result on quadratic residues

Theorem. Let $p$ be an odd prime. Then $-3$ is a quadratic residue modulo $p$ if and only if $p\equiv 1\pmod{3}$.

Proof. Preliminary to the proof, we remark first that $-1$ is a quadratic residue modulo $p$, where $p$ is an odd prime, if and only if $p\equiv 1\pmod{4}$.

If $p\equiv 1\pmod{4}$ then

 $\left(\frac{-3}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{3}{p}\right)=% \left(\frac{p}{3}\right).$

Now if $p\equiv 3\pmod{4}$, then

 $\left(\frac{-3}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{3}{p}\right)=(-% 1)(-1)\left(\frac{p}{3}\right)=\left(\frac{p}{3}\right).$

Thus, $\displaystyle\left(\frac{-3}{p}\right)=\left(\frac{p}{3}\right)$, and $\displaystyle\left(\frac{p}{3}\right)=1$ if and only if $p\equiv 1\pmod{3}$. $\Box$

Title result on quadratic residues ResultOnQuadraticResidues 2013-03-22 16:08:09 2013-03-22 16:08:09 gilbert_51126 (14238) gilbert_51126 (14238) 25 gilbert_51126 (14238) Theorem msc 11-00