My previous post asked you to take any Fibonacci number, square it, and also multiply the two adjacent Fibonacci numbers, and see if a pattern emerged. Here’s a table I made for the first 6 Fibonacci numbers:

(Hmm, the numbers in that last row sure look familiar…) It seems that the square of a Fibonacci number and the product of its two adjacent Fibonacci numbers always differ by exactly one. Moreover, which one is bigger alternates: the square is bigger for odd values of and the adjacent product is bigger for even values of . Algebraically,

.

This is actually true, and is known as *Cassini’s identity*, since it was first published by the Italian astronomer Gian Domenico Cassini in 1680. Let’s prove it!

First, we can check that it holds when :

.

Now we can assume it holds for some , and show that it also holds for :

So by induction, Cassini’s identity holds for all . (Actually, there is a sensible way to define negative Fibonacci numbers which makes Cassini’s identity true for *all* integers , but perhaps that can be the subject of another post!)

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## About Brent

Assistant Professor of Computer Science at Hendrix College. Functional programmer, mathematician, teacher, pianist, follower of Jesus.

This is a nice short algebraic proof but I think to get real insight into these things we need a combinatorial proof.

I could write it up but I’d basically be copying pp. 7-9 of

Proofs that Really Countby Benjamin and Quinn.Ah, I wasn’t aware of a combinatorial proof of Cassini’s Identity. I agree combinatorial proofs are much better for furnishing insight. I’ll have to look it up!

Here is a combinatorial proof A Bijective proof of Cassini’s Fibonacci identity by M. Werman and D. Zeilberger, Discrete Math. 58, 109 (1986).

Thanks!

The algebraic proof does provide certain insights as well. For example, it tells you that the property,

F(n+1)^2 – F(n)*F(n+2) = –[F(n)^2 – F(n–1)*F(n+1)] for all n,

is independent of the initial values F(1) = 1, F(2) = 1, and therefore that something like Cassini’s identity will hold for any choice of initial conditions.

This property can be rewritten F(n+1)^2 + F(n)^2 = F(n)*F(n+2) + F(n–1)*F(n+1), which has a simple geometric interpretation:

The area of the two squares

is the same as that of the two rectangles

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This identity plays a role in a famous “paradox”:

http://virtualmathtutor.blogspot.com/2011/01/math-problem-84.html

Right, the proof works for any Fibonacci-like sequence with a constant multiplier on the right-hand side. For example, in the Lucas sequence (L(1)=1, L(2)=3), the constant is -5.