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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 Dec 2 '19 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 Dec 3 '19 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 Dec 3 '19 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 Dec 3 '19 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 Dec 4 '19 at 15:21
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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 Dec 3 '19 at 18:36
  • $\begingroup$ @Seva probably $1$ for all of them. $\endgroup$ – Will Jagy Dec 3 '19 at 18:38

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