Quadratic reciprocity

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

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

The prime factors ${\displaystyle p}$ dividing ${\displaystyle f(n)}$ are ${\displaystyle p=2,5}$, and every prime whose final digit is ${\displaystyle 1}$ or ${\displaystyle 9}$; no primes ending in ${\displaystyle 3}$ or ${\displaystyle 7}$ ever appear. Now, ${\displaystyle p}$ is a prime factor of some ${\displaystyle n^{2}-5}$ whenever ${\displaystyle n^{2}-5\equiv 0{\bmod {p}}}$, i.e. whenever ${\displaystyle n^{2}\equiv 5{\bmod {p}},}$ i.e. whenever 5 is a quadratic residue modulo ${\displaystyle p}$. This happens for ${\displaystyle p=2,5}$ and those primes with ${\displaystyle p\equiv 1,4{\bmod {5}},}$ and the latter numbers ${\displaystyle 1=(\pm 1)^{2}}$ and ${\displaystyle 4=(\pm 2)^{2}}$ are precisely the quadratic residues modulo ${\displaystyle 5}$. Therefore, except for ${\displaystyle p=2,5}$, we have that ${\displaystyle 5}$ is a quadratic residue modulo ${\displaystyle p}$ iff ${\displaystyle p}$ is a quadratic residue modulo ${\displaystyle 5}$.

The law of quadratic reciprocity gives a similar characterization of prime divisors of ${\displaystyle f(n)=n^{2}-q}$ for any prime q, which leads to a characterization for any integer ${\displaystyle q}$.

Let p be an odd prime. A number modulo p is a quadratic residue whenever it is congruent to a square (mod p); otherwise it is a quadratic non-residue. ("Quadratic" can be dropped if it is clear from the context.) Here we exclude zero as a special case. Then as a consequence of the fact that the multiplicative group of a finite field of order p is cyclic of order p-1, the following statements hold:

• There are an equal number of quadratic residues and non-residues; and
• The product of two quadratic residues is a residue, the product of a residue and a non-residue is a non-residue, and the product of two non-residues is a residue.

For the avoidance of doubt, these statements do not hold if the modulus is not prime. For example, there are only 3 quadratic residues (1, 4 and 9) in the multiplicative group modulo 15. Moreover, although 7 and 8 are quadratic non-residues, their product 7x8 = 11 is also a quadratic non-residue, in contrast to the prime case.

Quadratic residues are entries in the following table:

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.

Another way to organize the data is to see which primes are residues mod which other primes, as illustrated in the following 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).

The reciprocity law states that these patterns hold for all p and q.

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 ${\displaystyle x^{2}\equiv -1{\bmod {p}}}$ is solvable if and only if ${\displaystyle p}$ is congruent to 1 modulo 4.

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 ${\displaystyle x^{2}\equiv 2{\bmod {p}}}$ is solvable if and only if ${\displaystyle p}$ 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 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 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).

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.

The value of the Legendre symbol of ${\displaystyle -1}$ (used in the proof above) follows directly from Euler's criterion:

${\displaystyle \left({\frac {-1}{p}}\right)\equiv (-1)^{\frac {p-1}{2}}{\bmod {p}}}$

by Euler's criterion, but both sides of this congruence are numbers of the form ${\displaystyle \pm 1}$, so they must be equal.

Whether ${\displaystyle 2}$ is a quadratic residue can be concluded if we know the number of solutions of the equation ${\displaystyle x^{2}+y^{2}=2}$ with ${\displaystyle x,y\in \mathbb {Z} _{p},}$ which can be solved by standard methods. Namely, all its solutions where ${\displaystyle xy\neq 0,x\neq \pm y}$ can be grouped into octuplets of the form ${\displaystyle (\pm x,\pm y),(\pm y,\pm x)}$, and what is left are four solutions of the form ${\displaystyle (\pm 1,\pm 1)}$ and possibly four additional solutions where ${\displaystyle x^{2}=2,y=0}$ and ${\displaystyle x=0,y^{2}=2}$, which exist precisely if ${\displaystyle 2}$ is a quadratic residue. That is, ${\displaystyle 2}$ is a quadratic residue precisely if the number of solutions of this equation is divisible by ${\displaystyle 8}$. And this equation can be solved in just the same way here as over the rational numbers: substitute ${\displaystyle x=a+1,y=at+1}$, where we demand that ${\displaystyle a\neq 0}$ (leaving out the two solutions ${\displaystyle (1,\pm 1)}$), then the original equation transforms into

