Editing Quadratic reciprocity (section)

===Other statements===
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.'''<ref>Ireland & Rosen, pp 60&ndash;61.</ref> If <math>p \equiv \pm q \bmod{4a}</math> then <math>\left(\tfrac{a}{p}\right)=\left(\tfrac{a}{q}\right).</math>

This can be proven using [[Gauss's lemma (number theory)|Gauss's lemma]].
 
:'''Quadratic Reciprocity (Gauss; Fourth Proof).'''<ref>Gauss, "Summierung gewisser Reihen von besonderer Art", reprinted in ''Untersuchumgen uber hohere Arithmetik'', pp.463&ndash;495</ref> 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&nbsp;R&nbsp;3;&nbsp;3×7×11&nbsp;R&nbsp;5;&nbsp;3×5×11&nbsp;R&nbsp;7;&nbsp; and&nbsp; 3×5×7&nbsp;N&nbsp;11, so there are an odd number of nonresidues.

:'''Eisenstein's Formulation of Quadratic Reciprocity.'''<ref>Lemmermeyer, Th. 2.28, pp 63&ndash;65</ref> Assume
::<math>p\ne q, \quad p'\ne q', \quad p \equiv p' \bmod{4}, \quad q \equiv q' \bmod{4}.</math>
:Then
::<math> \left(\frac{p}{q}\right) \left(\frac{q}{p}\right) =\left(\frac{p'}{q'}\right) \left(\frac{q'}{p'}\right).</math>

:'''Mordell's Formulation of Quadratic Reciprocity.'''<ref>Lemmermeyer, ex. 1.9, p. 28</ref> Let ''a'', ''b'' and ''c'' be integers. For every prime, ''p'', dividing ''abc'' if the congruence 
::<math>ax^2 + by^2 + cz^2 \equiv 0 \bmod{\tfrac{4abc}{p}}</math>
:has a nontrivial solution, then so does:
::<math>ax^2 + by^2 + cz^2 \equiv 0 \bmod{4abc}.</math>

:'''Zeta function formulation'''
:As mentioned in the article on [[Dedekind zeta function]]s, 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