If $p^4 | z^2$, then $p^2 | z$ for $p$ a prime and $z$ some positive integer. Problem: Suppose that $p$ is a prime, and $z$ is some positive integer. If  $p^4 | z^2$, then $p^2 | z$.
Thoughts: 
If for some positive integer $a$, that $p^4 a = z^2$, then necessarily 
$p^2 \sqrt{a} = z$, so that if my desired conclusion is correct, then $a$ would have to be a perfect square. How can I show this is true?
Edit: I know that if $gcd(p^4, a) = 1$, then necessarily $a$ is a square.

$\dfrac{z^2}{p^4} = n\in\Bbb Z\ $ so $\ x = \dfrac{z}{p^2}\Rightarrow\,x^2= n\,$ $\Rightarrow\,x\in\Bbb Z\,$ by the Rational Root Test.
