Cassini’s identity

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:

$\begin{array}{lrclrcl} n \quad & & F_n^2 & & & F_{n-1}F_{n+1} & \\ 1 & 1^2 & = & 1 \qquad & 0 \cdot 1 & = & 0 \\ 2 & 1^2 & = & 1 & 1 \cdot 2 & = & 2 \\ 3 & 2^2 & = & 4 & 1 \cdot 3 & = & 3 \\ 4 & 3^2 & = & 9 & 2 \cdot 5 & = & 10 \\ 5 & 5^2 & = & 25 & 3 \cdot 8 & = & 24 \\ 6 & 8^2 & = & 64 & 5 \cdot 13 & = & 65\end{array}$

(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 $n$ and the adjacent product is bigger for even values of $n$ . Algebraically,

$F_{n+1}F_{n-1} - F_n^2 = (-1)^n \qquad (n > 0)$ .

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 $n = 1$ :

$F_2 F_0 - F_1^2 = 1 \cdot 0 - 1^2 = -1 = (-1)^1$ .

Now we can assume it holds for some $k > 0$ , and show that it also holds for $k+1$ :

$\begin{array}{rcl} F_{k+1}F_{k-1} - F_k^2 & = & (-1)^k \\ F_{k+1} (F_{k+1} - F_k) - F_k^2 & = & (-1)^k \\ F_{k+1}^2 - F_{k+1}F_k - F_k^2 & = & (-1)^k \\ F_{k+1}^2 - (F_{k+1} + F_k)F_k & = & (-1)^k \\ F_{k+1}^2 - F_{k+2}F_k & = & (-1)^k \\ F_{k+2}F_k - F_{k+1}^2 & = & (-1)^{k+1} \end{array}$

So by induction, Cassini’s identity holds for all $n > 0$ . (Actually, there is a sensible way to define negative Fibonacci numbers which makes Cassini’s identity true for all integers $n$ , but perhaps that can be the subject of another post!)

About Brent

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

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8 Responses to Cassini’s identity

Joshua Zucker says:

May 15, 2011 at 10:32 pm

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 Count by Benjamin and Quinn.
- Brent says:
  
  May 16, 2011 at 8:53 am
  
  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!
fibonacci says:

May 18, 2011 at 1:03 pm

Here is a combinatorial proof A Bijective proof of Cassini’s Fibonacci identity by M. Werman and D. Zeilberger, Discrete Math. 58, 109 (1986).
- Brent says:
  
  May 18, 2011 at 2:44 pm
  
  Thanks!
Will Orrick says:

May 23, 2011 at 11:02 am
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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parzan says:

June 21, 2011 at 7:30 am

This identity plays a role in a famous “paradox”:
http://virtualmathtutor.blogspot.com/2011/01/math-problem-84.html
J Andrew Lipscomb says:

June 22, 2011 at 10:53 am

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.