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!
If $p=3$ then $a=b=1$.

If $p\equiv 1\pmod 3$ then
$$(-3/p)=(3/p)(-1)^{(p-1)/2}=(p/3)(-1)^{(p-1)/2}(-1)^{(p-1)/2}=(1/3)=1$$

(Here, $(a/b)$ means the Legendre symbol).

Then, $-3$ is a square mod $p$. Assume that $m^2\equiv -3\pmod p$. Let $x=(m-1)(p+1)/2$. We see that
$$x^2+x+1\equiv 0\pmod p$$
or
$$(x-\zeta)(x+\zeta)=kp$$
Note that $\Bbb Z[\zeta]$ is an UFD. If $p$ were prime in $\Bbb Z[\zeta]$ then $p$ would divide $x+\zeta$ or $x-\zeta$. This is clearly impossible, so $p$ is not prime in $\Bbb Z[\zeta]$.

Then
$$p=(a+b\zeta)(c+d\zeta)$$

Taking norms
$$p^2=(a^2+ab+b^2)(c^2+cd+d^2)$$
and both factors in RHS must be $p$.

Example: Take $p=97$. We see that $26^2+3=679=97\cdot 7$. And $25/2\equiv 61\pmod {97}$ so $61^2+61+1=97\cdot 39$. Then $97$ divide $(61+\zeta)(61-\zeta)$. Since 
$$\frac{61\pm\zeta}{97}\notin\Bbb Z[\zeta]$$
then $97$ can be factored in $\Bbb Z[\zeta]$. This method does not give the factorization, but we can try by brute force. 
$$97=8^2+3\cdot 8+3^2$$
$a^2 + ab+ b^2 \equiv \{0_{a\equiv 0}|1_{a\equiv\pm 1}\} + \{-1_{\text{if }a\equiv -b\equiv \pm 1}|0_{a\equiv 0\lor b\equiv 0}|1_{b\equiv \pm 1}\} + \{0_{b\equiv 0}|1_{b\equiv\pm 1}\} \mod 3$.

If one of $a$ or $b$ is equiv to $0$ and the other isnt then the result is $0+0 + 1$ (or $1+0+1$)$\equiv 1 \mod 3$.

If $a \equiv b$ then the result is $\{0|1\} + \{0|1\} + \{0|1\}\equiv \{0|3\}\equiv 0 \mod 3$.

And if $a \equiv -b \equiv \pm 1$ the result is $1 -1 +1\equiv 1 \mod 3$.

So the result follows precisely when $a\not \equiv b \mod 3$.

What am I missing about the question?

If $p = 3$.  possibilities are $(\pm 1,\pm 1),(\pm 2, \mp 1)$.