Quadratic Reciprocity

Question

For which prime numbers p does the congruence $x^2+x+1\equiv0$ mod p have solutions?

I am new to the topic of quadratic reciprocity and I know how to answer this question had it been for which prime numbers p does the congruence $x^2\equiv-6$ mod p have solutions?

Can I perhaps split the congruence into two parts, solve them individually and then combine solutions?

Thanks in advance.

Answer 1

Hint. For $p\ne2$ we have $$x^2+x+1\equiv0\pmod p\quad\Leftrightarrow\quad (2x+1)^2\equiv-3\pmod p\ .$$

Answer 2

If $p\neq 2$ we have equivalence with $4(x^2+x+1)\equiv 0 \pmod p$. We ca now complete the square $(2x+1)^2\equiv {-3} \pmod p$ this means $-3$ is a quadratic residue. For $p=2$ we have $\forall x, x^2+x+1\equiv 1\pmod 2$