Let $p$ be a prime number such that $p \equiv 1 \mod 3$. Prove that there is an $x \in \mathbb{Z}$ such that $ x^2 \equiv -3 \mod p$. Any hints or solutions? Thanks in advance.
(I already proved that $\mathbb{Z}\backslash p \mathbb{Z}$ has an element of order $3$ and that $a^2 + a + 1 = 0$ in $(\mathbb{Z}\backslash p \mathbb{Z})^{\times}$. So maybe that is useful.)