# Cassini's Identity

## Contents

## Theorem

Let $F_k$ be the $k$th Fibonacci number.

Then $F_{n+1}F_{n-1} - F_n^2 = \left({-1}\right)^n$.

This is also sometimes reported (slightly less elegantly) as $F_{n+1}^2 - F_n F_{n+2} = \left({-1}\right)^n$

## Proof 1

We see that:

- $F_2 F_0 - F_1^2 = 1 \times 0 - 1 = -1 = \left({-1}\right)^1$

so the proposition holds for $n=1$.

We also see that:

- $F_3 F_1 - F_2^2 = 2 \times 1 - 1 = \left({-1}\right)^2$

so the proposition holds for $n=2$.

Suppose the proposition is true for $n=k$, that is:

- $F_{k+1}F_{k-1} - F_k^2 = \left({-1}\right)^k$

It remains to be shown that it follows from this that the proposition is true for $n=k+1$, that is:

- $F_{k+2}F_k - F_{k+1}^2 = \left({-1}\right)^{k+1}$

So:

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle F_{k+2} F_k - F_{k+1}^2\) | \(=\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle \left({F_k + F_{k+1} }\right) F_k - F_{k+1}^2\) | \(\displaystyle \) | \(\displaystyle \) | |||

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle F_k^2 + F_k F_{k+1} - F_{k+1}^2\) | \(\displaystyle \) | \(\displaystyle \) | |||

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle F_k^2 + F_k F_{k+1} - F_{k+1} \left({F_k + F_{k-1} }\right)\) | \(\displaystyle \) | \(\displaystyle \) | |||

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle F_k^2 + F_k F_{k+1} - F_k F_{k+1} - F_{k+1} F_{k-1}\) | \(\displaystyle \) | \(\displaystyle \) | |||

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle F_k^2 - F_{k+1} F_{k-1}\) | \(\displaystyle \) | \(\displaystyle \) | |||

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle \left({-1}\right) \left({F_{k+1} F_{k-1} - F_k^2}\right)\) | \(\displaystyle \) | \(\displaystyle \) | |||

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle \left({-1}\right) \left({-1}\right)^k\) | \(\displaystyle \) | \(\displaystyle \) | |||

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle \left({-1}\right)^{k+1}\) | \(\displaystyle \) | \(\displaystyle \) |

By the Principle of Mathematical Induction, the proof is complete.

$\blacksquare$

Note that from the above we have that:

- $F_{k+2} F_k - F_{k+1}^2 = \left({-1}\right)^{k+1}$

from which:

- $F_{n+1}^2 - F_n F_{n+2} = \left({-1}\right)^n$

follows immediately.

## Proof 2

First this identity is proved:

- $\begin{bmatrix} F_{n+1} & F_n \\ F_n & F_{n-1} \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^n$

then the determinant of both sides is taken.

### Basis for the Induction

- $\begin{bmatrix} F_2 & F_1 \\ F_1 & F_0 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^1$

### Induction Hypothesis

For $k \in \Z_{>0}$, it is assumed that:

- $\begin{bmatrix} F_{k+1} & F_k \\ F_k & F_{k-1} \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^k$

It remains to be shown that:

- $\begin{bmatrix} F_{k+2} & F_{k+1} \\ F_{k+1} & F_k \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^{k+1}$

### Induction Step

The induction step follows from conventional matrix multiplication:

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^{k+1}\) | \(=\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle \begin{bmatrix} F_{k+1} & F_k \\ F_k & F_{k-1} \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}\) | \(\displaystyle \) | \(\displaystyle \) | by the induction hypothesis | ||

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle \begin{bmatrix} F_{k+1} + F_k & F_{k+1} \\ F_k + F_{k-1} & F_k \end{bmatrix}\) | \(\displaystyle \) | \(\displaystyle \) | by conventional matrix multiplication | ||

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle \begin{bmatrix} F_{k+2} & F_{k+1} \\ F_{k+1} & F_k \end{bmatrix}\) | \(\displaystyle \) | \(\displaystyle \) |

So by induction:

- $\begin{bmatrix} F_{n+1} & F_n \\ F_n & F_{n-1} \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^n$

### Determinant

Now we calculate the determinants:

The LHS follows directly from the order 2 determinant:

- $\begin{bmatrix} F_{n+1} & F_n \\ F_n & F_{n-1} \end{bmatrix} = F_{n+1} F_{n-1} - F_n^2$

Now for the RHS:

#### Basis for the Induction

- $\begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix} = 1 \times 0 - 1 \times 1 = -1 = \left({-1}\right)^1$

#### Induction Hypothesis

For $k \in \Z_{>0}$, it is assumed that:

- $\begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix}^k = \left({-1}\right)^k$

It remains to be shown that:

- $\begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix}^{k+1} = \left({-1}\right)^{k+1}$

#### Induction Step

The induction step follows from Determinant of Matrix Product:

- $\begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix}^{k+1} = \begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix}^k \begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix} = \left({-1}\right)^k \left({-1}\right) = \left({-1}\right)^{k+1}$

Hence by induction:

- $\forall n \in \Z_{>0}: \begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix}^n = \left({-1}\right)^n$

$\blacksquare$

## Source of Name

This entry was named for Giovanni Domenico Cassini.

## Sources

- George E. Andrews:
*Number Theory*(1971)... (previous)... (next): $\S 1.1$: Exercise $10$