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!

Write $ax+by=1$, $m=aa'$, $m=bb'$. Let $t=b'x+a'y$.

Then $abt=abb'x+baa'y=m(ax+by)=m$ and so $ab \mid m$.

Edit: Perhaps this order is more natural and less magical:

$m = m(ax+by) = max+mby = bb'ax+aa'by = ab(b'x+a'y)$.

Hint $\rm\qquad a,b\mid m\iff ab\mid am,bm \iff ab\mid \overbrace{(am,bm)}^{\large \color{#c00}{ (a,\,b)\,m}}\iff ab/(a,b)\mid m$

Remark $\ $ 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). Note the key role played by the gcd distributive law, i.e. $\rm\:\color{#c00}{(a,b)\,c} = (ac,bc).$

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 $\rm\,(a,b)= 1.$

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 $ab|m$.