Quadratic reciprocity

In number theory, the law of quadratic reciprocity is a theorem about modular arithmetic that gives conditions for the solvability of quadratic equations modulo prime numbers. There are a number of equivalent statements of the theorem. One version of the law states that for p and q odd prime numbers,

$\left(\frac{p}{q}\right) \left(\frac{q}{p}\right) = (-1)^{\frac{p-1}{2}\frac{q-1}{2}}$

where

$\left(\frac{p}{q}\right)$

denotes the Legendre symbol.

This law, combined with the properties of the Legendre symbol, means that any Legendre symbol can be calculated. This makes it possible to determine, for any quadratic equation, $x^2\equiv a \pmod{p},$ where p is an odd prime, if it has a solution. However, it does not provide any help at all for actually finding the solution. The solution can be found using quadratic residues.

The theorem was conjectured by Euler and Legendre and first proved by Gauss.^[1] He refers to it as the "fundamental theorem" in the Disquisitiones Arithmeticae and his papers, writing

The fundamental theorem must certainly be regarded as one of the most elegant of its type. (Art. 151)

Privately he referred to it as the "golden theorem."^[2] He published six proofs, and two more were found in his posthumous papers. There are now over 200 published proofs.^[3]

The first section of this article gives a special case of quadratic reciprocity that is representative of the general case. The second section gives the formulations of quadratic reciprocity found by Legendre and Gauss.

Motivating example

Consider the polynomial $f(n) = n^2 - 5$ and its values for $n \in \N.$ The prime factorizations of these values are given as follows:

n	f(n)		n	f(n)		n	f(n)
1	−4	−2^2	16	251	251	31	956	2^2⋅239
2	−1	−1	17	284	2^2⋅71	32	1019	1019
3	4	2^2	18	319	11⋅29	33	1084	2^2⋅271
4	11	11	19	356	2^2⋅89	34	1151	1151
5	20	2^2⋅5	20	395	5⋅79	35	1220	2^2⋅5⋅61
6	31	31	21	436	2^2⋅109	36	1291	1291
7	44	2^2⋅11	22	479	479	37	1364	2^2⋅11⋅31
8	59	59	23	524	2^2⋅131	38	1439	1439
9	76	2^2⋅19	24	571	571	39	1516	2^2⋅379
10	95	5⋅19	25	620	2^2⋅5⋅31	40	1595	5⋅11⋅29
11	116	2^2⋅29	26	671	11⋅61	41	1676	2^2⋅419
12	139	139	27	724	2^2⋅181	42	1759	1759
13	164	2^2⋅41	28	779	19⋅41	43	1844	2^2⋅461
14	191	191	29	836	2^2⋅11⋅19	44	1931	1931
15	220	2^2⋅5⋅11	30	895	5⋅179	45	2020	2^2⋅5⋅101

The prime numbers that appear as factors are 2,5, and all the prime numbers whose final digit is 1 or 9. No primes ending in 3 or 7 ever appear. Another way of phrasing this is that the primes p for which there exists an n such that n² ≡ 5 (mod p) are precisely 2, 5, and those primes p that are ≡ 1 or 4 (mod 5). Or in other words, when p is a prime that is neither 2 nor 5, 5 is a quadratic residue modulo p iff p is 1 or 4 modulo 5. In other words, 5 is a quadratic residue modulo p iff p is a quadratic residue modulo 5.

The law of quadratic reciprocity gives a similar characterization of prime divisors of f(n) = n² − c for any integer c.

Terminology, data, and two statements of the theorem

A quadratic residue (mod n) is any number congruent to a square (mod n). A quadratic nonresidue (mod n) is any number that is not congruent to a square (mod n). The adjective "quadratic" can be dropped if the context makes it clear that it is implied. When working modulo primes (as in this article), it is usual to treat zero as a special case. By doing so, the following statements become true:

• Modulo a prime, there are an equal number of quadratic residues and nonresidues.
• Modulo a prime, the product of two quadratic residues is a residue, the product of a residue and a nonresidue is a nonresidue, and the product of two nonresidues is a residue.

Table of quadratic residues

Squares mod primes

n	1	2	3	4	5	6	7	8	9	10	11	12	13	14	15	16	17	18	19	20	21	22	23	24	25
n^2	1	4	9	16	25	36	49	64	81	100	121	144	169	196	225	256	289	324	361	400	441	484	529	576	625
mod 3	1	1	0	1	1	0	1	1	0	1	1	0	1	1	0	1	1	0	1	1	0	1	1	0	1
mod 5	1	4	4	1	0	1	4	4	1	0	1	4	4	1	0	1	4	4	1	0	1	4	4	1	0
mod 7	1	4	2	2	4	1	0	1	4	2	2	4	1	0	1	4	2	2	4	1	0	1	4	2	2
mod 11	1	4	9	5	3	3	5	9	4	1	0	1	4	9	5	3	3	5	9	4	1	0	1	4	9
mod 13	1	4	9	3	12	10	10	12	3	9	4	1	0	1	4	9	3	12	10	10	12	3	9	4	1
mod 17	1	4	9	16	8	2	15	13	13	15	2	8	16	9	4	1	0	1	4	9	16	8	2	15	13
mod 19	1	4	9	16	6	17	11	7	5	5	7	11	17	6	16	9	4	1	0	1	4	9	16	6	17
mod 23	1	4	9	16	2	13	3	18	12	8	6	6	8	12	18	3	13	2	16	9	4	1	0	1	4
mod 29	1	4	9	16	25	7	20	6	23	13	5	28	24	22	22	24	28	5	13	23	6	20	7	25	16
mod 31	1	4	9	16	25	5	18	2	19	7	28	20	14	10	8	8	10	14	20	28	7	19	2	18	5
mod 37	1	4	9	16	25	36	12	27	7	26	10	33	21	11	3	34	30	28	28	30	34	3	11	21	33
mod 41	1	4	9	16	25	36	8	23	40	18	39	21	5	32	20	10	2	37	33	31	31	33	37	2	10
mod 43	1	4	9	16	25	36	6	21	38	14	35	15	40	24	10	41	31	23	17	13	11	11	13	17	23
mod 47	1	4	9	16	25	36	2	17	34	6	27	3	28	8	37	21	7	42	32	24	18	14	12	12	14

This table is complete for odd primes less than 50. To check whether a number m is a quadratic residue mod one of these primes p, find a ≡ m (mod p) and 0 ≤ a < p. If a is in row p, then m is a residue (mod p); if a is not in row p of the table, then m is a nonresidue (mod p).

