<s>[INST] To prove $a^2+ab+b^2 \Big|(a+b)^{2n}+a^{2n}+b^{2n}$ whenever $3$ does not divide $n$ If $3$ does not divide a positive integer $n$ , then how to prove that $a^2+ab+b^2 \Big|(a+b)^{2n}+a^{2n}+b^{2n}$ ?                
 [/INST] I will stay in the reals. Note that:
$$(a+b)^2\equiv ab \mod{a^2+ab+b^2}\implies (a+b)^{2n}\equiv (ab)^n \mod{a^2+ab+b^2}$$
So $(a+b)^n+a^{2n}+b^{2n}\equiv a^nb^n+a^{2n}+b^{2n}$. Also, we know that $a^3\equiv b^3$. So setting $n=3q+1$, we have:
$$a^nb^n+a^{2n}+b^{2n}\equiv ab^{6q+1}+a^2b^{6q}+b^{6q+2} \equiv b^{6q}(ab+a^2+b^2)\equiv 0$$
Similarly, if $n=3q+2$:
$$a^nb^n+a^{2n}+b^{2n}\equiv a^2b^{6q+2}+ab^{6q+3}+b^{6q+4} \equiv b^{6q+2}(a^2+ab+b^2)\equiv 0$$
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