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 $(b_1,\ldots,b_n)$.
 A: 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. 
A: 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. $$a=4n(n+1)(n^2+n-1),\ b=(2n+1)(2n^2+2n-1),\ c=2(2n+1)(n^2+n-1)$$
A: 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:
$(a,b,c)=(9,8,6)$
