Consider some smaller number, x, such that P = 2x^21, and P^2+1 = (2x^21)^2, when (2x^21)^2+1= 2y^2.
Then 4x^44x^22y^2+2 = 0. We can substitute X=x^2 to obtain 4X^24X2y^2+2=0, when X is in {1,4,21,120,687,...}A046090 in Sloane and y is in {1,5,29,169,985,...} A001653 in Sloane. If we can now show that only 1 and 4 are perfect squares for X, we are done, with P=7 as the sole solution.
With modern technology, the proof isn't particularly hard, only tedious. A046090 is computed to a thousand terms in Sloane, some of which are very large. As it is a recurrence relation, it is periodic mod any integer, which makes it possible to compute whether each term is a quadratic nonresidue mod a number, say N.
There is a list of quadratic residues in Wikipedia. If we take A046090 under the list of quadratic residues, we find that nonresidue solutions (cases where A046090(n) cannot be square) fill up very quickly. For example, N=11, with period 12, has nonresidues for n=12k+3,4,8,9,10,11; 19, with period 20, has nonresidues for n=20k+3,5,6,8,13,14,17,18; and 23, with period 11, has nonresidues for n=11k+3,4,5,7,8,10,14,15,16,18,19, producing a table of forms, something like this:
1
2
3 4k+3 12k+3 10k+3 20k+3 14k+3
4 12k+4 5k+4
5 20k+5 14k+5
6 8k+6 20k+6 12k+6
7 4k+3 8k+7 10k+7 12k+7
8 12k+8 20k+8
9 12k+9 10k+9 5k+4
10 12k+10 14k+10
11 4k+3 12k+11
12 14k+12
13 10k+3 20k+13
14 8k+6 20k+14 5k+4
15 4k+3 8k+7 12k+3
16 12k+4
17 10k+7 20k+17 14k+3
18 20k+18 12k+6
19 4k+3 10k+9 12k+7 14k+5 5k+4
20 12k+8
21 12k+9
22 8k+6 12k+10
23 4k+3 12k+11 8k+7 10k+3
24 14k+10 5k+4
25 20k+5
26 20k+6 14k+12
27 4k+3 10k+7 12k+3
28 20k+8 12k+4
29 10k+9 5k+4
30 8k+6 12k+6
31 4k+3 8k+7 12k+7 14k+3
32 12k+8
33 10k+3 20k+13 12k+9 14k+5
34 20k+14 5k+4 12k+10
35 4k+3 12k+11
36 11k+3
37 10k+7 20k+17
38 20k+18 14k+10
39 4k+3 10k+9 5k+4
40 14k+12
etc. There are probably neater ways of completing the table using ideal candidates, i.e. those with short periods and a low common multiple (e.g, 41 has a period of only 6, taken with A046090).
Hence n=1,2, implying that A046090(n) =1, 4 are indeed the only squares in A046090, and P=7 is the sole solution to the puzzle.
In fact, I've suspected for a while that there is a large class of these sequences exhibiting the same property  only two square terms; e.g. A001653. I suspect there is a more elegant and more generalised solution.
Edited on October 24, 2016, 1:15 am

Posted by broll
on 20161023 22:37:56 