<s>[INST] On Ramanujan's curious equality for $\sqrt{2\,(1-3^{-2})(1-7^{-2})(1-11^{-2})\cdots} $ In Ramanujan's Notebooks, Vol IV, p.20, there is the rather curious relation for primes of form $4n-1$,
$$\sqrt{2\,\Big(1-\frac{1}{3^2}\Big) \Big(1-\frac{1}{7^2}\Big)\Big(1-\frac{1}{11^2}\Big)\Big(1-\frac{1}{19^2}\Big)} = \Big(1+\frac{1}{7}\Big)\Big(1+\frac{1}{11}\Big)\Big(1+\frac{1}{19}\Big)$$
Berndt asks: if this is an isolated result, or are there others? After some poking with Mathematica, it turns out that, together with $p= 2$, we can use the primes  of form $4n+1$,
$$\sqrt{2\,\Big(1-\frac{1}{2^6}\Big) \Big(1-\frac{1}{5^2}\Big)\Big(1-\frac{1}{13^2}\Big)\Big(1-\frac{1}{17^2}\Big)} = \Big(1+\frac{1}{5}\Big)\Big(1+\frac{1}{13}\Big)\Big(1+\frac{1}{17}\Big)$$
(Now why did Ramanujan miss this $4n+1$ counterpart?) More generally, given,
$$\sqrt{m\,\Big(1-\frac{1}{n^2}\Big) \Big(1-\frac{1}{a^2}\Big)\Big(1-\frac{1}{b^2}\Big)\Big(1-\frac{1}{c^2}\Big)} = \Big(1+\frac{1}{a}\Big)\Big(1+\frac{1}{b}\Big)\Big(1+\frac{1}{c}\Big)$$

Q: Let $p =a+b+c,\;q = a b + a c + b c,\;r =abc$. For the special case $m = 2$, are there infinitely many integers $1<a<b<c$ such that,
  $$n =\sqrt{\frac{2(p-q+r-1)}{p-3q+r-3}}$$
  and $n$ is an integer? (For general $m$, see T. Andrew's comment below.)

Note: A search with Mathematica reveals numerous solutions, even for prime $a,b,c$. It is highly suggestive there may be in fact parametric solutions.
 [/INST] It can be transformed to next equation.
$$\displaystyle m=\frac{n^2}{n^2-1}\frac{a+1}{a-1}\frac{b+1}{b-1}\frac{c+1}{c-1}$$
so, function m decrease monotonously ,when all values are bigger than 2. If whatever m is, this equation has only finite solutions(when all are integer), $2\leq m\leq12(n=2,a=2,b=3,c=5)$.
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