How do we know that $c|a$ if $c=\gcd(a,b)$

Question

Prove that if $c = \gcd(a, b)$ then $c^2| ab$.

Proof: If $c = \gcd(a,b)$ then $c|a$ and $c|b$, therefore $c^2|ab$.

I do not understand how $c|a$ and $c|b$ if $c = \gcd(a,b)$

Answer 1

Stating the solution in a very rigorous way, we can say :

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 $$c^2 | ab$$