Help with a proof I came across I came across this in my textbook and was wondering how it could be proven.

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

It's near some Euclid and Extended Euclid proofs so I was wondering if it had to do with that.

A proof involving Euclidian Algorithm goes as follows: There is a theorem called Bezuts Lemma that follows from E.A. that states that if $(a,b)=1$, then $\exists x,y$ st $ax+by=1$ (take set of numbers of the form, take the smallest one, and use E.A to show both $a,b$ divide it). Then $axc+byc=c$. Let $ak=bc$ then $c=axc+aky=a(xc+ky)$
