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Let $x$ be a nonzero element in a commutative ring, then $\exists y\neq0(xy=0)$ ($x$ is a zero divisor) iff $\exists y\neq 0(x^2y=0)$. $(\rightarrow)$ part is pretty trivial. How to prove the other way?

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    $\begingroup$ $x^2y = x(xy)$. If $xy = 0$ then you are still done. $\endgroup$ – Tobias Kildetoft Oct 16 '13 at 12:49
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Hint: $x^2y=x(xy)$. Consider the cases $xy = 0$ and $xy \ne 0$.

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