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! 

Hint $\rm\, \ a,b\mid m\iff ab\mid am,bm
 \!\!\overset{\rm\color{#0a0} U\!\!}\iff ab\mid \overbrace{(am,bm)}^{\large \color{#c00}{ (a,\,b)\,m}}\!\iff ab/(a,b)\mid m$
Remark $\ $ We used $\rm\color{#0a0} U$= gcd Universal Property and  the gcd distributive law $\rm\:\color{#c00}{(a,b)\,c} = (ac,bc).\ $ If above we use Bezout's Identity to replace the gcd $\rm\:(a,b)\:$ by  $\rm\:j\,a + k\,b\:$ (its linear representation) then we obtain the proof by Bezout in lhf's answer (here in divisibility language).
The above proof is more general than  Bezout-based proofs since there are rings with gcds not of linear (Bezout) form, e.g. $\,\rm \Bbb Z[x,y]\,$ the ring of polynomials in $\,\rm x,y\,$ with integer coefficients, where $\,\rm gcd(x,y) = 1\,$ but  $\rm\, x\, f + y\, g\neq 1\,$ (else evaluating at $\rm\,x,y = 0\,$ yields $\,0 = 1).\,$
The proof shows that $\rm\ a,b\mid m\iff ab/(a,b)\mid m,\ $  i.e. $\ \rm lcm(a,b) = ab/(a,b)\ $  using the universal definition of lcm. $ $ The OP is the special case $\rm\,(a,b)= 1.$
