<s>[INST] If $\gcd(a,b)=1$, then $\gcd(a+b,a^2 -ab+b^2)=1$ or $3$. Hint:  $a^2 -ab +b^2 = (a+b)^2 -3ab.$
I know we can say that there exists an $x,y$ such that $ax + by = 1$.  So in this case,
$(a+b)x + ((a+b)^2 -3ab)y =1.$
I thought setting $x = (a+b)$ and $y = -1$ would help but that gives me $3ab =1.$  Any suggestions?
 [/INST] We have $\gcd(a+b,a^2-ab+b^2)=\gcd(a+b,3ab)$.
If $p$ is a prime with $p|\gcd(a+b,3ab)$, then $p=3$ or $p|a$ or $p|b$.
If $p|a$, then $p\nmid b$, hence $p\nmid a+b$. Similarly, if $p|b$, then $b\nmid a+b$.
Thus we conclude $p=3$.
But then if $9|\gcd(a+b,3ab)$, clearly $3|a$ or $3|b$ and again $3\nmid a+b$.
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