Editing Quadratic reciprocity (section)

===Proofs of the supplements===
The value of the Legendre symbol of <math>-1</math> (used in the proof above) follows directly from [[Euler's criterion]]:

:<math> \left(\frac{-1}{p}\right)\equiv (-1)^{\frac{p-1}{2}} \bmod p</math>

by Euler's criterion, but both sides of this congruence are numbers of the form <math>\pm 1</math>, so they must be equal.

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

:<math>a=-\frac{2(t+1)}{(t^2+1)}.</math>

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

:<math>p-\left(1+\left(\frac{-1}{p}\right)\right)-1</math>

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