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^2-1)^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^2-1)^2+1=3q^2; but then 3p^4-2p^2-q^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 2016-11-12 20:53:58