When can a prime integer $p$ be represented as the norm of an Eisenstein integer? Consider the Eisenstein integers $\Bbb{Z}[\omega]$, where $\omega:=e^{2\pi i/3}$. Let $p$ be an odd prime. By working on the prime factorization in $Z[\omega]$ using the norm function, I get that if $$p\mid(a^2+3)$$ for some integer $a$, then $$p=N(x+y\omega)$$ for some $x,y\in\Bbb{Z}$ and $N$ is the norm of $Z[\omega]$, which is similar to the result for the Gaussian integers: if $z^2+1=0$ is solvable over ${\Bbb F}_p$, then 
$$
p=N'(x+yi)
$$ 
for some $x,y\in\Bbb{Z}$ and $N'$ is the norm of ${\Bbb Z}[i]$. We know that the converse is also true for ${\Bbb Z}[i]$: if $p=x^2+y^2$ for some $x,y\in\Bbb{Z}$, then $z^2+1=0$ is solvable in $\Bbb{F}_p$. 
Is this also true for the Eisenstein integers? Namely, if $p=N(x+y\omega)$ for some $x,y\in{\Bbb Z}$, then $z^2+3=0$ is solvable over ${\Bbb F}_p$?
 A: Yes. I think it's much easier to understand if you think of $\omega$ as being $$\frac{-1 + \sqrt{-3}}{2}.$$ If $\mathcal{O}_{\mathbb{Q}(\sqrt{d})}$ is a unique factorization domain, then the Legendre symbol being $\left(\frac{d}{p}\right) = 1$ is a standard test for $p$ being a split prime in $\mathcal{O}_{\mathbb{Q}(\sqrt{d})}$. The Legendre symbol tells you whether $d$ is a quadratic residue modulo $p$ or not.
Take $13$ for example. We have $\left(\frac{-3}{13}\right) = 1$. We check that $$13 = \left(\frac{7 - \sqrt{-3}}{2}\right)\left(\frac{7 + \sqrt{-3}}{2}\right) = (3 - \omega)(4 + \omega).$$ The squares modulo $13$ are $0, 1, 4, 9, 3, 12, 10, 10, 12, 3, 9, 4, 1$. You see for example that $6^2 = 3 \times 13 - 3$ and $7^2 = 4 \times 13 - 3$. And since this is modular, there are infinitely more solutions. Neat, huh?
A: Suppose $p=N(x+y\omega)=x^2+xy+y^2$ for some integer $x$ and $y$. Then
$$
4p=(2x-y)^2+3y^2
$$
which implies that
$$
(2x-y)^2+3y^2=0\pmod{p}
$$
To show $z^2+3=0$ is solvable in ${\Bbb F}_p$, it suffices to show that
$
y
$
is invertible in ${\Bbb F}_p$ (which implies that $y^2$ is also invertible). If $y=0\pmod{p}$, then it follows form the chosen of $x$ and $y$ that $x=0\pmod{p}$, which is impossible since we would have $p^2\mid p$. Consequently, we have
$$
(2x/y-1)^2+3=0\pmod{p}.
$$
