I'm stuck at proving:
Let $a, b, c \in \mathbb{Z}, a\cdot b \neq 0$. Prove: $$ a \mid m \hspace{15px}\text{and}\hspace{15px} b \mid m \implies \frac{a \cdot b}{\text{gcd}(a, b)} \mid m $$
My current attempt looks like this:
Let $t := \text{gcd}(a, b)$ donate the greatest common divisor of $a$ and $b$.
$$ \implies\exists\, a', b': a = t\cdot a' \hspace{15px}\text{and}\hspace{15px} b = t \cdot b' $$
$$ \implies \frac{a b}{\text{gcd}(a, b)} = \frac{a b}{t} = \frac{t a'b}{t} = a' b $$
I know that since b \mid m: $a'b \mid m$ iff $\text{gcd}(a', b) = 1$ and $a' \mid m$, but I'm not sure whether this is the right approach to this problem.
Currently I don't know about identities involving the least common multiple.