Difference between revisions of "Quadratic reciprocity"

Revision as of 16:58, 3 December 2007

Let $p$ be a prime, and let $a$ be any integer not divisible by $p$ . Then we can define the Legendre symbol $\left(\frac{a}{p}\right)=\begin{cases} 1 & a\mathrm{\ is\ a\ quadratic\ residue\ modulo\ } p, \\ -1 & \mathrm{otherwise}.\end{cases}$

We say that $a$ is a quadratic residue modulo $p$ if there exists an integer $n$ so that $n^2\equiv a\pmod p$ . We can then define $\left(\frac{a}{p}\right)=0$ if $a$ is divisible by $p$ .

Quadratic Reciprocity Theorem

There are three parts. Let $p$ and $q$ be distinct odd primes. Then the following hold:

$\left(\frac{-1}{p}\right)=(-1)^{(p-1)/2}$ .
$\left(\frac{2}{p}\right)=(-1)^{(p^2-1)/8}$ .
$\left(\frac{p}{q}\right)\left(\frac{q}{p}\right)=(-1)^{(p-1)/2\ (q-1)/2}$ .

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

If $a\equiv b\pmod{p}$ , then $\left(\frac{a}{p}\right)=\left(\frac{b}{p}\right)$ .
$\left(\frac{ab}{p}\right)=\left(\frac{a}{p}\right)\left(\frac{b}{p}\right)$ .

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

@@ Line 7: / Line 7: @@
 There are three parts. Let <math>p</math> and <math>q</math> be distinct [[odd integer | odd]] primes. Then the following hold:
-* <math>\left(\frac{-1}{p}\right)=(-1)^{(p-1)/4}</math>.
+* <math>\left(\frac{-1}{p}\right)=(-1)^{(p-1)/2}</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>.
+* <math>\left(\frac{p}{q}\right)\left(\frac{q}{p}\right)=(-1)^{(p-1)/2\ (q-1)/2}</math>.
 This theorem can help us evaluate Legendre symbols, since the following laws also apply:

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Difference between revisions of "Quadratic reciprocity"

Revision as of 16:58, 3 December 2007

Quadratic Reciprocity Theorem