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
Prove it!
(In reply to
re(2): Long solution. by Jer)
I was thinking in terms of writing up the solution.
Your line 5 as given is in terms a statement limited to Lucas identities.
In fact, at that point, they no longer need to be Lucas identities; any integers we choose that satisfy the equality a^2-3ab+b^2 = -5 (and no others) are necessarily of the form {(L2n+2), (L2n)} (or vice versa) simply because those numbers, and no others, satisfy the equality.
But if so, we no longer need to specify that they are Lucas identities as such. So I would say that the proof was complete at line 5.
That is the point I was trying to get at.
|
|
Posted by broll
on 2013-12-17 03:15:58 |
re(3): Long solution.
