Prove: If $a\mid m$ and $b\mid m$ and $\gcd(a,b)=1$ then $ab\mid m\,$ [LCM = product for coprimes] 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! 

If $\gcd (a,b) =1$, then $a$ and $b$ have no prime factors in common. This means if we divide $m$ by $a$, the result is still divisible by $b$. So $b | \frac{m}{a}$, thus $ab|m$.
