Proof on gcd divisor

Question

Proof on gcd divisor.

Having some trouble even understanding where to start for this example question:

Let $a, b, c \in\mathbb Z^+$. Prove that if $\gcd(a,b) = c$ then $c^2|ab$

So I know what a $\gcd$ is and how to calculate it but I'm at a bit of a loss how to prove that $c^2|ab$ , and can't find any similar examples to compare the process too.

Answer 1

If $\gcd(a,b)=c$, this implies that $a=CA$ and that $b=CB$ where $A$ and $B$ are coprime integers.

So $ab=c^2AB$. So $c^2|ab$.

Answer 2

It also follows if you look at their prime factorizations. Suppose that $a=p_1^{a_1}\cdots p_m^{a_m}$ and $b=p_1^{b_1}\cdots p_m^{b_m}.$ Then $c=p_1^{\min\{a_1,b_1\}}\cdots p_m^{\min\{a_m,b_m\}}.$ Now $ab=p_1^{a_1+b_1}\cdots$ and as $\min\{a_i,b_i\}\leqslant a_i,b_i$ for each $i$ the conclusion follows.