Proof by counter example: does it work?

Question

Let $n \in \mathbb{N}_{\geq 2}$. Prove that the following statements are equivalent

(1) For every $x \in \mathbb{Z}_n,$ if $x^2 \equiv 0,$ then $x \equiv 0.$

(2) There is no prime $p$ such that $p^2 | n.$

To prove (1) $\implies$ (2), I attempted $\lnot (2) \implies \lnot (1).$

So, I assumed that there does exist some prime $p$ such that $p^2|n.$ Since the statement should be true for all $n \geq 2,$ it seemed ok to me to specify a particular case: $n = 5.$

Here, $5^2|25.$ Also, $5^2 \equiv 0$ but $5 \not\equiv 0.$

Does such a counter example work to prove this? I have a lot of doubt.

Thank you

Answer 1

A correct way to word your idea would be as follows.

Suppose that $n$ is divisible by the square of the prime $p$. Then we can write $n=p^aq$ where $a\ge 2$ and $q$ is coprime to $p$.

Now consider $x=p^{a-1}q$.

$x^2$ is divisible by $n$ but $x$ is not.

Answer 2

More generally $\,d^2\mid n\,\Rightarrow\, d^2n\mid n^2\Rightarrow n\mid (n/d)^2\,\overset{(1)}\Rightarrow\,n\mid n/d\,\Rightarrow\,d \mid 1.\,$ OP is case $\,d = p\,$ prime.