If $c\mid ab$ and $\gcd(c,a)=d$, then $c\mid db$ I came across this problem in my number theory text and am having a bit of trouble with it:
Prove if $c\mid ab$ and $\gcd(c,a)=d$, then $c\mid db$.
Here's what I have so far:
If $c\mid ab$, then there exists an integer $x$ such that $cx=ab$.
Because $\gcd(c,a)=d$, $d\mid c$ and $d\mid a$. Let $y$ be such that $dy=a$. 
Then,
$$cx=ab=dyb,$$
so $$cx \frac{d}{a} = db.$$
But I don't know how to show that $\frac{dx}{a}$ is an integer. Can anyone please offer ideas on what I've done so far?

There are no mistakes in what you've done so far, but I agree that it's not immediately apparent why $dx/a$ is an integer.
I would start by noting that the answer to this problem is well known when $d = 1$, so maybe I should try to turn it into a problem about relatively prime integers. So write $a = da'$ and $c = dc'$, and consider what the gcd of $a'$ and $c'$ is. Then see if you can carry on from there by reformulating the problem using $a'$ and $c'$ rather than $a$ and $c$.
Just a note: It is possible to do this problem without appealing to Bezout's Theorem. In other words, for those familiar with rings, the fact is valid in factorial rings, not just principal ones.
