TITLE: Prove: If $a\mid m$ and $b\mid m$ and $\gcd(a,b)=1$ then $ab\mid m\,$ [LCM = product for coprimes]
QUESTION [5 upvotes]: 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!

REPLY [6 votes]: 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 employ 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 (but using divisibility language).
This proof is more general than the Bezout proof 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 $\rm\,(a,b)= 1.$