${\displaystyle a=-{\frac {2(t+1)}{(t^{2}+1)}}.}$

Here ${\displaystyle t}$ can have any value that does not make the denominator zero – for which there are ${\displaystyle 1+\left({\frac {-1}{p}}\right)}$ possibilities (i.e. ${\displaystyle 2}$ if ${\displaystyle -1}$ is a residue, ${\displaystyle 0}$ if not) – and also does not make ${\displaystyle a}$ zero, which excludes one more option, ${\displaystyle t=-1}$. Thus there are

${\displaystyle p-\left(1+\left({\frac {-1}{p}}\right)\right)-1}$

possibilities for ${\displaystyle t}$, and so together with the two excluded solutions there are overall ${\displaystyle p-\left({\frac {-1}{p}}\right)}$ solutions of the original equation. Therefore, ${\displaystyle 2}$ is a residue modulo ${\displaystyle p}$ if and only if ${\displaystyle 8}$ divides ${\displaystyle p-(-1)^{\frac {p-1}{2}}}$. This is a reformulation of the condition stated above.

Fermat proved[5] (or claimed to have proved)[6] a number of theorems about expressing a prime by a quadratic form:

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

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

${\displaystyle pq=x^{2}+5y^{2}.}$

Euler conjectured, and Lagrange proved, that[7]

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

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

Translated into modern notation, Euler stated [8] that for distinct odd primes p and q:

1. If q ≡ 1 (mod 4) then q is a quadratic residue (mod p) if and only if there exists some integer b such that p ≡ b² (mod q).
2. If q ≡ 3 (mod 4) then q is a quadratic residue (mod p) if and only if there exists some integer b which is odd and not divisible by q such that p ≡ ±b² (mod 4q).

This is equivalent to quadratic reciprocity.

He could not prove it, but he did prove the second supplement.[9]

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

${\displaystyle a^{p}\equiv a{\bmod {p}}.}$

Thus if p does not divide a, using the non-obvious fact (see for example Ireland and Rosen below) that the residues modulo p form a field and therefore in particular the multiplicative group is cyclic, hence there can be at most two solutions to a quadratic equation:

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

Legendre[10] 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	${\displaystyle b^{\frac {a-1}{2}}\equiv 1{\bmod {a}}}$	${\displaystyle a^{\frac {b-1}{2}}\equiv 1{\bmod {b}}}$
II	${\displaystyle a^{\frac {b-1}{2}}\equiv -1{\bmod {b}}}$	${\displaystyle b^{\frac {a-1}{2}}\equiv -1{\bmod {a}}}$
III	${\displaystyle a^{\frac {A-1}{2}}\equiv 1{\bmod {A}}}$	${\displaystyle A^{\frac {a-1}{2}}\equiv 1{\bmod {a}}}$
IV	${\displaystyle a^{\frac {A-1}{2}}\equiv -1{\bmod {A}}}$	${\displaystyle A^{\frac {a-1}{2}}\equiv -1{\bmod {a}}}$
V	${\displaystyle a^{\frac {b-1}{2}}\equiv 1{\bmod {b}}}$	${\displaystyle b^{\frac {a-1}{2}}\equiv 1{\bmod {a}}}$
VI	${\displaystyle b^{\frac {a-1}{2}}\equiv -1{\bmod {a}}}$	${\displaystyle a^{\frac {b-1}{2}}\equiv -1{\bmod {b}}}$
VII	${\displaystyle b^{\frac {B-1}{2}}\equiv 1{\bmod {B}}}$	${\displaystyle B^{\frac {b-1}{2}}\equiv -1{\bmod {b}}}$
VIII	${\displaystyle b^{\frac {B-1}{2}}\equiv -1{\bmod {B}}}$	${\displaystyle B^{\frac {b-1}{2}}\equiv 1{\bmod {b}}}$

He says that since expressions of the form

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

will come up so often he will abbreviate them as:

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

This is now known as the Legendre symbol, and an equivalent[11][12] definition is used today: for all integers a and all odd primes p

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

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

He notes that these can be combined:

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

A number of proofs, especially those based on Gauss's Lemma,[13] explicitly calculate this formula.

