I've tried out a few pairs of numbers for $a$,$b$ where I have combinations of prime-prime, prime-non prime, non-prime-non-prime, and I know the statement is true, but how should I go about proving this?
I know that if $\gcd(a,b)=1$ and $a|mb$, then $m=a$, $m=2a$, $m=3a$ etc. Does this then prove that $a|m$?
If the $\gcd(a,b)=1$ then there exist $x,y$ s.t $$ax+by=1$$ and so if we multiply both sides by $m$ we get $$amx+bmy=m$$
LHS is divisible by $a$ and so RHS.
Bezout's lemma is a good place to start. Since ${\rm gcd}(a,b) = 1$ there exist integers $x$ and $y$ such that $ax + by = 1$.
Because $a | mb$ there exists an integer $t$ such that $at = mb$. Multiply the first equation by $m$
$$amx + bmy = m$$
then use $at = mb$
$$amx + aty = m$$
and finally factor out $a$
$$a(mx + ty) = m$$
to discover $a|m$.
