Quadratic reciprocity

Let $p$ be a prime, and let $a$ be any integer. Then we can define the Legendre symbol $\genfrac{(}{)}{}{}{a}{p} =\begin{cases} 1 & \text{if } a \text{ is a quadratic residue modulo } p, \\ 0 & \text{if } p \text{ divides } a, \\ -1 & \text{otherwise}.\end{cases}$

We say that $a$ is a quadratic residue modulo $p$ if there exists an integer $n$ so that $n^2\equiv a\pmod p$ .

Equivalently, we can define the function $a \mapsto \genfrac{(}{)}{}{}{a}{p}$ as the unique nonzero multiplicative homomorphism of $\mathbb{F}_p$ into $\mathbb{R}$ .

Quadratic Reciprocity Theorem

There are three parts. Let $p$ and $q$ be distinct odd primes. Then the following hold: $\begin{align*} \genfrac{(}{)}{}{}{-1}{p} &= (-1)^{(p-1)/2} , \\ \genfrac{(}{)}{}{}{2}{p} &= (-1)^{(p^2-1)/8} , \\ \genfrac{(}{)}{}{}{p}{q} \genfrac{(}{)}{}{}{q}{p} &= (-1)^{(p-1)(q-1)/4} . \end{align*}$ This theorem can help us evaluate Legendre symbols, since the following laws also apply:

If $a\equiv b\pmod{p}$ , then $\genfrac{(}{)}{}{}{a}{p} = \genfrac{(}{)}{}{}{b}{p}$ .
$\genfrac{(}{)}{}{}{ab}{p}\right) = \genfrac{(}{)}{}{}{a}{p} \genfrac{(}{)}{}{}{b}{p}$ (Error compiling LaTeX. Unknown error_msg).

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

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Quadratic reciprocity

Quadratic Reciprocity Theorem