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 $d=1$?
If $p$ is a prime such that $p\mid (a^2+b^2)$ and $p\mid a^2b^2$, then either $p\mid a$ or $p\mid b$. Assume that $p\mid a$, then $p\mid a^2$. But then $p\mid a^2+b^2$ implies that $p\mid b^2$, which means that $p\mid b$. This would imply that $(a,b) \neq 1$ - a contradiction.
We have $d|a^2(a^2+b^2)-a^2b^2$, $d|a^4$ and by symmetry, $d|b^4$. So $d|gcd(a^4,b^4)=gcd(a,b)^4=1$
