Prove that the ring $\mathbb{Z}_n$ contains a nonzero element $a$, such that $a^2=0$ if and only if $p^2|n$ for some prime p

Question

I'm completely lost on this one. The best I could come up with is that $pn=p^2q$ for some integer $q$, but this doesn't imply $p^2=nq$. Hints would be appreciated.

Answer 1

It is clear that it holds when $p^2|n$ for some prime $p$, because then we can consider the number $a=\frac{n}{p}$.

Conversely, suppose that such $a$ exists. Then $$n|a^2$$ Then every prime factor of $n$ is a factor of $a^2$. Now, if $n$ is not divisible by the square of any prime number, then it is a product of distinct primes. But all such distinct primes divide $a^2$ and hence $a$. So, $n|a$ and thus $a=0$ which is a contradiction to the assumption that $n$ is a product of distinct primes.

Hope it helps:)