If $a$ and $b $ are two integers, and $a \mid b$, then gcd$(a^2,b^2)$ =
I think the answer is $a^2$. Is it correct?
Certainly $a^2\mid b^2$, and $n \nmid a^2$ for $n>a^2$.
If $a\mid b$, then $b = ka$ for some integer $k$.
Then we aim to find $\gcd(a^2, (ka)^2)=\gcd(a^2, k^2a^2)$. So, yes, $a^2$ is the greatest common divisor.