# How you prove that $p^2 \mid m$ and $p^2 \mid n \Rightarrow p^2 \mid mn$?

We know that (Characterization of prime number) $p \mid ab \Rightarrow p \mid a$ or $p \mid b$, where $p$ is prime number. How you prove that $p^2 \mid m$ and $p^2 \mid n \Rightarrow p^2 \mid mn$?

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you probably meant $p\vert m$ and $p\vert n$. –  tomasz Jun 17 '12 at 20:46

You surely do not mean to ask this, since if $p^2$ divides $m$, then automatically $p^2$ divides $mn$, whether or not $p$ is prime.

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$m=ap^2$, $n=bp^2$, $mn=abp^4$,

$p^2\mid p^4$

Note that also $p^4\mid mn$

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For any integers $a$, $b$, and $c$ whatsoever, if $a\mid b$, then $a\mid bc$. This is because if $b=ka$ for some $k\in\mathbb{Z}$, then we obviously have $bc=(kc)a$.

Letting $a=p^2$ and either $b=m$, $c=n$ or $b=n$, $c=m$ produces your claim.

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