Prove that $\mathbb{Z}_p[x]/(x^2+x+1)$ is not a field I have to prove the following:
Let $p$ be a prime and denote $\mathbb{Z}_p=\mathbb{Z}/p\mathbb{Z}$. Show that $\mathbb{Z}_p[x]/(x^2+x+1)$ is not a field if and only if $p=3$ or $p \equiv 1 \pmod 3$.
First I consider showing that $x^2+x+1$ is reducible, We may write $x^2+x+1 \equiv 0$ which implies that $x^2 \equiv -x -1$ which implies $x^2 \equiv -x-1 \equiv 2x+2 \pmod 3$. Thus, finally showing that $x^2+x+1 \equiv x^2-2x-2$. Now, $x^2-2x-2$ is reducible. Thus our original polynomial can be written as a reducible polynomial, thus we do not have a field?
Any help would be greatly appreciated.

By the quadratic formula, $x^2 + x + 1$ has roots $\frac{-1 \pm \sqrt{-3}}{2}$.
Therefore, we consider when $-3$ has a square root mod $p$. If it does, then $x^2+x+1$ is not prime and $\mathbb{F}_p [x] / (x^2 + x + 1)$ is not a field. Otherwise, $x^2+x+1$  is irreducible and $\mathbb{F}_p [x] / (x^2 + x + 1)$ is a field.
If $ p =3$ then $-3=0$ which has a square root $0$. For $p = 2$, we can check that neither $0$ nor $1$ is a root, so $x^2+x+1$ is irreducible.
Otherwise, use the law of quadratic reciprocity. Let $p \geq 5$ be a prime number. We have $\left( \frac{3}{p} \right) = \left( \frac{-1}{p} \right) \left( \frac{-3}{p} \right)$ and $\left( \frac{p}{3} \right) \left( \frac{3}{p} \right) = (-1)^{\frac{p-1}{2}}$. Rearranging, we get $\left( \frac{-3}{p} \right) = \left( \frac{p}{3} \right)\left( \frac{-1}{p} \right)(-1)^{\frac{p-1}{2}}$.
$\left( \frac{-3}{p} \right)$ is what we are trying to find. By Euler's criterion, $\left( \frac{-1}{p} \right)$ is $1$ if $p = 1 \mod 4$ and $-1$ if $p = 3 \mod 4$.
$(-1)^{\frac{p-1}{2}}$ is $1$ if $p = 1 \mod 4$ and $-1$ if $p = 3 \mod 4$. Therefore $\left( \frac{-1}{p} \right)$ and $(-1)^{\frac{p-1}{2}}$ cancel out, so $\left( \frac{p}{3} \right) = \left( \frac{-3}{p} \right)$
By periodicity of the Legendre symbol, $\left( \frac{p}{3} \right)$ is $1$ if $p = 1  \mod 3$, and $-1$ if $p = 2 \mod 3$.
