$a, b, c, d$ are fixed positive integers. If $(ad - bc) \mid a$ and $(ad - bc) \mid c$, show that $\gcd(an + b, cn + d)= 1$ for any $n \in \mathbb{N}$ I've tried a couple of things trying to solve this problem but I get no answer.
These are one of the few things I know about “Gcd” and division:
If $a\mid b$ and $a \mid c$, then $a \mid b \cdot x + c \cdot y$ for any $x, y \in \mathbb{Z}$.
I also know that if $a$ divides $b$ and $c$, also divides its sum and it’s subtraction.
“gcd” also represents the minor linear combination.
I’d really appreciate your help, thank you.

Suppose that $a = x(ad-bc)$ and $c = y(ad-bc)$.  Then
$$ad-bc = [x(ad-bc)] \cdot d - b[y(ad-bc)] = (dx - by) (ad - bc).$$
However, since $(ad-bc) \mid a$, and $a$ is positive, it follows that $ad - bc \ne 0$.  Thus, we must have that $dx - by = 1$, from which we can conclude that $\gcd(b,d) = 1$ (which proves the case of $n = 0$).
Now, for the general case, note that $a(cn+d) - (an+b)c = ad-bc$ so it divides both $a$ and $c$.  Therefore, by the previous paragraph with $a' := a$, $b' := an+b$, $c' := c$, $d' := cn+d$, we get $\gcd(b', d') = 1$, i.e. $\gcd(an+b, cn+d) = 1$.
