Prove that if $\gcd(a,b)=1$, then $\gcd(a^2+b^2, a^2b^2)=1$.
My attempt:
If $a$ is prime to $b$ the $gcd(a,b)=1$. Assume that $a^2+b^2$ and $a^2b^2$ are not prime to each other. Let $d=gcd(a^2+b^2, a^2b^2)$. Then $d|a^2+b^2, ~~ \&~~d|a^2b^2$. We shall have to prove that $d=1$.
How to prove that $d=1$?