I looked at this a little bit. A132592 has the definition: a(0)=0, a(1)=8 and a(n) = 34*a(n-1) - a(n-2) + 16. but you can also use every 4th term of the sequence defined by: A[0] = 0; A[1] = 0; A[N] = 2 A[N-1] + A[N-2] + 1 (which is A48739 with two extra leading 0 terms). Naturally, that means that if a is a term of A48739, you can let r = 2a s = a+1 and 2^r + 2^s + 1 = (2^a + 1)^2. Of course, the r and s are not squares, except every 4th one, but maybe the others are of interest for some reason. For example, 3 is a member of A48739, so we have r=6, s=4, and 2^6+2^4+1 = 81 = (2^3+1)^2. Then we have the first case where r and s are squares: a=8, r=16, s=9; 2^16+2^9+1 = (2^8+1)^2. And then the a=20 case: r=40, s=21, 2^40+2^21+1 = 1099513724929 = (2^20+1)^2. On Sun, Mar 24, 2013 at 9:17 PM, Jeffrey Shallit <shallit@uwaterloo.ca>wrote:

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.

Victor

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