Ah for the heady days when I still had not heard of the Pell equation!
Using trusty Dario Alpern supplies the recurrence relation
Xn+1 = P Xn + Q Yn
Yn+1 = R Xn + S Yn
P = 127
Q = 672
R = 24
S = 127
The first few terms for P are {1,127,32257,...} etc. WolframAlpha resolves this to P = ±1/2 ((12748 sqrt(7))^n(127+48 sqrt(7))^n), and we need to take the negative value, multiply by 2 and add 2:
(12748 sqrt(7))^n+(127+48 sqrt(7))^n+2 (I)
Next we look at the values of the square roots of this: {2,16,254,4048,64514 ...}, A090727 in Sloane with formula (8+sqrt(63))^n + (8sqrt(63))^n.
Squaring this gives: (83 sqrt(7))^(2n)+(8+3 sqrt(7))^(2n)+2((83 sqrt(7))(8+3 sqrt(7)))^n.
More conveniently:
(83 sqrt(7))^(2n)+(8+3 sqrt(7))^(2n)+2 (II)
Now, (83 sqrt(7))^(2) = 12748 sqrt(7); while (8+3 sqrt(7))^(2) =127+48 sqrt(7) (III)
So at last we have the equality:
(12748 sqrt(7))^n+(127+48 sqrt(7))^n+2 = (12748 sqrt(7))^n + (127+48 sqrt(7))^n+2. (I),(III)
This number is always square since it is the square of (8+sqrt(63))^n + (8sqrt(63))^n, see (II).
And the mystery is solved at last.
Edited on September 30, 2013, 2:14 am

Posted by broll
on 20130930 02:09:06 