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So I am trying to make some way on this proof,but I cannot see how any of these relate. Here is the full question: Let $a,b,m$ be integers such that $a\mid m$, $b\mid m$ and $\gcd(a,b)=1$, then $ab\mid m$.
Listing what I know I can see that by definition
$m = ap$ where $p$ is some integer,
$m= bq$ where $q$ is some integer,
$\gcd(a,b)=1=ax+by$ where $x,y$ are some integers and we are looking to prove
$ m = ab(r) $ where $r$ is some integer.
What I thought about doing is either setting $ap$ = $bp$ and then doing something with that for noticing the fact that $m= m\cdot 1$ so by replacement you can do $m= m(ax+by)$ and then
$$m(ax+by) = abr.$$
But I feel like this is leading nowhere. Am I on the right track? I can't seem to find any relevant propositions.