On Ramanujan’s curious equality for √2(1−3−2)(1−7−2)(1−11−2)⋯\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 4n1,

2(1132)(1172)(11112)(11192)=(1+17)(1+111)(1+119)

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,

2(1126)(1152)(11132)(11172)=(1+15)(1+113)(1+117)

(Now why did Ramanujan miss this 4n+1 counterpart?) More generally, given,

m(11n2)(11a2)(11b2)(11c2)=(1+1a)(1+1b)(1+1c)

Q: Let p=a+b+c,q=ab+ac+bc,r=abc. For the special case m=2, are there infinitely many integers 1<a<b<c such that,
n=2(pq+r1)p3q+r3
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.

Answer

It can be transformed to next equation.

m=n2n21a+1a1b+1b1c+1c1

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), 2m12(n=2,a=2,b=3,c=5).

Attribution
Source : Link , Question Author : Tito Piezas III , Answer Author : Takahiro Waki

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