A quadratic residue question

Question

Let $p>3$ be a prime number. Is it true that $$ \left(\frac{-3}{p}\right)= \begin{cases} 1 & p\equiv1(\bmod{\,3})\\ -1 & p\equiv-1(\bmod{\,3})\\ \end{cases}\quad? $$ Thanks!

Answer 1

Using quadratic reciprocity $$ \begin{align} \left(\frac{-3}{p}\right)&=\left(\frac{-1}{p}\right)\left(\frac{3}{p}\right) \\ &=\left(\frac{-1}{p}\right)\left(\frac{-1}{p}\right)\left(\frac{p}{3}\right) \\ &= \left(\frac{p}{3}\right) \end{align} $$ And it's easy to check that squares are $\equiv 1\bmod{3}$.

Answer 2

If $p\equiv 1\pmod{3}$, by Cauchy's theorem there is an order-$3$ element of $\mathbb{Z}/(p\mathbb{Z})^*$.
If we call such element $\omega$, we have $\omega^3\equiv 1\pmod{p}$ and $\omega\not\equiv 1\pmod{p}$, hence $$ \omega^2+\omega+1 \equiv 0\pmod{p} \tag{1}$$ $$ 4\omega^2+4\omega+1 \equiv -3\pmod{p} \tag{2}$$ $$ (2\omega+1)^2 \equiv -3\pmod{p}\tag{3} $$ and $-3$ is a quadratic residue. Conversely, if $-3$ is a quadratic residue for some prime $p>3$, there must be an order-$3$ element of $\mathbb{Z}/(p\mathbb{Z})^*$, hence $3\mid(p-1)$ by Lagrange's theorem and $p\equiv 1\pmod{3}$.