Difference between revisions of "Quadratic reciprocity"

m (Quadratic Reciprocity Theorem)
m
Line 1: Line 1:
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>.
+
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 ==
 
== Quadratic Reciprocity Theorem ==

Revision as of 19:46, 18 October 2006

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:

  • $\left(\frac{-1}{p}\right)=(-1)^{(p-1)/4}$.
  • $\left(\frac{2}{p}\right)=(-1)^{(p^2-1)/8}$.
  • $\left(\frac{p}{q}\right)\left(\frac{q}{p}\right)=(-1)^{(p-1)/4\ (q-1)/4}$.

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.)

Invalid username
Login to AoPS