Fibonacci Identity

Date: 12/10/2001 at 17:49:43
From: Doug Moll
Subject: Form of Fibonacci numbers

I am trying to create an inductive proof for the particular identity 
of Fibonacci numbers that:

F(n-1) * F(n+1) = (-1)^n + (Fn)^2

I know I do a base case of n being 0, and then an inductive step of 
n being n+1, but beyond that I just can't get the math to work out. 
Is this a valid identity, and if so, how would I work out the math 
to inductively show this?

Thanks much.
Doug

Date: 12/11/2001 at 06:28:07
From: Doctor Floor
Subject: Re: Form of Fibonacci numbers

Hi,

Thanks for your question.

Well, I suppose you have been able to show that this works for n = 0.

Now we suppose that F(n-1)*F(n+1) = (-1)^n + (F(n))^2 holds for n. 
We can rewrite this to

  F(n-1) * F(n+1) - (F(n))^2 = (-1)^n [*]

Now we will replace all n by n+1 at the left-hand side of the 
equation, and show that it yields (-1)^(n+1). For that we use truth 
we suppose of [*] as well as application of the Fibonacci rule 
F(m+2) = F(m)+F(m+1) for all m:

  F(n)*F(n+2) - (F(n+1))^2 =
  F(n)*[F(n+1) + F(n)] - F(n+1)*[F(n) + F(n-1)] =
  F(n)*F(n+1) + (F(n))^2 - F(n)*F(n+1) - F(n+1)*F(n-1) =
  (F(n))^2 - F(n+1)*F(n-1) = 
  -(-1)^n = (-1)^(n+1)

And we get where we want to be.

See for an approach without induction, from the Dr. Math archive

   Fibonacci Trick
   http://mathforum.org/dr.math/problems/roberts.6.01.99.html   

If you need more help, just write back.

Best regards,
- Doctor Floor, The Math Forum
  http://mathforum.org/dr.math/