# Prove: If $a\mid m$ and $b\mid m$ and $\gcd(a,b)=1$ then $ab\mid m$

Prove: If $$a\mid m$$ and $$b\mid m$$ and $$\gcd(a,b)=1$$ then $$ab\mid m$$

I thought that $$m=ab$$ but I was given a counterexample in a comment below.

So all I really know is $$m=ax$$ and $$m=by$$ for some $$x,y \in \mathbb Z$$. Also, $$a$$ and $$b$$ are relatively prime since $$\gcd(a,b)=1$$.

One of the comments suggests to use Bézout's identity, i.e., $$aq+br=1$$ for some $$q,r\in\mathbb{Z}$$. Any more hints?

New to this divisibility/gcd stuff. Thanks in advance!

• Hint: Bézout's identity. – wj32 Sep 12 '12 at 23:22
• No, you can’t conclude that $ab=m$: consider the example $a=2,b=3$, and $m=12$. – Brian M. Scott Sep 12 '12 at 23:22