Trying to work through a Fibonacci problem using Binet's Formula

I am trying to show that the square of any Fibonacci Number differs by one from the product of its neighbors:

$\displaystyle (F_n)^2 - ((F_{n-1}) \times (F_{n+1})) = \pm 1 $

My best attempt is to substitute Binet's Formula:

$\displaystyle \left( \frac { ( 1 + \sqrt{5} )^n - (1 - \sqrt{5} )^n } { 2^n \sqrt{5} } \right)^2 - \left( \frac { ( 1 + \sqrt{5} )^{n-1} - (1 - \sqrt{5} )^{n-1} } { 2^{n-1} \sqrt{5} } \times \frac { ( 1 + \sqrt{5} )^{n+1} - (1 - \sqrt{5} )^{n+1} } { 2^{n+1} \sqrt{5} } \right) $

and simplify, but keep getting buggered near the end. I get down to:

$\displaystyle \frac { ( ( 1 + \sqrt{5} )^{n-1} ( 1 - \sqrt{5} )^{n+1} ) + ( ( 1 - \sqrt{5} )^{n-1} ( 1 + \sqrt{5} )^{n+1} ) - 2 ( -4 )^n } { 5 ( 4^n ) } $

and can go no further due to the unequal bases. I know (because even values of "n" yield -1) that:

$\displaystyle { ( ( 1 + \sqrt{5} )^{n-1} ( 1 - \sqrt{5} )^{n+1} ) + ( ( 1 - \sqrt{5} )^{n-1} ( 1 + \sqrt{5} )^{n+1} ) = - 3 ( -4 )^n } $

which would make the final formula:

$\displaystyle \frac { - 5 ( -4 )^n } { 5 ( 4^n ) } = \pm 1 $

but I seem to be missing something. I could swear I've seen this done, but can't find it anywhere. Any pointers as to what I'm missing?

- Stephen

Re: Trying to work through a Fibonacci problem using Binet's Formula

I realized this morning that I had overlooked some of my algebra:

$\displaystyle x^{n-1} = \frac{x^n}{x}$ and $\displaystyle x^{n+1} = x^n \times x$

So, moving through a few more steps I get to:

$\displaystyle \frac{ \left( \frac{{(-4)^n-(-4)^n \sqrt{5}}}{(1+ \sqrt{5})} \right)+\left( \frac{{(-4)^n+(-4)^n \sqrt{5}}}{(1- \sqrt{5})} \right)-2(-4)^n}{5(4^n)}$

I'm not sure if it's any better, but it's something ...

Re: Trying to work through a Fibonacci problem using Binet's Formula

Wikipedia has a nice proof of this theorem (Cassini's identity):

Cassini and Catalan identities - Wikipedia, the free encyclopedia

Re: Trying to work through a Fibonacci problem using Binet's Formula

That **is** nice - I was hoping for something algebraic, but this is about as straightforward as I could ask for. I have two concerns with this solution though:

1. Using a matrix looks an awful lot like we simply restated the problem and called it solved

2. While creating the matrix using $\displaystyle F_n$ with an odd value of *n *works well, won't using an even value of *n* yield a determinant of $\displaystyle 1^n$ which will always be positive?

It looks like I'll have to brush up on matrices so I can explain it. Wikipedia to the rescue once again!

Thank you!

- Stephen

Re: Trying to work through a Fibonacci problem using Binet's Formula

1. But you are definitely allowed to write $\displaystyle F_{n-1}F_{n+1} - F_{n}^2$ as the determinant of a 2x2 matrix, right?

2. Yes, but $\displaystyle 1^n = 1$ for even n.

Re: Trying to work through a Fibonacci problem using Binet's Formula

Yes on both counts. My "concerns" were rather poorly worded - what I meant to convey is that I'm worried my unfamiliarity with matrices leaves me unprepared to answer what should be a couple of simple questions, not that there was a problem with the proof. Hence my need to "brush up."

- Stephen