The quadratic reciprocity law is the statement that certain patterns found in the table are true in general.

In this article, p and q always refer to distinct positive odd prime numbers.

±1 and the first supplement

Trivially 1 is a quadratic residue for all primes. The question becomes more interesting for −1. Examining the table, we find −1 in rows 5, 13, 17, 29, 37, and 41 but not in rows 3, 7, 11, 19, 23, 31, 43 or 47. The former set of primes are all congruent to 1 modulo 4, and the latter are congruent to 3 modulo 4.

First Supplement to Quadratic Reciprocity. The congruence $x^2 \equiv -1 \pmod{p}$ is solvable if and only if $p$ is congruent to 1 modulo 4.

±2 and the second supplement

Examining the table, we find 2 in rows 7, 17, 23, 31, 41, and 47, but not in rows 3, 5, 11, 13, 19, 29, 37, or 43. The former primes are all ≡ ±1 (mod 8), and the latter are all ≡ ±3 (mod 8). This leads to

Second Supplement to Quadratic Reciprocity. The congruence $x^2 \equiv 2 \pmod{p}$ is solvable if and only if $p$ is congruent to ±1 modulo 8.

−2 is in rows 3, 11, 17, 19, 41, 43, but not in rows 5, 7, 13, 23, 29, 31, 37, or 47. The former are ≡ 1 or ≡ 3 (mod 8), and the latter are ≡ 5, 7 (mod 8).

±3

3 is in rows 11, 13, 23, 37, and 47, but not in rows 5, 7, 17, 19, 29, 31, 41, or 43. The former are ≡ ±1 (mod 12) and the latter are all ≡ ±5 (mod 12).

−3 is in rows 7, 13, 19, 31, 37, and 43 but not in rows 5, 11, 17, 23, 29, 41, or 47. The former are ≡ 1 (mod 3) and the latter ≡ 2 (mod 3).

Since the only residue (mod 3) is 1, we see that −3 is a quadratic residue modulo every prime which is a residue modulo 3.

±5

5 is in rows 11, 19, 29, 31, and 41 but not in rows 3, 7, 13, 17, 23, 37, 43, or 47. The former are ≡ ±1 (mod 5) and the latter are ≡ ±2 (mod 5).

Since the only residues (mod 5) are ±1, we see that 5 is a quadratic residue modulo every prime which is a residue modulo 5.

−5 is in rows 3, 7, 23, 29, 41, 43, and 47 but not in rows 11, 13, 17, 19, 31, or 37. The former are ≡ 1, 3, 7, 9 (mod 20) and the latter are ≡ 11, 13, 17, 19 (mod 20).

Gauss's version

The observations about −3 and 5 continue to hold: −7 is a residue modulo p if and only if p is a residue modulo 7, −11 is a residue modulo p if and only if p is a residue modulo 11, 13 is a residue (mod p) if and only if p is a residue modulo 13, etc. The more complicated-looking rules for the quadratic characters of 3 and −5, which depend upon congruences modulo 12 and 20 respectively, are simply the ones for −3 and 5 working with the first supplement.

Example. For −5 to be a residue (mod p), either both 5 and −1 have to be residues (mod p) or they both have to be non-residues: i.e., p ≡ ±1 (mod 5) and p ≡ 1 (mod 4) or p ≡ ±2 (mod 5) and p ≡ 3 (mod 4). Using the Chinese remainder theorem these are equivalent to p ≡ 1, 9 (mod 20) or p ≡ 3, 7 (mod 20).

The generalization of the rules for −3 and 5 is Gauss's statement of quadratic reciprocity:

Quadratic Reciprocity (Gauss's Statement). If $q \equiv 1 \pmod{4}$ then the congruence $x^2 \equiv p \pmod{q}$ is solvable if and only if $x^2 \equiv q \pmod{p}$ is. If $q \equiv 3 \pmod{4}$ then the congruence $x^2 \equiv p \pmod{q}$ is solvable if and only if $x^2 \equiv -q \pmod{p}$ is.

These statements may be combined:

Quadratic Reciprocity (Combined Statement). Define:

$q^* = (-1)^{\frac{q-1}{2}}q$

Then the congruence $x^2 \equiv p \pmod{q}$ is solvable if and only if $x^2 \equiv q^* \pmod{p}$ is.

Table of quadratic character of primes

Legend

R	q is a residue (mod p)	q ≡ 1 (mod 4) or p ≡ 1 (mod 4) (or both)
N	q is a nonresidue (mod p)
R	q is a residue (mod p)	both q ≡ 3 (mod 4) and p ≡ 3 (mod 4)
N	q is a nonresidue (mod p)

