If $\gcd(a, ab) = p^3$ then $\gcd(a^2, b^2) =?$

Question

I'm trying to solve the following problem:

$\gcd(a, ab) = p^3 $, with $p$ prime, so what is $\gcd(a^2,b^2)$?

I was able to conclude that $ab = p^3$, using gcd property.

Thank you in advance.

Answer 1

As $a\mid ab$, we have $\gcd(a, ab)=a$, i.e. $a=p^3$. Let $k$ be the highest factor of $p$ dividing $b$, i.e. $b=p^kq$ where $p\not\mid q$. Then $b^2=p^{2k}q^2$ and $p\not\mid q^2$ ($2k$ is the highest factor of $p$ dividing $b^2$). Thus: $\gcd(a^2,b^2)=\gcd(p^6,p^{2k}q^2)=p^{\min(2k,6)}\in\{p^0,p^2,p^4,p^6\}$

One can prove that all four results are actually reached for $b=1,p,p^2,p^3$.