If gcd$(a,c) = 1$ and $b\mid c$, prove that $\gcd(a,b)=1$.
I just need someone to check my proof. My proof is as follows:
Proof: Assume $\gcd(a,c)=1$ and $b\mid c$. Then we know that $c = bk$ and by theorem $ax+cy=1$; $k,x,y \in \mathbb{Z}$. By substitution we obtain,
$ax+bky=1$
$ax+bp=1$, $p=ky$ is an integer.
Thus, $\gcd(a,b)=1$.
Yes it is correct, the theorem which you are referring to is Bézout's identity which guarantees that if $\gcd(a,b)=d$ then exist $x,y\in \mathbb{Z}$ such that
$$ax+by=d$$
and also that if we can find $x,y\in \mathbb{Z}$ such that $$ax+by=1$$
then $\gcd(a,b)=1$.