		q
		3	5	7	11	13	17	19	23	29	31	37	41	43	47	53	59	61	67	71	73	79	83	89	97
p	3		N	R	N	R	N	R	N	N	R	R	N	R	N	N	N	R	R	N	R	R	N	N	R
	5	N		N	R	N	N	R	N	R	R	N	R	N	N	N	R	R	N	R	N	R	N	R	N
	7	N	N		R	N	N	N	R	R	N	R	N	R	N	R	N	N	R	R	N	R	N	N	N
	11	R	R	N		N	N	N	R	N	R	R	N	N	R	R	R	N	R	R	N	N	N	R	R
	13	R	N	N	N		R	N	R	R	N	N	N	R	N	R	N	R	N	N	N	R	N	N	N
	17	N	N	N	N	R		R	N	N	N	N	N	R	R	R	R	N	R	N	N	N	R	R	N
	19	N	R	R	R	N	R		R	N	N	N	N	R	R	N	N	R	N	N	R	N	R	N	N
	23	R	N	N	N	R	N	N		R	R	N	R	N	R	N	R	N	N	R	R	N	N	N	N
	29	N	R	R	N	R	N	N	R		N	N	N	N	N	R	R	N	R	R	N	N	R	N	N
	31	N	R	R	N	N	N	R	N	N		N	R	N	R	N	R	N	R	R	N	N	N	N	R
	37	R	N	R	R	N	N	N	N	N	N		R	N	R	R	N	N	R	R	R	N	R	N	N
	41	N	R	N	N	N	N	N	R	N	R	R		R	N	N	R	R	N	N	R	N	R	N	N
	43	N	N	N	R	R	R	N	R	N	R	N	R		R	R	R	N	R	N	N	R	R	N	R
	47	R	N	R	N	N	R	N	N	N	N	R	N	N		R	R	R	N	R	N	R	R	R	R
	53	N	N	R	R	R	R	N	N	R	N	R	N	R	R		R	N	N	N	N	N	N	R	R
	59	R	R	R	N	N	R	R	N	R	N	N	R	N	N	R		N	N	R	N	R	N	N	N
	61	R	R	N	N	R	N	R	N	N	N	N	R	N	R	N	N		N	N	R	N	R	N	R
	67	N	N	N	N	N	R	R	R	R	N	R	N	N	R	N	R	N		R	R	N	R	R	N
	71	R	R	N	N	N	N	R	N	R	N	R	N	R	N	N	N	N	N		R	R	R	R	N
	73	R	N	N	N	N	N	R	R	N	N	R	R	N	N	N	N	R	R	R		R	N	R	R
	79	N	R	N	R	R	N	R	R	N	R	N	N	N	N	N	N	N	R	N	R		R	R	R
	83	R	N	R	R	N	R	N	R	R	R	R	R	N	N	N	R	R	N	N	N	N		N	N
	89	N	R	N	R	N	R	N	N	N	N	N	N	N	R	R	N	N	R	R	R	R	N		R
	97	R	N	N	R	N	N	N	N	N	R	N	N	R	R	R	N	R	N	N	R	R	N	R

Legendre's version

Another way to organize the data is to see which primes are residues mod which other primes, as illustrated in the above table. The entry in row p column q is R if q is a quadratic residue (mod p); if it is a nonresidue the entry is N.

If the row, or the column, or both, are ≡ 1 (mod 4) the entry is blue or green; if both row and column are ≡ 3 (mod 4), it is yellow or orange.

The blue and green entries are symmetric around the diagonal: The entry for row p, column q is R (resp N) if and only if the entry at row q, column p, is R (resp N).

The yellow and orange ones, on the other hand, are antisymmetric: The entry for row p, column q is R (resp N) if and only if the entry at row q, column p, is N (resp R).

Quadratic Reciprocity (Legendre's Statement). If p or q are congruent to 1 modulo 4 then: $x^2 \equiv q \pmod{p}$ is solvable if and only if $x^2 \equiv p \pmod{q}$ is solvable. If p and q are congruent to 3 modulo 4 then: $x^2 \equiv q \pmod{p}$ is solvable if and only if $x^2 \equiv p \pmod{q}$ is not.

It is a simple exercise to prove that Legendre's and Gauss's statements are equivalent – it requires no more than the first supplement and the facts about multiplying residues and nonresidues.

Connection with cyclotomy

The early proofs of quadratic reciprocity are relatively unilluminating. The situation changed when Gauss used Gauss sums to show that quadratic fields are subfields of cyclotomic fields, and implicitly deduced quadratic reciprocity from a reciprocity theorem for cyclotomic fields. His proof was cast in modern form by later algebraic number theorists. This proof served as a template for class field theory, which can be viewed as a vast generalization of quadratic reciprocity

Robert Langlands formulated the Langlands program, which gives a conjectural vast generalization of class field theory. He wrote:^[4]

I confess that, as a student unaware of the history of the subject and unaware of the connection with cyclotomy, I did not find the law or its so-called elementary proofs appealing. I suppose, although I would not have (and could not have) expressed myself in this way that I saw it as little more than a mathematical curiosity, fit more for amateurs than for the attention of the serious mathematician that I then hoped to become. It was only in Hermann Weyl's book on the algebraic theory of numbers^[5] that I appreciated it as anything more.

History and alternative statements

There are a number of ways to state the theorem. Keep in mind that Euler and Legendre did not have Gauss's congruence notation, nor did Gauss have the Legendre symbol.

In this article p and q always refer to distinct positive odd primes.

Fermat

Fermat proved^[6] (or claimed to have proved)^[7] a number of theorems about expressing a prime by a quadratic form:

$\begin{align} p=x^2+ y^2 \qquad &\Longleftrightarrow \qquad p=2 \quad \text{ or } \quad p\equiv 1 \pmod{4} \\ p=x^2+2y^2 \qquad &\Longleftrightarrow \qquad p=2 \quad \text{ or } \quad p\equiv 1, 3 \pmod{8} \\ p=x^2+3y^2 \qquad &\Longleftrightarrow \qquad p=3 \quad \text{ or } \quad p\equiv 1 \pmod{3} \\ \end{align}$

He did not state the law of quadratic reciprocity, although the cases −1, ±2, and ±3 are easy deductions from these and other of his theorems.

He also claimed to have a proof that if the prime number p ends with 7, (in base 10) and the prime number q ends in 3, and p ≡ q ≡ 3 (mod 4), then

$pq=x^2+5y^2.$

Euler conjectured, and Lagrange proved, that^[8]

$\begin{align} p &\equiv 1, 9 \pmod{20}\quad \Longrightarrow \quad p = x^2+5y^2 \\ p, q &\equiv 3, 7 \pmod{20} \quad \Longrightarrow \quad pq=x^2+5y^2 \end{align}$

Proving these and other statements of Fermat was one of the things that led mathematicians to the reciprocity theorem.

Euler

Translated into modern notation, Euler stated:^[9]

1. If q ≡ 1 (mod 4) then q is a quadratic residue (mod p) if and only if p ≡ r (mod q), where r is a quadratic residue of q.
2. If q ≡ 3 (mod 4) then q is a quadratic residue (mod p) if and only if p ≡ ±b² (mod 4q), where b is odd and not divisible by q.

