WolframAlpha gives integer p = +1/2 ((12748 sqrt(7))^n(127+48 sqrt(7))^n):
Hence 2p+2=((12748 sqrt(7))^n+(127+48 sqrt(7))^n)+2 needs to be square, for the conjecture to hold.
The first term is a+b+ab+2 which is 256, or 16^2. The second is a^2+2304*b^2+a^2+2304*b*2+2 = 64516, or 254^2 etc.
With a little tinkering it will be apparent that the square roots of these values correspond to integer values of some x such that x^n=(8+3*7^0.5)^n+(83*7^0.5)^n:
16,254,4048,64514,1028176,16386306,261152656, 4162056194, 66331746448, 1057145886974 etc.<o:p></o:p>
Thus we are asked to show that the square of the integer (x^n) is a perfect square; which is true.

Posted by broll
on 20100502 12:13:31 