# Natural number solutions for equations of the form $\frac{a^2}{a^2-1} \cdot \frac{b^2}{b^2-1} = \frac{c^2}{c^2-1}$

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 $$(b_1,\ldots,b_n)$$.

• "I wonder" means "I haven't put any work into this myself"? May 5, 2020 at 23:49
• What does like mean in "solutions ... like $(a,b,c)=(9,8,6)$"? May 6, 2020 at 12:29
• 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),... May 6, 2020 at 14:28
• @Dmitry do all $c$ of the form $n^2-3$ give solutions? Most of the solutions you give have $c$ of the form. May 7, 2020 at 1:45
• @Gerry yes, for $c$ in 2..1000 give all $n$ in 3..31 May 7, 2020 at 5:03

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.

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. $$a=4n(n+1)(n^2+n-1),\ b=(2n+1)(2n^2+2n-1),\ c=2(2n+1)(n^2+n-1)$$

• I've alerted the moderators to your "account multiplicity" issue. May 13, 2020 at 13:30
• Please follow the directions at mathoverflow.net/help/merging-accounts to merge accounts. May 18, 2020 at 12:48

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:

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

• 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 $p$. Jun 8, 2020 at 23:36
• 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 $(2295,2024,1518)$. Jun 9, 2020 at 3:17