$x^2+3$ 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.)  

The map $f\colon x\mapsto 2x+1$ maps roots of $x^2+x+1$ to roots of $x^2+3$. Indeed, $$f(x)^2+3=(2x+1)^2+3=4(x^2+x+1)=0$$ whenever $x^2+x+1=0$. 
On the other hand, if $2$ is invertible, then the map $g\colon x\mapsto \frac12(x-1)$ maps roots of $x^2+3$ to roots of $x^2+x+1$. Indeed, $$g(x)^2+g(x)+1=\frac14(x-1)^2+\frac12(x-1)+1=\frac14(x^2+3)=0$$ whenever $x^2+3=0$. 
Since $f$ and $g$ are inverse maps (if $2$ is invertible), they show that the sets of roots of $x^2+3$ and $x^2+x+1$ are in bijection, hence of the same cardinality.
Of course, the claim is trivially true for $\mathbb F_2$ (though it would be wrong for $\mathbb F_4$, which has two roots of $2^2+x+1$ and only one root of $x^2+3$)
