Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $p^2 \mid n \Rightarrow p^2 \mid mn$?

share|cite|improve this question
you probably meant $p\vert m$ and $p\vert n$. – tomasz Jun 17 '12 at 20:46
up vote 3 down vote accepted

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.

share|cite|improve this answer

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

$ p^2\mid p^4$

Note that also $p^4\mid mn$

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.