If $c = \gcd(a, b)$ then $c^2\mid ab$

Question

I was given this question below in class today but I'm unsure on how to do it and where to start. We learnt about this in class today but it was with numbers rather than letters so it has thrown me off a bit. Can someone please give me a push in the right direction? Thank you.

Let $a, b, c \in \mathbb{Z}^+$ with $c = \gcd(a, b)$. Prove that $c^2 \; \mid \; ab$.

Answer 1

If $c=gcd(a,b)$, then in particular $c$ is a divisor of both $a$ and $b$, so $a=cd$ and $b=cd'$. Now if you multiply these last two equations, you see it!