View source for Quadratic reciprocity

Let <math>p</math> be a [[prime number|prime]], and let <math>a</math> be any integer not divisible by <math>p</math>. Then we can define the [[Legendre symbol]] <math>\left(\frac{a}{p}\right)=\begin{cases} 1 & a\mathrm{\ is\ a\ quadratic\ residue\ modulo\ } p, \\ -1 & \mathrm{otherwise}.\end{cases}</math> We say that <math>a</math> is a '''quadratic residue''' modulo <math>p</math> if there exists an integer <math>n</math> so that <math>n^2\equiv a\pmod p</math>. We can then define <math>\left(\frac{a}{p}\right)=0</math> if <math>a</math> is divisible by <math>p</math>.

== Quadratic Reciprocity Theorem ==

There are three parts. Let <math>p</math> and <math>q</math> be distinct odd primes. The the following hold:

* <math>\left(\frac{-1}{p}\right)=(-1)^{(p-1)/4}</math>.
* <math>\left(\frac{2}{p}\right)=(-1)^{(p^2-1)/8}</math>.
* <math>\left(\frac{p}{q}\right)\left(\frac{q}{p}\right)=(-1)^{(p-1)/4\ (q-1)/4}</math>.

This theorem can help us evaluate Legendre symbols, since the following laws also apply:

* If <math>a\equiv b\pmod{p}</math>, then <math>\left(\frac{a}{p}\right)=\left(\frac{b}{p}\right)</math>.
* <math>\left(\frac{ab}{p}\right)=\left(\frac{a}{p}\right)\left(\frac{b}{p}\right)</math>.

There also exist quadratic reciprocity laws in other [[ring of integers|rings of integers]]. (I'll put that here later if I remember.)