<s>[INST] 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.
 [/INST] By lil' Fermat, $a^{3k+1}\cong b^{3k+1}\bmod p\implies a\cdot
b^{3k}\cong b\cdot b^{3k}\bmod p$.  Thus $\begin{align} a\cong b\bmod p\implies a=b+pk \implies a^2+ab+b^2=(b+pk)^2+(b+pk)b+b^2=3b^2+3bpk+p^2k^2\end{align}$.  But this expression is divisible by $p$.  So $3b^2=-3bpk-p^2k^2+pl\implies p|3b^2\implies p|b\implies p|a$. 
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