# Prove that $mn|a$ implies $m|a$ and $n|a$

I am trying to prove this statement about divisibility: $mn|a$ implies $m|a$ and $n|a$.

I cannot start the proof. I need to prove either the right or left side. I don't know how to use divisibility theorems here. Generally, I have problems in proving mathematical statements.

This is my attempt: $m$ divides $a$ implies that $mn$ also divides $a$. How do I show that $n$ also divides $a$?

-
Hint: $\: a = (mn)k\, =\, m(nk),\:$ i.e. it follows by associativity of multiplication. – Bill Dubuque Nov 4 '12 at 16:46
great, thanks Bill! – doniyor Nov 4 '12 at 18:29
And, by commutativity, $\: a = (mn)k\, =\, (nm)k\, =\, n(mk),\:\Rightarrow\:n\mid a\ \$ – Bill Dubuque Nov 4 '12 at 20:19

If $mn|a$ then $a=kmn$ for some integer $k$. Then $a=(km)n$ where $km$ is an integer so that $n|a$. Similarly, $a=(kn)m$ where $kn$ is an integer so that $m|a$.