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In some of my research, I found multiple equations of this form: $$\frac{ax+b}{cx+d}=k$$ where $a,b,c,d$ are all non-zero integers. Is there a way (that doesn't include factoring or checking within a range), if given the values of $a,b,c,d$ to determine what integer values of $x$ makes $k$ (the ratio) an integer as well? For example use $$a=-6,b=3,c=1,d=-6$$But I am more interested in a general algorithm/method.

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    $\begingroup$ This is probably unrelated but I thought I should comment that this is the formula ( in $\mathbb{C}$ of course) of the Mobius Transform. $\endgroup$
    – Ryan Shesler
    Apr 29, 2019 at 1:27
  • $\begingroup$ Suppose $0 = b + ax - dy - cxy.$ What is the locus of all points $(x,y)$ on the curve? $\endgroup$
    – Somos
    Apr 29, 2019 at 1:33
  • $\begingroup$ @Somos I don't know actually. Could you elaborate? $\endgroup$
    – DUO Labs
    Apr 29, 2019 at 14:26
  • $\begingroup$ @Somos What is the condition all the points meet? $\endgroup$
    – DUO Labs
    Apr 29, 2019 at 20:52

2 Answers 2

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You really can't avoid factoring, I think.

You can rewrite the equation as $$ (cx+d)(cy-a) = bc-ad $$ Whenever $bc-ad$ has a divisor $A$ such that $A \equiv d \bmod c$ and $(bc-ad)/A \equiv -a \bmod c$, you get a solution with $x = (A-d)/c$ and $y = ((bc-ad)/A + a)/c$.

Conversely, if you have an solution $(x,y)$, then $cx+d$ and $cy-a$ are divisors of $bc-ad$. So any solution where neither $cx+d$ nor $cy-a$ is $\pm 1$ will give you a way to factor $bc-ad$.

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Above equation shown below:

$\frac{ax+b}{cx+d}=k$ --------$(1)$

For, $(a,b,c,d)=(3,25,5,2)$ equation $(1)$ has solution shown below:

$x=[(-7m-2)/5]$

$k=[(3m-17)/5m]$

For, $m=(-17/7)$ we get:

$(x,k)=(3,2)$

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  • $\begingroup$ Are those the only solutions? $\endgroup$
    – DUO Labs
    Apr 29, 2019 at 15:36
  • $\begingroup$ Interesting, the only other solution is $(1,4)$, and both add up to $5$. $\endgroup$
    – DUO Labs
    Apr 29, 2019 at 15:38
  • $\begingroup$ And how did you get $(-17/7)$? $\endgroup$
    – DUO Labs
    Apr 29, 2019 at 19:59
  • $\begingroup$ @Quote Dave. For, variable [k= (3m-17)/(5m)], let [(3m-17)=(10m)] & we get [(x,k)=(3,2)] for m=(-17/7). $\endgroup$
    – Sam
    Apr 30, 2019 at 9:41
  • $\begingroup$ Ok now I understand, but why 3m-17=10m. Why not 3m-17=13m? $\endgroup$
    – DUO Labs
    May 8, 2019 at 20:39

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