In

Ramanujan’s Notebooks, Vol IV, p.20, there is the rather curious relation for primes of form 4n−1,√2(1−132)(1−172)(1−1112)(1−1192)=(1+17)(1+111)(1+119)

Berndt asks:

if this is an isolated result, or are there others?After some poking withMathematica, it turns out that, together with p=2, we can use the primes of form 4n+1,√2(1−126)(1−152)(1−1132)(1−1172)=(1+15)(1+113)(1+117)

(

Now why did Ramanujan miss this 4n+1 counterpart?) More generally, given,√m(1−1n2)(1−1a2)(1−1b2)(1−1c2)=(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 theremany integers 1<a<b<c such that,infinitely

n=√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 withMathematicareveals 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=n2n2−1a+1a−1b+1b−1c+1c−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≤m≤12(n=2,a=2,b=3,c=5).

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