There is a solution when each bracket is k times a square. We need only consider the second bracket.
For k=2 we have for some integers p,q: (2p^21)^2+1=2q^2, with p=+/2, q=+/5, p=+/1, q=+/1, p=0, q=+/1. This is essentially the same problem as the one here.
Say k is greater than 2, for example (3p^21)^2+1=3q^2; but then 3p^42p^2q^2 = 2/3, a fraction, and so on with any larger k, thereby contradicting the choice of p and q as integers.
So there are no other solutions.
Edited on November 12, 2016, 9:12 pm

Posted by broll
on 20161112 20:53:58 