Solution to the Diophantine equation subject to particular conditions

Question

I am working on vector bundles on algebraic surfaces and all my work has brought me to a Diophantine equation that I need to know if it always has a solution. Said equation is:

$$\left(2+{c\over (c,d)}m\right)x+\left(4-{d\over (c,d)}m\right)y=1,$$

where

1. $c,d\geq 0$ fixed,
2. $(c,d)$ denotes the greatest common divisor of $c$ and $d$,
3. $m\in\mathbb{Z}$ is not fixed.

I know that the equation has a solution (and in this case infinitely many) if and only if

$$\left(2+{c\over (c,d)}m, 4-{d\over (c,d)}m\right)|1 \Leftrightarrow \left(2+{c\over (c,d)}m, 4-{d\over (c,d)}m\right)=1.$$

I have been doing cases for concrete values of $c$ and $d$ (if they are coprime and also if they are not) and I have always found an integer $m\in\mathbb{Z}$ such that the greatest common divisor of the resulting expression is $1$, which has made me wonder if said value $m$ always exists. Does anyone have a hint how to approach the problem?

I will thank you very much.

Answer 1

Let $a:=\frac{c}{(c,d)}$ and $b:=\frac{d}{(c,d)}$. Note that $(a,b)=1$. Let $x$ and $y$ be such that $ax-by=1$. Let $m=1-2x-4y$. Then $$(2+am)x+(4-bm)y=2x+4y+m(ax-by)=1.$$

Answer 2

Let $a=c/\gcd(c,d)$ and $b=d/(c,d)$, choose $x,y$ such that $ax-by=1$, such choice exists since $\gcd(a,b)=1$, then $m=1-2x-4y$