This is equivalent to quadratic reciprocity.

He could not prove it, but he did prove the second supplement.^[10]

Legendre and his symbol

Fermat proved that if p is a prime number and a is an integer,

$a^p\equiv a \pmod{p}.$

Thus, if p does not divide a,

$a^{\frac{p-1}{2}} \equiv \pm 1 \pmod{p}.$

Legendre^[11] lets a and A represent positive primes ≡ 1 (mod 4) and b and B positive primes ≡ 3 (mod 4), and sets out a table of eight theorems that together are equivalent to quadratic reciprocity:

Theorem	When	it follows that
I	$b^{\frac{a-1}{2}}\equiv 1 \pmod{a}$	$a^{\frac{b-1}{2}}\equiv 1 \pmod{b}$
II	$a^{\frac{b-1}{2}}\equiv -1 \pmod{b}$	$b^{\frac{a-1}{2}}\equiv -1 \pmod{a}$
III	$a^{\frac{A-1}{2}}\equiv 1 \pmod{A}$	$A^{\frac{a-1}{2}}\equiv 1 \pmod{a}$
IV	$a^{\frac{A-1}{2}}\equiv -1 \pmod{A}$	$A^{\frac{a-1}{2}}\equiv -1 \pmod{a}$
V	$a^{\frac{b-1}{2}}\equiv 1 \pmod{b}$	$b^{\frac{a-1}{2}}\equiv 1 \pmod{a}$
VI	$b^{\frac{a-1}{2}}\equiv -1 \pmod{a}$	$a^{\frac{b-1}{2}}\equiv -1 \pmod{b}$
VII	$b^{\frac{B-1}{2}}\equiv 1 \pmod{B}$	$B^{\frac{b-1}{2}}\equiv -1 \pmod{b}$
VIII	$b^{\frac{B-1}{2}}\equiv -1 \pmod{B}$	$B^{\frac{b-1}{2}}\equiv 1 \pmod{b}$

He says that since expressions of the form

$N^{\frac{c-1}{2}}\pmod{c}, \qquad \gcd(N, c) = 1$

will come up so often he will abbreviate them as:

$\left(\frac{N}{c}\right) \equiv N^{\frac{c-1}{2}} \pmod{c} = \pm 1.$

This is now known as the Legendre symbol, and an equivalent^[12][13] definition is used today: for all integers a and all odd primes p

$\left(\frac{a}{p}\right) = \begin{cases} 0 & a \equiv 0 \pmod{p} \\ 1 & a \not\equiv 0\pmod{p} \text{ and } \exists x : a\equiv x^2\pmod{p} \\-1 &a \not\equiv 0\pmod{p} \text{ and there is no such } x. \end{cases}$

Legendre's version of quadratic reciprocity

$\left(\frac{p}{q}\right) = \begin{cases} \left(\tfrac{q}{p}\right) & p\equiv 1 \pmod{4} \quad \text{ or } \quad q \equiv 1 \pmod{4} \\ -\left(\tfrac{q}{p}\right)\ & p \equiv 3 \pmod{4} \quad \text{ and } \quad q \equiv 3 \pmod{4} \end{cases}$

He notes that these can be combined:

$\left(\frac{p}{q}\right) \left(\frac{q}{p}\right) = (-1)^{\frac{p-1}{2}\frac{q-1}{2}}.$

A number of proofs, especially those based on Gauss's Lemma,^[14] explicitly calculate this formula.

The supplementary laws using Legendre symbols

$\begin{align} \left(\frac{-1}{p}\right) &= (-1)^{\frac{p-1}{2}} = \begin{cases} 1 &p \equiv 1 \pmod{4}\\ -1 & p \equiv 3 \pmod{4}\end{cases} \\ \left(\frac{2}{p}\right) &= (-1)^{\frac{p^2-1}{8}} = \begin{cases} 1 & p \equiv 1, 7 \pmod{8}\\ -1 & p \equiv 3, 5\pmod{8}\end{cases} \end{align}$

Legendre's attempt to prove reciprocity is based on a theorem of his:

Legendre's Theorem. Let a, b and c be integers where any pair of the three are relatively prime. Moreover assume that at least one of ab, bc or ca is negative (i.e. they don't all have the same sign). If

$\begin{align} u^2 &\equiv -bc \pmod{a} \\ v^2 &\equiv -ca \pmod{b} \\ w^2 &\equiv -ab \pmod{c} \end{align}$

are solvable then the following equation has a nontrivial solution in integers:

$ax^2 + by^2 + cz^2=0.$

Example. Theorem I is handled by letting a ≡ 1 and b ≡ 3 (mod 4) be primes and assuming that $\left (\tfrac{b}{a} \right) = 1$ and, contrary the theorem, that $\left (\tfrac{a}{b} \right ) = -1.$ Then $x^2+ay^2-bz^2=0$ has a solution, and taking congruences (mod 4) leads to a contradiction.

This technique doesn't work for Theorem VIII. Let b ≡ B ≡ 3 (mod 4), and assume

$\left (\frac{B}{b} \right ) = \left (\frac{b}{B} \right ) = -1.$

Then if there is another prime p ≡ 1 (mod 4) such that

$\left (\frac{p}{b} \right ) = \left (\frac{p}{B} \right ) = -1,$

the solvability of $Bx^2+by^2-pz^2=0$ leads to a contradiction (mod 4). But Legendre was unable to prove there has to be such a prime p; he was later able to show that all that is required is:

Legendre's Lemma. If a is a prime that is congruent to 1 modulo 4 then there exists a prime b such that $\left (\tfrac{a}{b} \right ) = -1.$

but he couldn't prove that either. Hilbert symbol (below) discusses how techniques based on the existence of solutions to $ax^2+by^2+cz^2=0$ can be made to work.

Gauss

Part of Article 131 in the first edition (1801) of the Disquisitiones, listing the 8 cases of quadratic reciprocity

Gauss first proves^[15] the supplementary laws. He sets^[16] the basis for induction by proving the theorem for ±3 and ±5. Noting^[17] that it is easier to state for −3 and +5 than it is for +3 or −5, he states^[18] the general theorem in the form:

If p is a prime of the form 4n + 1 then p, but if p is of the form 4n + 3 then −p, is a quadratic residue (resp. nonresidue) of every prime, which, with a positive sign, is a residue (resp. nonresidue) of p. In the next sentence, he christens it the "fundamental theorem" (Gauss never used the word "reciprocity").

