If $\gcd(ab,c)=d$ and $c|ab$ then $c=d$

Question

For all positive integers $a$, $b$, $c$ and $d$, if $\gcd(ab, c) = d$ and $c | ab$, then $c = d$.

Need help proving this question, I know that $abx + cy = d$ for integers $x,y$ and that $c|ab$ can be $c=q\cdot ab$ but I'm not sure how to apply these facts or if they're even useful in this proof.

Any help to get me started would be great.

Answer 1

$c|ab$ means that $ab=q\cdot c$, not the other way around!

Therefore, you have $ab=q\cdot c$ and $ab x + c y = d$ which you can rewrite into $$qcx + cy = d\\ c(qx+y) = d$$

Can you continue from here?

Answer 2

$ab$ is a red herring. When $y$ divides $x$, then $\gcd(x,y)=y$: the greatest common divisor is the largest divisor of both numbers; since $y$ is a divisor of $x$ and of course also a divisor of itself, it is eligible as common divisor. Then it's the greatest one.

Answer 3

You don't need $Bezout's \space Theorem$ to prove this. It follows directly from the definition of $gcd$. Clearly, the greatest divisor of $c$ is $c$ itself(find how?). So, $gcd(x,c)≤c$, for any $x$. Clearly equality holds if and only if $c$ is a divisor of $x$, too. Thus, we get the solution.