I'd start with https://oeis.org/A132592 and if a is a term of that sequence, I'd let x^2 = 2a y^2 = a+1 so that 2^(x^2) + 2^(y^2) + 1 = (2^a + 1)^2. Is that the same solution you found? On 3/22/13 10:44 PM, Victor Miller wrote:
I came across the following problem:
Show that there are an infinite number of pairs (x,y) of positive integers such that
2^(x^2) + 2^(y^2) + 1 is a square.
I would suspect that the answer that you'll come up with is the same as what I came up with, but I wondered if it's possible to characterize all (or all but a finite number) of the pairs which work. By this I mean to show that all (x,y) are in one of a finite set of specific sequences, with a possible finite set of exceptions.
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