I think I need to use the Fundamental Theorem of Arithmetic but that only applies to $a > 1$ so I think I need to do it by cases.
Case 1) $a = 0$.
Case 2) $a = 1$.
Case 3) $a > 1$.
Case 4) $a < 0$.
Cases 1 and 2 are easy. I think Case 4 will follow from Case 3 but I struggle with Case 3.
Case 3) $a > 1$. Then by Fundamental Theorem of Arithmetic, $a$ is equal to a unique product of primes. Then, $a^2$ is a product of those primes squared. Then, because $p$ divides $a^2$, $a^2 = pk$ for some $k$ (an integer). So that product of primes squared equals $pk$. Now how do I argue that one of those primes must be $p$? Once we see that one of those primes is $p$ (i.e. $p$ is a prime factor of $a$), we can say that $p$ divides $a$.
So my questions are: How do I argue that one of those primes must be $p$? Can the proof be done without resorting to four cases?