Calculate $\gcd(a^2b^2, a^2 + ab + b^2)$ Given $\gcd(a, b) = 1$, calculate $d =\gcd(a^2b^2, a^2 + ab + b^2)$ in terms of $a$ and $b$.
I have tried some manipulations of the terms arriving to some expressions such as that $d$ divides $a^4 + b^4$ or that $d$ divide s $(a+b)^4$ but those haven't given me much help. I have also tried dividing the problem in cases: $d$ divides $a$ but not $b$, $d$ divides $a$ and $b$, $d$ doesn't divide neither $a$ or $b$, etc
The first statements come from:
$d\vert a^2(a^2+ab+b^2) - (a^2b^2) = a^4 +a^3b$
$d\vert b^2(a^2+ab+b^2) - (a^2b^2) = ab^3 +b^4$
$d\vert ab(a^2+ab+b^2) - (a^2b^2) = a^3b+ab^3$
$d\vert a^4+a^3b - (a^3b+ab^3) + ab^3 +b^4= a^4 +b^4$
$d\vert a^4 +b^4= (a+b)^4-4(a^3b+ab^3)-6a^2b^2 \to d\vert (a+b)^4$

It is easier in such problems to consider first the case when $d=p$ is a prime number (instead of an arbitary postive integer).
Indeed, suppose that $p\mid\gcd(a^2b^2,a^2+ab+b^2)$. Then, $p\mid a^2b^2$ and $p\mid a^2+ab+b^2$. Can you now find out for which prime $p$ it is possible given that $\gcd(a,b)=1$?
