Existence of integers $a$ and $b$ such that $p = a^2 +ab+b^2$ for $p = 3 $ or $p\equiv 1 \mod 3$ **Eisenstein primes, existence of integers **
I am working in subring $$R = \{a + b\zeta : a,b \in \mathbb{Z}\}$$ of $\mathbb{C}$ where $\zeta = \frac{1 + \sqrt{-3}}{2} \in \mathbb{C}$. I want to show that there exist integers $a$ and $b$ with $p = a^2 + ab+ b^2$ where $a,b \in \mathbb{Z}$ and that $p =3$ or $p \cong 1 \mod 3$. In this problem, we have a norm defined by $$N(a+ b\zeta) = a^2 +ab +b^2.
$$ I have been stuck in this problem for a while now. I was wondering if anyone could give me a hint on how to find the existence of such integers. I have been thinking of using the norm somehow, like considering some $x \in R$ such that $N(x) = p$ and from there try to show something about $a$ and $b$, but I haven't really made any progress. Any help with this will be ggreatly appreciated!

Consider $p = a^2 +ab+b^2$.
If $p$ divides $a$, then $p$ divides $b$, and so $p^2$ divides the RHS but not the LHS.
So, $p$ does not divide either $a$ or $b$.
Then, $p$ divides $a^3-b^3=(a-b)(a^2 +ab+b^2)$, that is $a^3 \equiv b^3 \bmod p$.
If $bc \equiv 1 \bmod p$, then $(ac)^3 \equiv 1 \bmod p$.
If $ac \equiv 1 \bmod p$, then $a \equiv b \bmod p$ and so $3a^2 \equiv 0 \bmod p$. This implies $3 \equiv 0 \bmod p$ and $p=3$.
If $ac \not\equiv 1 \bmod p$, then $ac$ is an element of order $3$ in $U(p)$ and so $3$ divides $p-1$, by Lagrange's theorem of group theory.
All this shows that the conditions in the problem are necessary. But they also point the way to the existence of a solution.
Indeed, for $p=3$ we can take $a=b=1$.
For $p \equiv 1 \bmod 3$, take $g$ a primitive root mod $p$ and let $a=g^{\frac{p-1}{3}}$ and $b=1$.
