Prove that if $a|bc$ and $\gcd(a,b)|c$ , then $a|c^2$

Question

Honestly, I'm just not sure where to start at all. I've just been staring at this for a while

Answer 1

Hint

Let $a/A=b/B=(a,b)=d$(say)

So, $d|c$

$a|bc\iff A|Bc\implies A|c$ as $(A,B)=1$

Answer 2

You can think on it in terms of inequalities.

Let $p$ a prime, and $p^A || a, p^B||b, p^C||c$ (the notation $p^m||N$ is a shortcut for "$p^m|N$ but $p^{m+1}\not|M$").

A basic lemma here is: if $p^x||X,p^y||Y$, we have $X|Y$ if and only if $x \leq y$.

Also, if $D=GCD(X,Y)$ and $p^d||D$, then $D=min(A,B)$

If we have $a|bc$, then $A \leq B+C$. Also, if $gcd(a,b)|c$, then $min(A,B) \leq C$.

Or $min(A+C,B+C) \leq C+C=2C$

But $A \leq A+C$ and $A \leq B+C$. Then $A \leq min(A+C,B+C) \leq 2C$. And it is equivalent to say $a|c^2$,