If $a$ and $b$ are positive integers, and $p$ is prime then $p$ divides $\gcd(a^2,b)$ if and only if p divides $\gcd(a,b^2)$

Question

I am stuck in the follow problem:

State whether the following statement is true or false:

If $a$ and $b$ are positive integers, and $p$ is prime then $p$ divides $\gcd(a^2,b)$ if and only if p divides $\gcd(a,b^2)$

My work:

When $a=2$, and $b=5$, The $\gcd(4,5) = 1$ and $\gcd(2,25) = 1$, the only "number" that divides $1$, is $1$, but that is not a prime by convention.

But my feeling says: this is not the right answer.

Answer 1

Let $p$ be a prime.

$p \mid n \iff p\mid n^2 \tag*{$\dagger$}\label{eq1}$

Proof:
$\implies$

$p \mid n \implies p \mid n^2$ is trivially true.

$\impliedby$

Consider the contrapositive i.e $p\nmid n \implies p\nmid n^2$ which is again trivially true $\blacksquare$

Using this result, we have:

$$p \mid \gcd(a^2,b) \overset{\text{def}}{\Longleftrightarrow}(p \mid a^2) \land (p \mid b) \overset{\ref{eq1}}{\Longleftrightarrow} (p\mid a) \land (p\mid b^2) \overset{\text{def}}{\Longleftrightarrow} p\mid \gcd(a,b^2)$$