I have this really weird confusion with $gcds$ and and basic theory dividing numbers and at the moment, I am stuck at this.
If $gcd(a,b) = 1$, it means the biggest number that divides them evenly is only $1$. So if $a | mb$, then the smallest $m$ must be $a$ itself because anything smaller than $a$ would imply that….
I can't think what might be the problem. I've tested it out a few a numbers that were relatively prime, but I don't know how to complete the generalization.
On
$a$ and $b$ share a prime factor
If $a|bm$ then $bm$ has all of the prime factors that $a$ has (as least as many times). Suppose that $$a = \prod_{i=1}^n p_n$$where the $p_n$ are prime, but not necessarily distinct. Then $a$ is the lowest number such that $$a = c\prod_{i=1}^n p_n \implies m \ne c\prod_{i=1}^n p_n$$So $b$ must have at least one of the prime factors $p_n$. And if it does, then $\gcd(a,b) \ge p_k$ for some $k$.
Hint If gcd$(a,b)=1$ then there exists some integers $x,y$ so that $$ax+by=1$$
Since you know that $a|bm$ and you need to get $a|m$, you have to somehow get somewhere in the above equality both $bm$ and $m$.
It should be obvious how to do that.