If $\gcd (a, bc) = d$ and $\gcd(a, b) = 1$, then $\gcd(a, c) = d $

Question

I have proceeded to as far as declaring these with EEA but I do not know how to show that $d \mid c$ for the conclusion. Any help to go to this direction will be greatly appreciated.

Answer 1

Note that $d \mid a$ because $\gcd(a,bc) = d$.

We need $d \mid c$ as well, as you state. We know that $d \mid bc$ so that $bc = Ad$ for some $A$.

Then there are $C,D$ such that $Ca + Db = 1$, by $\gcd(a,b) = 1$.

So $Cac + Dbc = c$ and so $Cac + DAd = c$, using $bc = Ad$.

As $d \mid a$, $d$ divides $Cac$ and $DAd$, and hence their sum as well. So $d$ divides $c$, as required.

Answer 2

Use Gauß's lemma:

If $d\mid bc$ and $\gcd(d,b)=1$, then $d\mid c$.

Answer 3

We know that $d \mid a$. However, since $\gcd(a,b) = 1$, we also know that $b$ and $d$ share no common factors (that is, $\gcd(d,b) = 1$).

Then, since we also know that $d \mid bc$, it must also be that $d \mid c$.