1. Prime GCD

Suppose that $gcd(m,n)=p$, p prime, find $gcd(m^2,n^2)$.

I'm guessing that $p^2$ is the answer to this thing, and surely it divides both $m^2$ and $n^2$. But I'm having trouble afterward, thanks!

2. You can first look at prime divisors. If $q$ is a prime number dividing both $m^2$ and $n^2$, then $q$ divides both $m$ and $n$ (because of primality), hence $q$ divides $p$ and $q=p$ (again because of primality of $p$).

So $\gcd(m^2,n^2)$ is a power of $p$. It is at least $p^2$, as you said, so it suffices to rule out the case $\gcd(m^2,n^2)=p^3$. However, in the prime decomposition of $m^2$, the exponents are even (twice what they are for $m$), so that $p^3|m^2$ implies $p^2|m$ (you could also simply say that $p|\left({m\over p}\right)^2$ and $p$ is prime, so that $p|{m\over p}$, which means $p^2|m$). And the same holds with $n$, whereas $p^2$ does not divide both $m$ and $n$ (it is greater than their gcd). This concludes that $p^3$ can't be the gcd, hence it is $p^2$.

Laurent.