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Consider the equation $$\frac{a^2}{a^2-1} \cdot \frac{b^2}{b^2-1} = \frac{c^2}{c^2-1}.$$ Of course, there are solutions to this like $(a,b,c) = (9,8,6)$. Is there any known approximation for the number of solutions $(a,b,c)$, when $2 \leq a,b,c \leq k$ for some $k \geq 2.$

More generally, consider the equation $$\frac{a_1^2}{a_1^2-1} \cdot \frac{a_2^2}{a_2^2-1} \cdot \ldots \cdot \frac{a_n^2}{a_n^2-1} = \frac{b_1^2}{b_1^2-1} \cdot \frac{b_2^2}{b_2^2-1}\cdot \ldots \cdot \frac{b_m^2}{b_m^2-1}$$ for some natural numbers $n,m \geq 1$. Similarly to the above question, I ask myself if there is any known approximation to the number of solutions $(a_1,\ldots,a_n,b_1,\ldots,b_m)$, with natural numbers $2 \leq a_1, \ldots, a_n, b_1, \ldots, b_m \leq k$ for some $k \geq 2$. Of course, for $n = m$, all $2n$-tuples are solutions, where $(a_1,\ldots,a_n)$ is just a permutation of $(b_1,\ldots,b_n)$.

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    $\begingroup$ "I wonder" means "I haven't put any work into this myself"? $\endgroup$ – Gerry Myerson May 5 at 23:49
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    $\begingroup$ What does like mean in "solutions ... like $(a,b,c)=(9,8,6)$"? $\endgroup$ – Gerry Myerson May 6 at 12:29
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    $\begingroup$ Let $a^2\to u+v$, $b^2\to u-v$ and $c$ as parameter. Then solving difference of squares $\Bigl(2 (u - c^2)\Bigr)^2 - \Bigl(2 v\Bigr)^2 = (2 c^2 - 1)^2 - 1$. (a,b,c)=(9,8,6),(26,15,13),(55,24,22),(99,35,33),(50,49,35),(161,48,46),(120,55,50),(244,63,61),(351,80,78),(485,99,97),... $\endgroup$ – Dmitry Ezhov May 6 at 14:28
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    $\begingroup$ @Dmitry do all $c$ of the form $n^2-3$ give solutions? Most of the solutions you give have $c$ of the form. $\endgroup$ – Gerry Myerson May 7 at 1:45
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    $\begingroup$ @Gerry yes, for $c$ in 2..1000 give all $n$ in 3..31 $\endgroup$ – Dmitry Ezhov May 7 at 5:03
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It seems worth noting that the equation in the title does have infinitely many solutions in positive integers, as for all $n$ it is satisfied by $$a={n(n^2-3)\over2},\ b=n^2-1,\ c=n^2-3.$$ The number of solutions of this form with $a\le k$ will be on the order of $\root3\of{2k}$, but Dmitry has found solutions not of this form.

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Here's another infinite family. Let $x,y$ be positive integers such that $x^2-2y^2=\pm1$ – there are infinitely many such pairs. Let $a=x^2$, $b=2y^2$, $c=xy$, then a little algebra will show that $(a,b,c)$ satisfy the equation in the title.

E.g., $x=3$, $y=2$ leads to $(9,8,6)$, and $x=7$, $y=5$ yields $(49,50,35)$, two triples already found by Dmitry, while $x=17$, $y=12$ gets us $(289,288,204)$.

This infinite family is much thinner than the one in the other answer.

[I seem to have become disconnected from the account under which I posted the other answer.]

EDIT: A third infinite family. $$a=4n(n+1)(n^2+n-1),\ b=(2n+1)(2n^2+2n-1),\ c=2(2n+1)(n^2+n-1)$$

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Above equation shown below, has solution:

$\frac{a^2}{a^2-1} \cdot \frac{b^2}{b^2-1} = \frac{c^2}{c^2-1}$

$a=9w(2p-1)(18p-7)$

$b=4w(72p^2-63p+14)$

$c=3w(72p^2-63p+14)$

Where, w=[1/(36p^2-7)]

For, $p=0$ we get:

$(a,b,c)=(9,8,6)$

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    $\begingroup$ For which values of $p$ does this lead to a solution in integers? Very few, I think. Your formula gives $b=8-28(7p-2)/(36p^2-7)$, and a linear over a quadratic can only be an integer for finitely many integers $p$. $\endgroup$ – Gerry Myerson Jun 8 at 23:36
  • $\begingroup$ There are infinitely many integer solutions with $3b=4c$. Let $n$ be odd, and $(8+3\sqrt7)^n=x+y\sqrt7$, $x,y$ integers. Then $a=3y$, $b=x$, $c=3x/4$ is a solution in integers with $3b=4x$. $n=1$ gives $(9,8,6)$; $n=3$ gives $(2295,2024,1518)$. $\endgroup$ – Gerry Myerson Jun 9 at 3:17

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