Find $x$ such that $[x] \neq [0]$, $[x]\in\mathbb{Z}_n$, but $[x]^2=0$.

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Find $x$ such that $[x] \neq [0]$, $[x]\in\mathbb{Z}_n$, but $[x]^2=0$. (Here $[x]$ denotes the equivalence class of $x$).

My goal here is to express $x$ in terms of $p$, a prime, and $m$, a natural number where $n=p^2m$.

However, it seems to me that the only number less than $n$ that squares to $n$ would be its square root. IE. $x=p \sqrt{m}$. Yet, this is an impossibility since $\sqrt{m}$ might very well be irrational. Any help is greatly appreciated.

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Hint: what happens for $x=pm{}{}$?