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

From these two supplements, we can obtain a third reciprocity law for the quadratic character -2 as follows:

For -2 to be a quadratic residue, either -1 or 2 are both quadratic residues, or both non-residues :${\displaystyle {\bmod {p}}}$.

So either :${\displaystyle {\frac {p-1}{2}}{\text{ or }}{\frac {p^{2}-1}{8}}}$ are both even, or they are both odd. The sum of these two expressions is

${\displaystyle {\frac {p^{2}+4p-5}{8}}}$ which is an integer. Therefore,

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

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

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

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

${\displaystyle ax^{2}+by^{2}+cz^{2}=0.}$

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

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

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

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

the solvability of ${\displaystyle Bx^{2}+by^{2}-pz^{2}=0}$ 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 p is a prime that is congruent to 1 modulo 4 then there exists an odd prime q such that ${\displaystyle \left({\tfrac {p}{q}}\right)=-1.}$

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

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

Gauss first proves[14] the supplementary laws. He sets[15] the basis for induction by proving the theorem for ±3 and ±5. Noting[16] that it is easier to state for −3 and +5 than it is for +3 or −5, he states[17] 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:

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):

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,[18] 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:

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

The proof of quadratic reciprocity uses complete induction.

Gauss's Version in Legendre Symbols.

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

These can be combined:

Gauss's Combined Version in Legendre Symbols. Let

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

In other words:

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

Then:

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

A number of proofs of the theorem, especially those based on Gauss sums derive this formula.[19] or the splitting of primes in algebraic number fields,[20]

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.[21] If ${\displaystyle p\equiv \pm q{\bmod {4a}}}$ then ${\displaystyle \left({\tfrac {a}{p}}\right)=\left({\tfrac {a}{q}}\right).}$

This can be proven using Gauss's lemma.

Quadratic Reciprocity (Gauss; Fourth Proof).[22] 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.[23] Assume

${\displaystyle p\neq q,\quad p'\neq q',\quad p\equiv p'{\bmod {4}},\quad q\equiv q'{\bmod {4}}.}$

Then

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

Mordell's Formulation of Quadratic Reciprocity.[24] Let a, b and c be integers. For every prime, p, dividing abc if the congruence

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

has a nontrivial solution, then so does:

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

Zeta function formulation

As mentioned in the article on Dedekind zeta functions, quadratic reciprocity is equivalent to the zeta function of a quadratic field being the product of the Riemann zeta function and a certain Dirichlet L-function

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

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

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

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

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:

${\displaystyle \left({\frac {M}{p}}\right)=(-1)^{\frac {(p-1)(M-1)}{4}}\left({\frac {p}{M}}\right),}$

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

${\displaystyle \left({\frac {a}{m}}\right)=\left({\frac {a}{m\pm 4an}}\right),\qquad n\in \mathbb {Z} ,m\pm 4an>0.}$

Example.

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

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

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

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 ${\displaystyle \left({\tfrac {a}{m}}\right)=-1,}$ 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.[25]

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

Let ${\displaystyle a,b,a',b'}$ be positive odd integers such that:

${\displaystyle {\begin{aligned}\gcd &(a,b)=\gcd(a',b')=1\\&a\equiv a'{\bmod {4}}\\&b\equiv b'{\bmod {4}}\end{aligned}}}$

Then

${\displaystyle \left({\frac {a}{b}}\right)\left({\frac {b}{a}}\right)=\left({\frac {a'}{b'}}\right)\left({\frac {b'}{a'}}\right).}$

The quadratic reciprocity law can be formulated in terms of the Hilbert symbol ${\displaystyle (a,b)_{v}}$ 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 ${\displaystyle (a,b)_{v}}$ is 1 or −1. It is defined to be 1 if and only if the equation ${\displaystyle ax^{2}+by^{2}=z^{2}}$ has a solution in the completion of the rationals at v other than ${\displaystyle x=y=z=0}$. The Hilbert reciprocity law states that ${\displaystyle (a,b)_{v}}$, for fixed a and b and varying v, is 1 for all but finitely many v and the product of ${\displaystyle (a,b)_{v}}$ 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.