Introducing the notation a R b (resp. a N b) to mean a is a quadratic residue (resp. nonresidue) (mod b), and letting a, a′, etc. represent positive primes ≡ 1 (mod 4) and b, b′, etc. positive primes ≡ 3 (mod 4), he breaks it out into the same 8 cases as Legendre:

Case	If	Then
1)	±a R a′	±a′ R a
2)	±a N a′	±a′ N a
3)	+a R b −a N b	±b R a
4)	+a N b −a R b	±b N a
5)	±b R a	+a R b −a N b
6)	±b N a	+a N b −a R b
7)	+b R b′ −b N b′	−b′ N b +b′ R b
8)	−b N b′ +b R b′	+b′ R b −b′ N b

In the next Article he generalizes this to what are basically the rules for the Jacobi symbol (below). Letting A, A′, etc. represent any (prime or composite) positive numbers ≡ 1 (mod 4) and B, B′, etc. positive numbers ≡ 3 (mod 4):

Case	If	Then
9)	±a R A	±A R a
10)	±b R A	+A R b −A N b
11)	+a R B	±B R a
12)	−a R B	±B N a
13)	+b R B	−B N b +N R b
14)	−b R B	+B R b −B N b

All of these cases take the form "if a prime is a residue (mod a composite), then the composite is a residue or nonresidue (mod the prime), depending on the congruences (mod 4)". He proves that these follow from cases 1) - 8).

Gauss needed, and was able to prove,^[19] a lemma similar to the one Legendre needed:

Gauss's Lemma. If p is a prime congruent to 1 modulo 8 then there exists an odd prime q such that:

$q <2\sqrt p+1 \quad \text{and} \quad \left(\frac{p}{q}\right) = -1.$

The proof of quadratic reciprocity uses complete induction.

Gauss's Version in Legendre Symbols.

$\left(\frac{p}{q}\right) = \begin{cases} \left(\frac{q}{p}\right) & q \equiv 1 \pmod{4} \\ \left(\frac{-q}{p}\right) & q \equiv 3 \pmod{4} \end{cases}$

These can be combined:

Gauss's Combined Version in Legendre Symbols. Let

$q^* = (-1)^{\frac{q-1}{2}}q.$

In other words:

$|q^*|=|q| \quad \text{and} \quad q^*\equiv 1 \pmod{4}.$

Then:

$\left(\frac{p}{q}\right) = \left(\frac{q^*}{p}\right).$

A number of proofs of the theorem, especially those based on Gauss sums derive this formula.^[20] or the splitting of primes in algebraic number fields,^[21]

Other statements

Note that the statements in this section are equivalent to quadratic reciprocity: if, for example, Euler's version is assumed, the Legendre-Gauss version can be deduced from it, and vice versa.

Euler's Formulation of Quadratic Reciprocity.^[22] If $p \equiv \pm q \pmod{4a}$ then $\left(\tfrac{a}{p}\right)=\left(\tfrac{a}{q}\right).$

This can be proven using Gauss's lemma.

Quadratic Reciprocity (Gauss; Fourth Proof).^[23] Let a, b, c, ... be unequal positive odd primes, whose product is n, and let m be the number of them that are ≡ 3 (mod 4); check whether n/a is a residue of a, whether n/b is a residue of b, .... The number of nonresidues found will be even when m ≡ 0, 1 (mod 4), and it will be odd if m ≡ 2, 3 (mod 4).

Gauss's fourth proof consists of proving this theorem (by comparing two formulas for the value of Gauss sums) and then restricting it to two primes. He then gives an example: Let a = 3, b = 5, c = 7, and d = 11. Three of these, 3, 7, and 11 ≡ 3 (mod 4), so m ≡ 3 (mod 4). 5×7×11 R 3; 3×7×11 R 5; 3×5×11 R 7;  and  3×5×7 N 11, so there are an odd number of nonresidues.

Eisenstein's Formulation of Quadratic Reciprocity.^[24] Assume

$p\ne q, \quad p'\ne q', \quad p \equiv p' \pmod{4}, \quad q \equiv q' \pmod{4}.$

Then

$\left(\frac{p}{q}\right) \left(\frac{q}{p}\right) =\left(\frac{p'}{q'}\right) \left(\frac{q'}{p'}\right).$

Mordell's Formulation of Quadratic Reciprocity.^[25] Let a, b and c be integers. For every prime, p, dividing abc if the congruence

$ax^2 + by^2 + cz^2 \equiv 0 \pmod{\tfrac{4abc}{p}}$

has a nontrivial solution, then so does:

$ax^2 + by^2 + cz^2 \equiv 0 \pmod{4abc}.$

Jacobi symbol

The Jacobi symbol is a generalization of the Legendre symbol; the main difference is that the bottom number has to be positive and odd, but does not have to be prime. If it is prime, the two symbols agree. It obeys the same rules of manipulation as the Legendre symbol. In particular

$\begin{align} \left(\frac{-1}{n}\right) = (-1)^{\frac{n-1}{2}} &= \begin{cases} 1 & n \equiv 1 \pmod{4}\\ -1 & n \equiv 3 \pmod{4}\end{cases} \\ \left(\frac{2}{n}\right) = (-1)^{\frac{n^2-1}{8}} &= \begin{cases} 1 & n \equiv 1, 7 \pmod{8}\\ -1 & n \equiv 3, 5\pmod{8}\end{cases} \end{align}$

and if both numbers are positive and odd (this is sometimes called "Jacobi's reciprocity law"):

$\left(\frac{m}{n}\right) = (-1)^{\frac{(m-1)(n-1)}{4}}\left(\frac{n}{m}\right).$

However, if the Jacobi symbol is 1 but the denominator is not a prime, it does not necessarily follow that the numerator is a quadratic residue of the denominator. Gauss's cases 9) - 14) above can be expressed in terms of Jacobi symbols:

$\biggl(\frac{M}{p}\biggr) = (-1)^{\frac{(p-1)(M-1)}{4}} \biggl (\frac{p}{M}\biggr),$

and since p is prime the left hand side is a Legendre symbol, and we know whether M is a residue modulo p or not.

