Let n be an integer, and let (Ln) signify the nth Lucas Number.
((Ln)2+(Ln+1)2)2 - 5((L2n+2)*(L2n)-1) = 0
(In reply to re: Long solution.
If they are I never would have realized it. I was just playing with the Lucas Numbers and the identities on mathworld until the solution popped out. It may be just by luck that
You are better at this type of problem than I. Why does the substitution a^2-3ab+b^2 = -5 even matter?
Posted by Jer
on 2013-12-16 15:46:27