result on quadratic residues

Theorem.Let $p$ be an odd prime. Then $-3$ is a quadratic residue modulo $p$ if and only if $p\equiv 1(mod3)$.

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 $p\equiv 1(mod4)$.

If $p\equiv 1(mod4)$ then

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

Now if $p\equiv 3(mod4)$, 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, $\left({\displaystyle \frac{-3}{p}}\right)=\left({\displaystyle \frac{p}{3}}\right)$, and $\left({\displaystyle \frac{p}{3}}\right)=1$ if and only if $p\equiv 1(mod3)$. $\mathrm{\square }$