Under which conditions is $\gcd(a+bx,c)=1$ solvable and what is the solution?

Question

Let $a,b,c\in\mathbb{Z}$, $c\neq0$. When is $\gcd(a+bx,c)=1$ solvable and what is $\{x\in\mathbb{Z}\mid\gcd(a+bx,c)=1\}$? A sufficient condition appears to be $\gcd(a,b)=1$ but it is not necessary as $a=65$, $b=40$, $c=52$, and $x=1$ shows.

Answer 1

There exists $x$ with $\,\gcd(a+bx,c)=1\iff \gcd(a,b,c) = 1.\,$ The direction $(\Leftarrow)$ is in this answer, and $(\Rightarrow)$ is clear, by $\,\gcd(a,b,c)\mid a+bx,c\,$ so also $\,\gcd(a+bx,c)$