Let $a$ and $b$ be relatively prime integers. Prove $a^2$ and $b^2$ are prime as well.

Question

Prime means the greatest divisor of that number is $1$ and itself. But where do I go from here?

Answer 1

If a prime $p$ exists with $p|a^2$ and $p|b^2$ then we must have that $p|a$ and $p|b$...which is impossible by comprimality of $a$ and $b$.

Answer 2

Hint: look at the primes that divide $a$ and $a^2$.

Answer 3

Let $gcd(a^2,b^2)=d $ , and let $p$ be a prime which divides d. Then $p|a^2$ and $p|b^2$ which implies that $p|a$ and $p|b$ a contradiction since $a$ and $b$ are relatively prime. So $d=1$.