This is a classic 'Pell' Equation, of the form x^2ny^2=1.
There is a good resource for solving them here: http://www.alpertron.com.ar/QUAD.HTM
The fundamental solution is then {(p=9, q=20) u (r=4 s=9) v} which can be used as explained in the above website to generate all other solutions of the equation. There is also a very good article here http://www.ams.org/notices/200202/fealenstra.pdf explaining the concept in more detail (by coincidence it also refers to the same equation x^{2}–5y^{2}=1).
The recurrence relation could then be used to confirm the claim; but there is a more elementary method:
Assume x is even and=2n. Then 4n^21=5y^2 and 5y^2 is worth 3(mod4). But 5y^2 is worth {0,1}(mod 4) for all y, a contradiction. So x is odd.
(x+1)(x1) = 5y^2, and either 3 divides x, or it divides y, since one of (x1), x, or (x+1) is divisible by 3.
(x+1)(x1) = 5y^2. Since x is odd, (x+1) is even and (x1) is even, so y is divisible by 4.
Hence x*y is divisible by 12. QED
Edited on December 4, 2010, 1:03 am

Posted by broll
on 20101203 04:49:42 