Editing Quadratic reciprocity (section)

====The supplementary laws using Legendre symbols====

:<math>\begin{align}
\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{align}</math>

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 :<math>\bmod p</math>.

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

:<math>\frac{p^2+4p-5}{8}</math> which is an integer. Therefore,

:<math>\begin{align}
\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{align}</math>

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 
::<math>\begin{align}
u^2 &\equiv -bc \bmod{a} \\
v^2 &\equiv -ca \bmod{b} \\
w^2 &\equiv -ab \bmod{c} 
\end{align}</math>
:are solvable then the following equation has a nontrivial solution in integers:
::<math>ax^2 + by^2 + cz^2=0.</math>

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

:<math>\left (\frac{B}{b} \right ) = \left (\frac{b}{B} \right ) = -1.</math>

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

:<math>\left (\frac{p}{b} \right ) = \left (\frac{p}{B} \right ) = -1,</math>

the solvability of <math>Bx^2+by^2-pz^2=0</math> 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 <math>\left (\tfrac{p}{q} \right ) = -1.</math>

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