also be stated like this:
for any square

x

2

∈

𝔽

p

*

, there is a unique

y

∈

S

such
that

x

2

=

y

2

. Apart from the usual choice

S

=

{

1

,

2

,

…

,

(

p

-

1

)

/

2

}

,

the set

{

2

,

4

,

…

,

p

-

1

}

has also been used, notably by Eisenstein.
I think it was also Eisenstein who gave us this

trigonometric identity

,
which is closely related to Gauss’ Lemma:

(

n

p

)

=

∏

a

∈

S

sin

⁡

(

2

⁢

π

⁢

a

⁢

n

/

p

)

sin

⁡

(

2

⁢

π

⁢

a

/

p

)

It is possible to prove Gauss’ Lemma or Proposition 2 “from scratch”, without
leaning on Euler’s criterion, the existence of a

primitive
root

, or the fact that a

polynomial

over

𝔽

p

has no more zeros
than its degree.

Title

Gauss’ lemma

Canonical

name

GaussLemma

Date of creation

2013-03-22 12:19:46

Last modified on

2013-03-22 12:19:46

Owner

drini (3)

Last modified by

drini (3)

Numerical id

14

Author

drini (3)

Entry type

Theorem

Classification

msc 11A15

Related topic

EulersCriterion

Related topic

ZolotarevsLemma

Generated on Fri Feb 9 15:11:50 2018 by

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proof of Euler’s criterion

proof of Euler’s criterion

(All

congruences

are modulo

p

for the proof; omitted for clarity.)

Let

x

=

a

(

p

-

1

)

/

2

Then

x

2

≡

1

by Fermat’s Little

Theorem

. Thus:

x

≡

±

1

Now consider the two possibilities:

•

If

a

is a

quadratic residue

then by

definition

,

a

≡

b

2

for some

b

. Hence:

x

≡

a

(

p

-

1

)

/

2

≡

b

p

-

1

≡

1

•

It remains to show that

a

(

p

-

1

)

/

2

≡

-

1

if

a

is a quadratic
non-residue. We can proceed in two ways:

Proof (a)

Partition

the set

{

1

,

…

,

p

-

1

}

into
pairs

{

c

,

d

}

such that

c

⁢

d

≡

a

. Then

c

and

d

must
always be distinct since

a

is a non-residue. Hence, the

product

of
the union of the partitions is:

(

p

-

1

)

!

≡

a

(

p

-

1

)

/

2

≡

-

1

and the result follows by Wilson’s Theorem.

Proof (b)

The equation:

z

(

p

-

1

)

/

2

≡

1

has at most

(

p

-

1

)

/

2

roots. But we already know of

(

p

-

1

)

/

2

distinct roots of the above equation, these being the quadratic residues modulo

p

. So

a

can’t be a root, yet

a

(

p

-

1

)

/

2

≡

±

1

. Thus we must have:

a

(

p

-

1

)

/

2

≡

-

1

QED.

Title

proof of Euler’s criterion

Canonical

name

ProofOfEulersCriterion

Date of creation

2013-03-22 12:41:53

Last modified on

2013-03-22 12:41:53

Owner

Koro (127)

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Koro (127)

Numerical id

7

Author

Koro (127)

Entry type

Proof

Classification

msc 11A15

Generated on Fri Feb 9 15:11:52 2018 by

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proof of quadratic reciprocity rule

proof of quadratic reciprocity rule

The quadratic reciprocity law is:

Theorem

:

(Gauss) Let

p

and

q

be distinct

odd primes

,
and write

p

=

2

⁢

a

+

1

and

q

=

2

⁢

b

+

1

. Then

(

p

q

)

⁢

(

q

p

)

=

(

-

1

)

a

⁢

b

.

(

(

v

w

)

is the

Legendre symbol

.)

Proof:

Let

R

be the subset

[

-

a

,

a

]

×

[

-

b

,

b

]

of

ℤ

×

ℤ

. Let

S

be the interval

[

-

(

p

⁢

q

-

1

)

/

2

,

(

p

⁢

q

-

1

)

/

2

]

of

ℤ

.
By the

Chinese remainder theorem

, there exists a unique

bijection

f

:

S

→

R

such that, for any

s

∈

S

, if we
write

f

⁢

(

s

)

=

(

x

,

y

)

, then

x

≡

s

(

mod

p

)

and

y

≡

s

(

mod

q

)

.
Let

P

be the subset of

R

consisting of the values of

f

on

[

1

,

(

p

⁢

q

-

1

)

/

2

]

.

P

contains, say,

u

elements of the form

(

x

,

0

)

such that

x

<

0

, and

v

elements of the form

(

0

,

y

)

with

y

<

0

. Intending to apply Gauss’s lemma,
we seek some kind of comparison between

u

and

v

.

We define three subsets of

P

by

R

0

=

{

(

x

,

y

)

∈

P

|

x

>

0

,

y

>

0

}

R

1

=

{

(

x

,

y

)

∈

P

|

x

<

0

,

y

≥

0

}

R

2

=

{

(

x

,

y

)

∈

P

|

x

≥

0

,

y

<

0

}

and we let

N

i

be the

cardinal

of

R

i

for each

i

.

P

has

a

⁢

b

+

b

elements

in the region

y

>

0

, namely

f

⁢

(

m

)

for all

m

of the form

k

+

l

⁢

q

with

1

≤

k

≤

b

and

0

≤

l

≤

a

.
Thus

N

0

+

N

1

=

a

⁢

b

+

b

-

(

b

-

v

)

+

u

i.e.

