Question regardles primes and the fundamental theorem of arithmetic

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I have been reading through my book of practice proofs and came across this particular question which has stumped me.

$p$ and $q$ are primes. Prove $\forall p \in \mathbb{Z}, \forall k \in \mathbb{Z}, (p^2\mid q^2) \Rightarrow p=q$.

I assume that the proof starts with stating that $p^n$ and $q^k$ in their product of primes form but I am unsure on how division in these forms work or if I'm even correct in assuming that is the first step. Can anyone give me an explanation of the answer so that even a 16 year old can understand? Thanks!

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Hint: if $p^n|q^k$, then in particular $p|q^k$. Now use the fact that if $p$ is prime and $p|ab$, then $p|a$ or $p|b$.

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Hint $\ $ By the Rational Root Test, if $\,(p,q)=1\,$ then $\, (q/p)^2 \in\Bbb Z\,\Rightarrow\,q/p\in \Bbb Z.$