Question about greatest common divisor.

Question

Let $a,b\in \mathbb{N}$ and $\gcd(a,b) = 1$ Is it correct that $\gcd(a^2,b^2) = 1$ as well? And if so, how do I prove it?

Answer 1

Prove the contrapositive statement

If $\gcd(a^2, b^2)\neq 1$, then $\gcd(a, b)\neq 1$

by considering an arbitrary prime that divides $\gcd(a^2, b^2)$.

Answer 2

suppose that $p$ is a prime such that $p|\gcd(a^2,b^2)$

$p|a^2 \implies p|a$ (because $p$ is prime)

similarly $p|b^2 \implies p|b$

$p|\gcd(a,b)=1$ which is impossible.

Therefore $gcd(a^2,b^2)=1$