# Cassini's Identity

## Theorem

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

Then:

$F_{n + 1} F_{n - 1} - F_n^2 = \paren {-1}^n$

### Negative Indices

Let $n \in \Z_{<0}$ be a negative integer.

Let $F_n$ be the $n$th Fibonacci number (as extended to negative integers).

Then Cassini's Identity:

$F_{n + 1} F_{n - 1} - F_n^2 = \paren {-1}^n$

continues to hold.

## 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:

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

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 we use this lemma:

$\forall n \in \Z_{>1}: \begin{bmatrix} F_{n + 1} & F_n \\ F_n & F_{n - 1} \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^n$

Then the determinant of both sides is taken.

The left hand side 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 right hand side:

#### Basis for the Induction

$\begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix} = 1 \times 0 - 1 \times 1 = -1 = \paren {-1}^1$

#### Induction Hypothesis

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

$\begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix}^k = \paren {-1}^k$

It remains to be shown that:

$\begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix}^{k + 1} = \paren {-1}^{k + 1}$

#### Induction Step

The induction step follows from Determinant of Matrix Product:

$\begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix}^{k+1} = \begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix}^k \begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix} = \paren {-1}^k \paren {-1} = \paren {-1}^{k + 1}$

Hence by induction:

$\forall n \in \Z_{>0}: \begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix}^n = \paren {-1}^n$

$\blacksquare$

## Also reported as

This is also sometimes reported (slightly less elegantly) as:

$F_{n + 1}^2 - F_n F_{n + 2} = \paren {-1}^n$

## Source of Name

This entry was named for Giovanni Domenico Cassini.

## Historical Note

Some sources attribute this result to Robert Simson and not to Giovanni Domenico Cassini.

However, it was mentioned as far back as $1608$ in a letter by Johannes Kepler.