I’m in the beginning stages of an intro to proofs class, so please bear with me. I’ll write what I have thus far.
Proof:
Let us suppose a|b. Then there exists an integer b such that b = ak. For a|bc, there must be an integer m such that bc=am.
Thus we have:
bc = am = (ak)m
What I’m unclear with is that, if I solve for m, I’ll end up wit bc/ak, which doesn’t exactly help my case. What am I doing wrong?
You're right up until the final sentence (on the premise you meant to reference $k$ in the second sentence). A priori, there is no reason for such an integer $m$ to exist.
Instead, what you might find fruitful as an angle is that, by multiplying both sides by $c$,
$$a|b \implies b = ak \implies bc = akc \implies a|bc$$
Why? Since the product of integers are integers, which $a,b,c,k$ are assumed to be, $kc \in \Bbb Z$. Therefore, there exists an integer $m := kc$ such that $bc=am$. (Bear in mind the definition for divisibility, $x|y \implies x = yz$ for some integer $z$.)