N

0

+

N

1

=

a

⁢

b

+

u

+

v

.

(1)

Swapping

p

and

q

, we have likewise

N

0

+

N

2

=

a

⁢

b

+

u

+

v

.

(2)

Furthermore, for any

s

∈

S

, if

f

⁢

(

s

)

=

(

x

,

y

)

then

f

⁢

(

-

s

)

=

(

-

x

,

-

y

)

.
It follows that for any

(

x

,

y

)

∈

R

other than

(

0

,

0

)

, either

(

x

,

y

)

or

(

-

x

,

-

y

)

is
in

P

, but not both.
Therefore

N

1

+

N

2

=

a

⁢

b

+

u

+

v

.

(3)

Adding (1), (2), and (3) gives us

0

≡

a

⁢

b

+

u

+

v

(

mod

2

)

so

(

-

1

)

a

⁢

b

=

(

-

1

)

u

⁢

(

-

1

)

v

which, in view of Gauss’s lemma, is the desired

conclusion

.

For a bibliography of the more than 200 known proofs of
the QRL, see

http://www.rzuser.uni-heidelberg.de/ hb3/fchrono.html

Lemmermeyer
.

Title

proof of quadratic reciprocity rule

Canonical name

ProofOfQuadraticReciprocityRule

Date of creation

2013-03-22 13:16:15

Last modified on

2013-03-22 13:16:15

Owner

mathcam (2727)

Last modified by

mathcam (2727)

Numerical id

12

Author

mathcam (2727)

Entry type

Proof

Classification

msc 11A15

Generated on Fri Feb 9 15:11:55 2018 by

LaTeXML
quadratic character of 2

quadratic character of 2

For any

odd prime

p

, Gauss’s lemma quickly yields

(

2

p

)

=

1

⁢

if

⁢

p

≡

±

1

(

mod

8

)

(1)

(

2

p

)

=

-

1

⁢

if

⁢

p

≡

±

3

(

mod

8

)

(2)

But there is another way, which goes back to Euler, and is worth
seeing, inasmuch as it is the prototype of certain more general

arguments

about character sums.

Let

σ

be a

primitive

eighth

root of unity

in an

algebraic closure

of

ℤ

/

p

⁢

ℤ

, and write

τ

=

σ

+

σ

-

1

.
We have

σ

4

=

-

1

,
whence

σ

2

+

σ

-

2

=

0

, whence

τ

2

=

2

.

By the

binomial formula

, we have

τ

p

=

σ

p

+

σ

-

p

.

If

p

≡

±

1

(

mod

8

)

, this implies

τ

p

=

τ

.
If

p

≡

±

3

(

mod

8

)

, we get instead

τ

p

=

σ

5

+

σ

-

5

=

-

σ

-

1

-

σ

=

-

τ

.
In both cases, we get

τ

p

-

1

=

(

2

p

)

,
proving (1) and (2).

A variation of the argument, closer to Euler’s, goes as follows.
Write

σ

=

exp

⁡

(

2

⁢

π

⁢

i

/

8

)

τ

=

σ

+

σ

-

1

Both are

algebraic integers

. Arguing much as above, we end up with

τ

p

-

1

≡

(

2

p

)

(

mod

p

)

which is enough.

Title

quadratic character of 2

Canonical name

QuadraticCharacterOf2

Date of creation

2013-03-22 13:58:03

Last modified on

2013-03-22 13:58:03

Owner

mathcam (2727)

Last modified by

mathcam (2727)

Numerical id

5

Author

mathcam (2727)

Entry type

Theorem

Classification

msc 11A15

Related topic

ValuesOfTheLegendreSymbol

Generated on Fri Feb 9 15:11:57 2018 by

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quadratic reciprocity for polynomials

quadratic reciprocity for polynomials

Let

F

be a

finite field

of

characteristic

p

, and let

f

and

g

be distinct monic

irreducible

(non-constant)

polynomials

in the polynomial ring

F

⁢

[

X

]

. Define the

Legendre symbol

(

f

g

)

by

(

f

g

)

:=

{

1

if

f

is a square in the quotient ring

F

⁢

[

X

]

/

(

g

)

,

-

1

otherwise.

The

quadratic reciprocity theorem

for polynomials over a finite field states that

(

f

g

)

⁢

(

g

f

)

=

(

-

1

)

p

-

1

2

⁢

deg

⁡

(

f

)

⁢

deg

⁡

(

g

)

.

References

1

Feng, Ke Qin and Ying, Linsheng,

An

elementary proof

of the law of quadratic reciprocity in

F

q

⁢

(

T

)

.

Sichuan Daxue Xuebao

26

(1989), Special Issue, 36–40.

2

Merrill, Kathy D. and Walling, Lynne H.,

On

quadratic reciprocity

over

function fields

.

Pacific J. Math.

173

(1996), no. 1, 147–150.

Title

quadratic reciprocity for polynomials

Canonical name

QuadraticReciprocityForPolynomials

Date of creation

2013-03-22 12:11:42

Last modified on

2013-03-22 12:11:42

Owner

djao (24)

Last modified by

djao (24)

Numerical id

8

Author

djao (24)

Entry type

Theorem

Classification

msc 11A15

Classification

msc 11T55

Classification

msc 11R58

Related topic

QuadraticReciprocityRule

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