Let n be an integer, and let (L_n) signify the nth Lucas Number.

((L_n)²+(L_n+1)²)² - 5((L_2n+2)*(L_2n)-1) = 0

Prove it!

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