The formulas listed in the preceding section are true for Jacobi symbols as long as the symbols are defined. Euler's formula may be written

$\biggl(\frac{a}{m}\biggr) =\biggl(\frac{a}{m \pm 4an}\biggr), \qquad n \in \Z, m\pm4an>0.$

Example.

$\left (\frac{2}{7} \right ) = \left (\frac{2}{15} \right ) = \left (\frac{2}{23} \right ) = \left (\frac{2}{31} \right ) = \cdots = 1.$

2 is a residue modulo the primes 7, 23 and 31:

$3^2 \equiv 2 \pmod{7}, \quad 5^2 \equiv 2 \pmod{23}, \quad 8^2 \equiv 2 \pmod{31}.$

But 2 is not a quadratic residue modulo 5, so it can't be one modulo 15. This is related to the problem Legendre had: if $\left (\tfrac{a}{m} \right) = -1,$ then a is a non-residue modulo every prime in the arithmetic progression m + 4a, m + 8a, ..., if there are any primes in this series, but that wasn't proved until decades after Legendre.^[26]

Eisenstein's formula requires relative primality conditions (which are true if the numbers are prime)

Let $a, b, a', b'$ be positive odd integers such that:

$\begin{align} \gcd &(a,b) =\gcd(a',b')= 1 \\ &a \equiv a' \pmod{4} \\ &b \equiv b' \pmod{4} \end{align}$

Then

$\biggl(\frac{a}{b}\biggr) \biggl(\frac{b}{a}\biggr) =\biggl(\frac{a'}{b'}\biggr) \biggl(\frac{b'}{a'}\biggr).$

Hilbert symbol

The quadratic reciprocity law can be formulated in terms of the Hilbert symbol $(a,b)_v$ where a and b are any two nonzero rational numbers and v runs over all the non-trivial absolute values of the rationals (the archimedean one and the p-adic absolute values for primes p). The Hilbert symbol $(a,b)_v$ is 1 or −1. It is defined to be 1 if and only if the equation $ax^2+by^2=z^2$ has a solution in the completion of the rationals at v other than $x=y=z=0$. The Hilbert reciprocity law states that $(a,b)_v$, for fixed a and b and varying v, is 1 for all but finitely many v and the product of $(a,b)_v$ over all v is 1. (This formally resembles the residue theorem from complex analysis.)

The proof of Hilbert reciprocity reduces to checking a few special cases, and the non-trivial cases turn out to be equivalent to the main law and the two supplementary laws of quadratic reciprocity for the Legendre symbol. There is no kind of reciprocity in the Hilbert reciprocity law; its name simply indicates the historical source of the result in quadratic reciprocity. Unlike quadratic reciprocity, which requires sign conditions (namely positivity of the primes involved) and a special treatment of the prime 2, the Hilbert reciprocity law treats all absolute values of the rationals on an equal footing. Therefore, it is a more natural way of expressing quadratic reciprocity with a view towards generalization: the Hilbert reciprocity law extends with very few changes to all global fields and this extension can rightly be considered a generalization of quadratic reciprocity to all global fields.

Other rings

There are also quadratic reciprocity laws in rings other than the integers.

Gaussian integers

In his second monograph on quartic reciprocity^[27] Gauss stated quadratic reciprocity for the ring $\Z[\imath]$ of Gaussian integers, saying that it is a corollary of the biquadratic law in $\Z[\imath],$ but did not provide a proof of either theorem. Peter Gustav Lejeune Dirichlet^[28] showed that the law in$\Z[\imath]$ can be deduced from the law for $\Z$ without using biquadratic reciprocity.

For an odd Gaussian prime $\pi$ and a Gaussian integer $\alpha$ relatively prime to with $\pi,$ define the quadratic character for $\Z[\imath]$ by:

$\left[\frac{\alpha}{\pi}\right]_2 \equiv \alpha^\frac{\mathrm{N} \pi - 1}{2}\pmod{\pi} = \begin{cases} 1 & \exists \eta \in \Z[\imath]: \alpha \equiv \eta^2 \pmod{\pi} \\ -1 & \text{otherwise} \end{cases}$

Let $\lambda = a + b \imath, \mu = c + d \imath$ be distinct Gaussian primes where a and c are odd and b and d are even. Then^[29]

$\left [\frac{\lambda}{\mu}\right ]_2 = \left [\frac{\mu}{\lambda}\right ]_2, \qquad \left [\frac{\imath}{\lambda}\right ]_2 =(-1)^\frac{b}{2}, \qquad \left [\frac{1+\imath}{\lambda}\right ]_2 =\left(\frac{2}{a+b}\right).$

Eisenstein integers

Consider the following third root of unity:

$\omega = \frac{-1 + \sqrt{-3}}{2}=e^\frac{2\pi \imath}{3}.$

The ring of Eisenstein integers is $\Z[\omega].$^[30] For an Eisenstein prime $\pi, \mathrm{N} \pi \neq 3,$ and an Eisenstein integer $\alpha$ with $\gcd(\alpha, \pi) = 1,$ define the quadratic character for $\Z[\omega]$ by the formula

$\left[\frac{\alpha}{\pi}\right]_2 \equiv \alpha^\frac{\mathrm{N} \pi - 1}{2}\pmod{\pi} =\begin{cases} 1 &\exists \eta \in \Z[\omega]: \alpha \equiv \eta^2 \pmod{\pi} \\ -1 &\text{otherwise} \end{cases}$

Let λ = a + bω and μ = c + dω be distinct Eisenstein primes where a and c are not divisible by 3 and b and d are divisible by 3. Eisenstein proved^[31]

