Show that if $\gcd(a,b) = 1$ and $a|n$, $b|n$ then $ab|n$
What I have is as follows:
If $\gcd(a,b) = 1$ and $a|n$ and $a|n$ we know that:
$a=mn$ and $b=sn$ were $m,s \in \mathbb{Z}$
$ab|n = (mn)(sn)|n = n(ms)|n = \frac{(n)(ms)}{n} = ms$
This is were I am stuck. Im I done here or am I missing something?
Since $gcd(a,b)=1$ we have that $ar+bs=1$ for some integers $r,s$ (Bézout's identity)
Also, $a|n\implies ak=n$ for some $k\in\mathbb{Z}$and $b|n\implies bc=n$ for some $c\in\mathbb{Z}.$
If we multiply $ar+bs=1$ by $n$, we have $arn+bsn=n$, which looks complicated until we realize we can use $ak=bc=n$ to substitute into that equation, and rearrange into something that makes a tad more sense: $$arn+bsn=n$$ $$ar(bc)+bs(ak)=n$$ $$ab(rc)+ab(sk)=n$$ $$(ab)(rc+sk)=n$$ $$(ab)C=n$$ And the last line implies that $ab|n$, where $C=rc+sk\in\mathbb{Z}$.
