Difference between revisions of "Quadratic reciprocity"

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This theorem can help us evaluate Legendre symbols, since the following laws also apply:
 
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>.
+
* 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>.
 
* <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.)
 
There also exist quadratic reciprocity laws in other [[ring of integers|rings of integers]]. (I'll put that here later if I remember.)

Revision as of 17:15, 12 July 2006

Let $p$ be a prime, and let $a$ be any integer not divisible by $p$. Then we can define the Legendre symbol $\left(\frac{a}{p}\right)={1a is a quadratic residue modulo p,1otherwise.$ (Error compiling LaTeX. Unknown error_msg) 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. The 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.)