$\left[\frac{\lambda}{\mu}\right]_2 \left [\frac{\mu}{\lambda}\right ]_2 = (-1)^{\frac{\mathrm{N} \lambda - 1}{2}\frac{\mathrm{N} \mu-1}{2}}, \qquad \left [\frac{1-\omega}{\lambda}\right ]_2 =\left(\frac{a}{3}\right), \qquad \left [\frac{2}{\lambda}\right ]_2 =\left (\frac{2}{\mathrm{N} \lambda }\right).$

Imaginary quadratic fields

The above laws are special cases of more general laws that hold for the ring of integers in any imaginary quadratic number field. Let k be an imaginary quadratic number field with ring of integers $\mathcal{O}_k.$ For a prime ideal $\mathfrak{p} \subset \mathcal{O}_k$ with odd norm $\mathrm{N} \mathfrak{p}$ and $\alpha\in \mathcal{O}_k,$ define the quadratic character for $\mathcal{O}_k$ as

$\left[\frac{\alpha}{\mathfrak{p} }\right]_2 \equiv \alpha^{\frac{\mathrm{N} \mathfrak{p} - 1}{2}} \pmod{\mathfrak{p}} = \begin{cases} 1 &\alpha\not\in \mathfrak{p} \text{ and } \exists \eta \in \mathcal{O}_k \text{ such that } \alpha - \eta^2 \in \mathfrak{p} \\ -1 & \alpha\not\in \mathfrak{p} \text{ and there is no such } \eta \\ 0 & \alpha\in \mathfrak{p} \end{cases}$

for an arbitrary ideal $\mathfrak{a} \subset \mathcal{O}_k$ factored into prime ideals $\mathfrak{a} = \mathfrak{p}_1 \cdots \mathfrak{p}_n$ define

$\left [\frac{\alpha}{\mathfrak{a}}\right ]_2 = \left[\frac{\alpha}{\mathfrak{p}_1 }\right]_2\cdots \left[\frac{\alpha}{\mathfrak{p}_n }\right]_2,$

and for $\beta \in \mathcal{O}_k$ define

$\left [\frac{\alpha}{\beta}\right ]_2 = \left [\frac{\alpha}{\beta \mathcal{O}_k}\right ]_2.$

Let $\mathcal{O}_k = \mathbb{Z} \omega_1\oplus \mathbb{Z} \omega_2,$ i.e. $\left\{\omega_1, \omega_2\right\}$ is an integral basis for $\mathcal{O}_k.$ For $\nu \in \mathcal{O}_k$ with odd norm $\mathrm{N}\nu,$ define (ordinary) integers a, b, c, d by the equations,

$\begin{align} \nu\omega_1&=a\omega_1+b\omega_2\\ \nu\omega_2&=c\omega_1+d\omega_2 \end{align}$

and a function

$\chi(\nu) := \imath^{(b^2-a+2)c+(a^2-b+2)d+ad}.$

If m = Nμ and n = Nν are both odd, Herglotz proved^[32]

$\left [\frac{\mu}{\nu}\right ]_2 \left[\frac{\nu}{\mu}\right]_2 = (-1)^{\frac{m-1}{2}\frac{n-1}{2}} \chi(\mu)^{m\frac{n-1}{2}} \chi(\nu)^{-n\frac{m-1}{2}}.$

Also, if

$\mu \equiv\mu' \pmod{4} \quad \text{and} \quad \nu \equiv\nu' \pmod{4}$

Then^[33]

$\left [\frac{\mu}{\nu}\right ]_2 \left[\frac{\nu}{\mu}\right]_2 = \left [\frac{\mu'}{\nu'}\right ]_2 \left[\frac{\nu'}{\mu'}\right]_2.$

Polynomials over a finite field

Let F be a finite field with q = pⁿ elements, where p is an odd prime number and n is positive, and let F[x] be the ring of polynomials in one variable with coefficients in F. If $f,g \in F[x]$ and f is irreducible, monic, and has positive degree, define the quadratic character for F[x] in the usual manner:

$\left(\frac{g}{f}\right) = \begin{cases} 1 & \gcd(f,g)=1 \text{ and } \exists h,k \in F[x] \text{ such that }g-h^2 = kf \\ -1 & \gcd(f,g)=1 \text{ and } g \text{ is not a square}\pmod{f}\\ 0 & \gcd(f,g)\ne 1 \end{cases}$

If $f=f_1 \cdots f_n$ is a product of monic irreducibles let

$\left(\frac{g}{f}\right) = \left(\frac{g}{f_1}\right) \cdots \left(\frac{g}{f_n}\right).$

Dedekind proved that if $f,g \in F[x]$ are monic and have positive degrees,^[34]

$\left(\frac{g}{f}\right) \left(\frac{f}{g}\right) = (-1)^{\frac{q-1}{2}(\deg f)(\deg g)}.$

Higher powers

The attempt to generalize quadratic reciprocity for powers higher than the second was one of the main goals that led 19th century mathematicians, including Carl Friedrich Gauss, Peter Gustav Lejeune Dirichlet, Carl Gustav Jakob Jacobi, Gotthold Eisenstein, Richard Dedekind, Ernst Kummer, and David Hilbert to the study of general algebraic number fields and their rings of integers;^[35] specifically Kummer invented ideals in order to state and prove higher reciprocity laws.

The ninth in the list of 23 unsolved problems which David Hilbert proposed to the Congress of Mathematicians in 1900 asked for the "Proof of the most general reciprocity law [f]or an arbitrary number field".^[36] In 1923 Emil Artin, building upon work by Philipp Furtwängler, Teiji Takagi, Helmut Hasse and others, discovered a general theorem for which all known reciprocity laws are special cases; he proved it in 1927.^[37]

The links below provide more detailed discussions of these theorems.

