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There is a curious Diphantine equation showing up in my research:

$$ \frac{1}{a^2-1}+\frac{1}{b^2-1}=\frac{1}{c^2-1}+\frac{1}{d^2-1}. $$

I am trying to find its integer solutions where $a$, $b$, $c$ and $d\ge2$. Of course there are trivial ones satisfying $\{a, b\} = \{c, d\}$, and there are nontrivial ones. My question is then: for which positive integer $m$ is there a nontrivial solution such that $m$ divides $a$, $b$, $c$ and $d$?

Result of some experiments: For $m=3$ there is an infinite family constructed from solutions to the Pell equation $2x^2-y^2=1$. Let $(x,y)$ be a solution such that both $x$ and $y$ are odd (which makes up half the solutions). Then

$$ a=3x, b=\frac{3(3x^2-1)}{2}, c=d=3y $$

is a solution with common divisor 3. Of course then for $m=1$ there are infinitely many solutions (and many more additional ones to that family). But so far I haven't found any nontrivial even solutions (i.e., for $m=2$).

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    $\begingroup$ I'd start by seeing whether one can characterize all integer solutions to $r^{-1}+s^{-1}=t^{-1}+u^{-1}$. $\endgroup$
    – Gerry Myerson
    Commented Dec 2, 2019 at 21:44
  • $\begingroup$ @GerryMyerson I don't see how this helps. All I see is that it defines a surface in $\mathbb P^3$. $\endgroup$
    – WhatsUp
    Commented Dec 3, 2019 at 0:07
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    $\begingroup$ @What maybe it doesn't help. I don't know. The idea is, if you can get a parametrization of all solutions of the equation I suggested, then you can stick that into the equation you actually want to solve, and maybe something will come of it. Or, maybe not. If no one has any better ideas, mine might be worth a try. $\endgroup$
    – Gerry Myerson
    Commented Dec 3, 2019 at 0:12
  • $\begingroup$ For $m=2$ (for instance): How about replacing $a,b,c,d$ with, say, $2x,2y,2u,2v$ to get $$ 1 + 4(x^2+y^2)+16(x^4+y^4) + \dotsb = 1 + 4(u^2+v^2)+16(u^4+v^4) + \dotsb $$ ($2$-adically), rewriting this as $$ (x^2+y^2)+4(x^4+y^4)+16(x^6+y^6) + \dotsb = (u^2+v^2)+4(u^4+v^4)+16(u^6+v^6) + \dotsb $$ and investigating the last equality modulo $2,4,8,\dotsc$? $\endgroup$
    – Seva
    Commented Dec 3, 2019 at 15:49
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    $\begingroup$ @Seva The 2-adic approach is not that easy, because I found $(x,y,u,v)=(222,10,6,14)$ satisfying your second equation mod $2^{18}$. This can then be Hensel lifted to a solution in 2-adic integers. $\endgroup$
    – Fan Zheng
    Commented Dec 4, 2019 at 15:21

1 Answer 1

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There are two families with $a = b+c+d.$ One has $c = d,$ the other $c=d+2$

there are also "sporadic" solutions. In this list, the 307 appears out of the blue.

In the solutions below, $$ a_{n+4} = 4 a_{n+2} - a_n \; , \; $$ same for $b_n$ 

    a        b        c        d
   11        3        4        4
   19        5        8        6
   41       11       15       15
   71       19       27       25
  153       41       56       56
  265       71       98       96
  307       67      169       71
  571      153      209      209
  989      265      363      361
 2131      571      780      780
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  • $\begingroup$ What is $m$ here? $\endgroup$
    – Seva
    Commented Dec 3, 2019 at 18:36
  • $\begingroup$ @Seva probably $1$ for all of them. $\endgroup$
    – Will Jagy
    Commented Dec 3, 2019 at 18:38

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