Let $a, b, c \in Z$ such that $\gcd(a,c) = d$ for some integer $d$. Prove if $a\mid bc$ then $a\mid bd$.

Question

Here is what I have tried.

If $\gcd (a,c) = d$ then you can pick $x, y$ such that $d = ax + cy$

So to show $bd = la$, multiply $b$ into above to get $bd = bax + bcy$

And since $bc = ma$, $bd = bax + may$

Is this sufficient proof? I think I need to get rid of the $b$ in $bax$

Answer 1

This is a sufficient proof. You have shown $bd = a(bx+my)=al$, which is what you wanted.