See also

• Artin reciprocity
• Cubic reciprocity
• Eisenstein reciprocity
• Euler's criterion
• Proofs of quadratic reciprocity
• Quartic reciprocity
• Zolotarev's lemma

Notes

1. Gauss, DA § 4, arts 107–150
2. E.g. in his mathematical diary entry for April 8, 1796 (the date he first proved quadratic reciprocity). See facsimile page from Felix Klein's Development of Mathematics in the 19th century
3. See F. Lemmermeyer's chronology and bibliography of proofs in the external references
4. http://www.math.duke.edu/langlands/Three.pdf
5. http://www.amazon.com/Algebraic-Theory-Numbers-Hermann-Weyl/dp/0691059179
6. Lemmermeyer, pp. 2–3
7. Gauss, DA, art. 182
8. Lemmermeyer, p. 3
9. Lemmermeyer, p. 5, Ireland & Rosen, pp. 54, 61
10. Ireland & Rosen, pp. 69–70. His proof is based on what are now called Gauss sums.
11. This section is based on Lemmermeyer, pp. 6–8
12. The equivalence is Euler's criterion
13. The analogue of Legendre's original definition is used for higher-power residue symbols
14. E.g. Kronecker's proof (Lemmermeyer, ex. p. 31, 1.34) is to use Gauss's lemma to establish that

    $\left(\frac{p}{q}\right) =\sgn\prod_{i=1}^{\frac{q-1}{2}}\prod_{k=1}^{\frac{p-1}{2}}\left(\frac{k}{p}-\frac{i}{q}\right)$

    and then switch p and q.
15. Gauss, DA, arts 108–116
16. Gauss, DA, arts 117–123
17. Gauss, DA, arts 130
18. Gauss, DA, Art 131
19. Gauss, DA, arts. 125–129
20. Because the basic Gauss sum equals $\sqrt{q^*}.$
21. Because the quadratic field $\Q(\sqrt{q^*})$ is a subfield of the cyclotomic field $\Q(e^{\frac{2\pi i}{q}})$
22. Ireland & Rosen, pp 60–61.
23. Gauss, "Summierung gewisser Reihen von besonderer Art", reprinted in Untersuchumgen uber hohere Arithmetik, pp.463–495
24. Lemmermeyer, Th. 2.28, pp 63–65
25. Lemmermeyer, ex. 1.9, p. 28
26. By Peter Gustav Lejeune Dirichlet in 1837
27. Gauss, BQ § 60
28. Dirichlet's proof is in Lemmermeyer, Prop. 5.1 p.154, and Ireland & Rosen, ex. 26 p. 64
29. Lemmermeyer, Prop. 5.1, p. 154
30. See the articles on Eisenstein integer and cubic reciprocity for definitions and notations.
31. Lemmermeyer, Thm. 7.10, p. 217
32. Lemmermeyer, Thm 8.15, p.256 ff
33. Lemmermeyer Thm. 8.18, p. 260
34. Bach & Shallit, Thm. 6.7.1
35. Lemmermeyer, p. 15, and Edwards, pp.79–80 both make strong cases that the study of higher reciprocity was much more important as a motivation than Fermat's Last Theorem was
36. Lemmermeyer, p. viii
37. Lemmermeyer, p. ix ff

References

The Disquisitiones Arithmeticae has been translated (from Latin) into English and German. The German edition includes all of Gauss's papers on number theory: all the proofs of quadratic reciprocity, the determination of the sign of the Gauss sum, the investigations into biquadratic reciprocity, and unpublished notes. Footnotes referencing the Disquisitiones Arithmeticae are of the form "Gauss, DA, Art. n".

• Gauss, Carl Friedrich; Clarke, Arthur A. (translator into English) (1986), Disquisitiones Arithemeticae (Second, corrected edition), New York: Springer, ISBN 0-387-96254-9
• Gauss, Carl Friedrich; Maser, Hermann (translator into German) (1965), Untersuchungen über höhere Arithmetik (Disquisitiones Arithemeticae & other papers on number theory) (Second edition), New York: Chelsea, ISBN 0-8284-0191-8

The two monographs Gauss published on biquadratic reciprocity have consecutively numbered sections: the first contains §§ 1–23 and the second §§ 24–76. Footnotes referencing these are of the form "Gauss, BQ, § n".

• Gauss, Carl Friedrich (1828), Theoria residuorum biquadraticorum, Commentatio prima, Göttingen: Comment. Soc. regiae sci, Göttingen 6
• Gauss, Carl Friedrich (1832), Theoria residuorum biquadraticorum, Commentatio secunda, Göttingen: Comment. Soc. regiae sci, Göttingen 7

These are in Gauss's Werke, Vol II, pp. 65–92 and 93–148. German translations are in pp. 511–533 and 534–586 of Untersuchungen über höhere Arithmetik.

Every textbook on elementary number theory (and quite a few on algebraic number theory) has a proof of quadratic reciprocity. Two are especially noteworthy:

Franz Lemmermeyer's Reciprocity Laws: From Euler to Eisenstein has many proofs (some in exercises) of both quadratic and higher-power reciprocity laws and a discussion of their history. Its immense bibliography includes literature citations for 196 different published proofs for the quadratic reciprocity law.

Kenneth Ireland and Michael Rosen's A Classical Introduction to Modern Number Theory also has many proofs of quadratic reciprocity (and many exercises), and covers the cubic and biquadratic cases as well. Exercise 13.26 (p. 202) says it all

Count the number of proofs to the law of quadratic reciprocity given thus far in this book and devise another one.

• Bach, Eric; Shallit, Jeffrey (1966), Algorithmic Number Theory (Vol I: Efficient Algorithms), Cambridge: The MIT Press, ISBN 0-262-02405-5
• Edwards, Harold (1977), Fermat's Last Theorem, New York: Springer, ISBN 0-387-90230-9
• Lemmermeyer, Franz (2000), Reciprocity Laws, Springer Monographs in Mathematics, Berlin: Springer-Verlag, doi:10.1007/978-3-662-12893-0, ISBN 3-540-66957-4, MR 1761696
• Ireland, Kenneth; Rosen, Michael (1990), A Classical Introduction to Modern Number Theory (second edition), New York: Springer, ISBN 0-387-97329-X

External links

• Hazewinkel, Michiel, ed. (2001), "Quadratic reciprocity law", Encyclopedia of Mathematics, Springer, ISBN 978-1-55608-010-4
• Quadratic Reciprocity Theorem from MathWorld
• A play comparing two proofs of the quadratic reciprocity law
• A proof of this theorem at PlanetMath
• A different proof at MathPages
• F. Lemmermeyer's chronology and bibliography of proofs of the Quadratic Reciprocity Law (233 proofs)