If $p\mid a^2$ then $p^2\mid a^2$. Show that it is true or false.

Question

For natural numbers $a$ and $p$ with $p$ prime, if $p$ divides $a^{2}$ then $p^{2}$ also divides $a^{2}$.

My understanding to the is if $p\mid a$, then $\gcd(a,p)$ should not be equal to $1$ and so as $\gcd(a^2,p)$ and I do not know about $p^2\mid a^2$.

Please need clarification.Thank you for your help.

Answer 1

Hint. A fundamental property of primes is: if $p\mid ab$ then $p\mid a$ or $p\mid b$ (or both).

What do you get if you apply this to the statement $p\mid a^2$?

Can you then finish the problem?

Answer 2

If you know the Fundamental Theorem of Arithmetic, this is very simple. Write $$a=\prod_{i=1}^np_i^{\alpha_i},$$ for primes $p_i$ and positive integers $\alpha_i$, so that $$a^2=\prod_{i=1}^np_i^{2\alpha_i}.$$ Since $p\mid a^2$, this implies that there exists $k$ with $p_k=p$. But then, $a^2$ has a factor of $p^{2\alpha_k}\geq p^2$, and we're done. $\blacksquare$