Prove that greatest common divisor of two numbers multiplied with itself divides the product of those numbers

When we say $c \mid a$ we really mean that $\frac ac$ is an integer. Similarily, $c\mid b$ means that $\frac bc$ is an integer. What can you now say about $\frac{ab}{c^2}$? This really has nothing to do with the "$\operatorname{g}$" in the abbreviation "$\gcd$", only the "$\operatorname{cd}$" part is relevant.

It is very simple.

Since $c=\gcd(a,b)$, so we can write that there exist distinct integers $p,q$ such that $a=cp$ and $b=cq$ where $\gcd(p,q)=1$.

Hence $$ab=cp\cdot cq = c^2 \cdot pq$$

Thus we can conclude that $$c^2 | ab$$