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There is a question with the elementary number theory tag that is locked because the question doesn't really make sense.

My variation is:

Let $x=ar+b$ and $y=cr+d.$ Find all values $ay-cx=ad−bc$ in terms of $m$ such that there are infinitely many values of $r$ such that $\gcd(x,y)=m,$ where $m$ is a constant.

A simpler way of saying this is that there are infinitely many values $r$ such that $\gcd(ar+b,cr+d)=m.$ The question is asking to find all values of $ad-bc.$

$m$ is any fixed integer (like 5, or 134, or 2156), while the values of $a,b,c,d$ vary based on $m.$ There are infinitely many values or $r$ that make $\gcd(ar+b,cr+d)=m$ true.

This seems like it is an interesting problem without ambiguity, I don't even know how to start, though.

Edit: I found this question that seems similar to this problem, but I am not sure to what extent. I didn't have enough time to really see how similar the problems are. I will continue to look into this and keep this updated.

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    Yea, I saw you try to make that question into this. Please don't do that in the future, asking the question like you have here is the right thing to do. – Rushabh Mehta Nov 14 '19 at 04:47
  • Ok, I am new to the site, I am sorry. –  Nov 14 '19 at 05:11
  • It's not at all clear what's given and what varies here. $a,b,c,d$ can't all be given since then there's only one value of $ad-bc$. $r$ can't be given since you want there to be infinitely many values of $r$. $x,y$ can't be given since you want $\gcd(x,y)=m$. Please clarify. – Gerry Myerson Nov 15 '19 at 05:07
  • You have edited, but it's still not clear what is fixed and what is varying. By the way, you should also edit the title so that it gives some indication of what, mathematically, the question is about. – Gerry Myerson Nov 15 '19 at 11:44
  • I'm not sure what you are asking me to clarify here. Do you want me to edit the title? –  Nov 15 '19 at 21:09
  • Yes, I want you to edit the title: "A variant on a question that is locked" gives people no idea of what kind of mathematics is under discussion here. AND I want you to indicate WHAT IS FIXED, AND WHAT IS VARYING. – Gerry Myerson Nov 16 '19 at 11:17
  • Any thoughts about the answer I posted? – Gerry Myerson Nov 17 '19 at 21:36
  • Ok, I understood how $m$ divides $ad-bc$ but how does that imply that $ad-bc$ can be all multiples of $m?$ –  Nov 17 '19 at 22:24
  • Let $a=1$, $b=c=0$, $d=km$ with $k$ arbitrary. Then $\gcd(x,y)=m$ for $r=m,(k+1)m,(2k+1)m,\dots$, and $ad-bc=km$. – Gerry Myerson Nov 18 '19 at 12:40
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    oh i see, thank you! –  Nov 19 '19 at 02:02

1 Answers1

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My attempt to make sense of the question:

Anything that divides both $x$ and $y$ divides $ay-cx$ which is $ad-bc$, so $m$ must be a divisor of $ad-bc$ if there are to be infinitely many $r$ such that etc. etc.

Moreover, if there is one value of $r$ such that $\gcd(x,y)=m$ then there are infinitely many, since $r+(ad-bc)$ also works.

Given $m$, we can take $ad-bc$ to be any multiple of $m$.

Gerry Myerson
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