$x^2+3$ has two zeros over ${\Bbb F}_p$ provided that $x^2+x+1\in{\Bbb F}_p[x]$ has two? The following is an exercise in abstract algebra:

If $p=1\pmod{3}$, then $x^2+x+1\in\Bbb{F}_p[x]$ has two zeros. Prove in this case that $-3$ is a quadratic residue mod $p$. 


Showing that $(x^2+x+1)\mid (x^p-x)$, one can get the first part done. But I don't see how the second part, namely, $x^2+3\in\Bbb{F}_p[x]$ has two zeros, can be deduced from the first one. Could one come up with some hints of how to get the first part? (I was thinking one might relate $x^2+3$ to $x^2+x+1$ to make an argument about the zeros of $x^2+3$. But I don't see how I can go on.)  

Complete the square. We have $x^2+x+1\equiv 0\pmod{p}$ iff $4x^2+4x+4\equiv 0\pmod{p}$ iff $(2x+1)^2\equiv -3\pmod{p}$.
Remark: Let $F$ be any field that has characteristic $\ne 2$. Then we can make precisely the same argument: We have $x^2+x+1=0$ if and only if $4x^2+4x+4=0$ if and only if $(2x+1)^2=-3$. So if $x^2+x+1=0$ has a solution, then $w^2=-3$ has a solution. And if $w^2=-3$, then by solving $2x+1=w$ we obtain a solution of $x^2+x+1=0$.
