Let $p$ be a prime of the form $3k+2$ that divides $a^2+ab+b^2$ for some integers $a,b$. Prove that $a,b$ are both divisible by $p$. Let $p$ be a prime of the form $3k+2$ that divides $a^2+ab+b^2$ for some integers $a,b$. Prove that $a,b$ are both divisible by $p$.  
My attempt:
$p\mid a^2+ab+b^2 \implies p\mid (a-b)(a^2+ab+b^2)\implies p\mid a^3-b^3$
So, we have, $a^{3k}\equiv b^{3k}\mod p$ and by Fermat's Theorem we have, $a^{3k+1}\equiv b^{3k+1}\mod p$ as $p$ is of the form $p=3k+2$.  
I do not know what to do next. Please help. Thank you.

Use quadratic reciprocity.
$$4(a^2+ab+b^2)=(2a+b)^2+3b^2$$
so if 
$$a^2+ab+b^2\equiv 0 \pmod p$$ then 
$$(2a+b)^2\equiv -3b^2 \pmod p$$ and $-3$ is a quadratic residue so
$$\left(\frac{-3}{p}\right)=1.$$
However by reciprocity,
$$\left(\frac{-3}{p}\right)=\left(\frac{p}{-3}\right)=\left(\frac{2}{3}\right)=-1$$