In his second monograph on quartic reciprocity[28] Gauss stated quadratic reciprocity for the ring ${\displaystyle \mathbb {Z} [i]}$ of Gaussian integers, saying that it is a corollary of the biquadratic law in ${\displaystyle \mathbb {Z} [i],}$ but did not provide a proof of either theorem. Dirichlet[29] showed that the law in ${\displaystyle \mathbb {Z} [i]}$ can be deduced from the law for ${\displaystyle \mathbb {Z} }$ without using quartic reciprocity.

For an odd Gaussian prime ${\displaystyle \pi }$ and a Gaussian integer ${\displaystyle \alpha }$ relatively prime to ${\displaystyle \pi ,}$ define the quadratic character for ${\displaystyle \mathbb {Z} [i]}$ by:

${\displaystyle \left[{\frac {\alpha }{\pi }}\right]_{2}\equiv \alpha ^{\frac {\mathrm {N} \pi -1}{2}}{\bmod {\pi }}={\begin{cases}1&\exists \eta \in \mathbb {Z} [i]:\alpha \equiv \eta ^{2}{\bmod {\pi }}\\-1&{\text{otherwise}}\end{cases}}}$

Let ${\displaystyle \lambda =a+bi,\mu =c+di}$ be distinct Gaussian primes where a and c are odd and b and d are even. Then[30]

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

Consider the following third root of unity:

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

The ring of Eisenstein integers is ${\displaystyle \mathbb {Z} [\omega ].}$[31] For an Eisenstein prime ${\displaystyle \pi ,\mathrm {N} \pi \neq 3,}$ and an Eisenstein integer ${\displaystyle \alpha }$ with ${\displaystyle \gcd(\alpha ,\pi )=1,}$ define the quadratic character for ${\displaystyle \mathbb {Z} [\omega ]}$ by the formula

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

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[32]

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

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 ${\displaystyle {\mathcal {O}}_{k}.}$ For a prime ideal ${\displaystyle {\mathfrak {p}}\subset {\mathcal {O}}_{k}}$ with odd norm ${\displaystyle \mathrm {N} {\mathfrak {p}}}$ and ${\displaystyle \alpha \in {\mathcal {O}}_{k},}$ define the quadratic character for ${\displaystyle {\mathcal {O}}_{k}}$ as

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

for an arbitrary ideal ${\displaystyle {\mathfrak {a}}\subset {\mathcal {O}}_{k}}$ factored into prime ideals ${\displaystyle {\mathfrak {a}}={\mathfrak {p}}_{1}\cdots {\mathfrak {p}}_{n}}$ define

${\displaystyle \left[{\frac {\alpha }{\mathfrak {a}}}\right]_{2}=\left[{\frac {\alpha }{{\mathfrak {p}}_{1}}}\right]_{2}\cdots \left[{\frac {\alpha }{{\mathfrak {p}}_{n}}}\right]_{2},}$

and for ${\displaystyle \beta \in {\mathcal {O}}_{k}}$ define

${\displaystyle \left[{\frac {\alpha }{\beta }}\right]_{2}=\left[{\frac {\alpha }{\beta {\mathcal {O}}_{k}}}\right]_{2}.}$

Let ${\displaystyle {\mathcal {O}}_{k}=\mathbb {Z} \omega _{1}\oplus \mathbb {Z} \omega _{2},}$ i.e. ${\displaystyle \left\{\omega _{1},\omega _{2}\right\}}$ is an integral basis for ${\displaystyle {\mathcal {O}}_{k}.}$ For ${\displaystyle \nu \in {\mathcal {O}}_{k}}$ with odd norm ${\displaystyle \mathrm {N} \nu ,}$ define (ordinary) integers a, b, c, d by the equations,

${\displaystyle {\begin{aligned}\nu \omega _{1}&=a\omega _{1}+b\omega _{2}\\\nu \omega _{2}&=c\omega _{1}+d\omega _{2}\end{aligned}}}$

and a function

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

If m = Nμ and n = Nν are both odd, Herglotz proved[33]

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

Also, if

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

Then[34]

${\displaystyle \left[{\frac {\mu }{\nu }}\right]_{2}\left[{\frac {\nu }{\mu }}\right]_{2}=\left[{\frac {\mu '}{\nu '}}\right]_{2}\left[{\frac {\nu '}{\mu '}}\right]_{2}.}$

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 ${\displaystyle f,g\in F[x]}$ and f is irreducible, monic, and has positive degree, define the quadratic character for F[x] in the usual manner: