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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.

Fernando Mauricio Rivera Vega
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    So your question is: given two coprime integers $a,b$, can we find $m$ such that $2+ma$ and $4-mb$ are coprime? – Aphelli Jun 09 '23 at 20:39
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    @Aphelli Oh! That's right, forget that $\left({c\over (c,d)}, {d\over (c,d)}\right)=1$. So yes, my problem is to find $m\in\mathbb{Z}$ such that $2+ma$ and $4-mb$ are coprime. Thank you. – Fernando Mauricio Rivera Vega Jun 09 '23 at 22:38
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    @FernandoMauricioRiveraVega Note that if $d=\gcd(2+ma,4-mb)$, then $d\mid b(2 + ma) + a(4 - mb) = 2(b + 2a)$. – John Omielan Jun 10 '23 at 21:36
  • @JohnOmielan If I'm correct, $d|2(b+2a)$ just tells me that $d$ is equal to one or must be even, right? – Fernando Mauricio Rivera Vega Jun 11 '23 at 23:28
  • @JohnOmielan Particularly since what you say must be satisfied for any $a$ and $b$, it can be deduced that $d=1$ or $d=2$. I don't know if it is useful for me to know that in addition $(a,b)=1$. – Fernando Mauricio Rivera Vega Jun 11 '23 at 23:30
  • @FernandoMauricioRiveraVega As I wrote in an answer I deleted, "Thus, with $m$ being the odd part of $b + 2a$ (i.e., $b + 2a$ with all factors of $2$ divided out), or more generally with any odd $m$ having at least one factor of each distinct, odd prime factor of $b + 2a$, then $d = 1$." This is because then $\gcd(2+ma,b+2a)=\gcd(4-mb,b+2a)=1$. In particular, this works for any $a$ and $b$, i.e., it doesn't require that $\gcd(a,b)=1$. – John Omielan Jun 12 '23 at 01:34
  • @FernandoMauricioRiveraVega Actually, I should clarify that this does require that we don't have both $a$ and $b$ being even (which $\gcd(a,b)=1$ does provide) because, if so, then both $2+ma$ and $4-mb$ would always be even and, thus, will never be coprime. – John Omielan Jun 12 '23 at 02:01
  • @FernandoMauricioRiveraVega Sorry, but there's also one more issue with what I wrote earlier. If $b$ is even and $a$ is odd, then $\gcd(4-mb,b+2a)=2$ instead. Nonetheless, we would still have $\gcd(2+ma,4-mb)=1$ since $2+ma$ is odd. – John Omielan Jun 12 '23 at 02:20

2 Answers2

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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 $m=1-2x-4y$

sb945
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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 $$(2+am)x+(4-bm)y=2x+4y+m(ax-by)=1.$$

Servaes
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