# Honsberger's Identity/Negative Indices

## Theorem

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

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

$F_{m + n} = F_{m - 1} F_n + F_m F_{n + 1}$

continues to hold, whether $m$ or $n$ are positive or negative.

## Proof

The proof proceeds by induction.

For all $n \in \Z_{\le 0}$, let $\map P n$ be the proposition:

$F_{m + n} = F_{m - 1} F_n + F_m F_{n + 1}$

This can equivalently be written:

For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:

$F_{m - n} = F_{m - 1} F_{-n} + F_m F_{-n + 1}$

$\map P 0$ is the case:

 $\ds F_{m - 1} F_0 + F_m F_1$ $=$ $\ds F_m$ Definition of Fibonacci Number: $F_0 = 0, F_1 = 1$ $\ds$ $=$ $\ds F_{m + 0}$

Thus $\map P 0)$ is seen to hold.

### Basis for the Induction

$\map P 1$ is the case:

 $\ds F_{m - 1} F_{-1} + F_m F_0$ $=$ $\ds F_{m - 1} F_{-1}$ Definition of Fibonacci Number: $F_0 = 0$ $\ds$ $=$ $\ds F_{m - 1} \paren {-1}^0 F_1$ Fibonacci Number with Negative Index $\ds$ $=$ $\ds F_{m - 1} F_1$ $\ds$ $=$ $\ds F_{m - 1}$ Definition of Fibonacci Number: $F_1 = 1$

Thus $\map P 1$ is seen to hold.

This is the basis for the induction.

### Induction Hypothesis

Now it needs to be shown that, if $\map P k$ and $\map P {k - 1}$ are true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.

So this is the induction hypothesis:

$F_{m - k} = F_{m - 1} F_{-k} + F_m F_{-k + 1}$

and:

$F_{m - \paren {k - 1} } = F_{m - 1} F_{-\paren {k - 1} } + F_m F_{-\paren {k - 1} + 1}$

from which it is to be shown that:

 $\ds F_{m - \paren {k + 1} }$ $=$ $\ds F_{m - 1} F_{-\paren {k + 1} } + F_m F_{-\paren {k + 1} + 1}$ $\ds \leadsto \ \$ $\ds F_{m - k - 1}$ $=$ $\ds F_{m - 1} F_{-k - 1} + F_m F_{-k}$

### Induction Step

First we note that:

 $\ds F_{m - k}$ $=$ $\ds F_{m - 1} F_{-k} + F_m F_{-k + 1}$ $\ds \leadstoandfrom \ \$ $\ds F_{m - k}$ $=$ $\ds F_{m - 1} F_{-k} + F_m F_{-\paren {k - 1} }$

and:

 $\ds F_{m - \paren {k - 1} }$ $=$ $\ds F_{m - 1} F_{-\paren {k - 1} } + F_m F_{-\paren {k - 1} + 1}$ $\ds \leadstoandfrom \ \$ $\ds F_{m - k + 1}$ $=$ $\ds F_{m - 1} F_{-k + 1} + F_m F_{-k + 2}$

This is the induction step:

 $\ds F_{m - k - 1}$ $=$ $\ds F_{m - k + 1} - F_{m - k}$ Definition of Fibonacci Number for Negative Index $\ds$ $=$ $\ds F_{m - \paren {k - 1} } - F_{m - k}$ $\ds$ $=$ $\ds \paren {F_{m - 1} F_{-\paren {k - 1} } + F_m F_{-\paren {k - 1} + 1} } - \paren {F_{m - 1} F_{-k} + F_m F_{-k + 1} }$ Induction Hypothesis $\ds$ $=$ $\ds F_{m - 1} \paren {F_{-\paren {k - 1} } - F_{-k} } + F_m \paren {F_{-\paren {k - 1} + 1} - F_{-k + 1} }$ $\ds$ $=$ $\ds F_{m - 1} \paren {F_{-k + 1} - F_{-k} } + F_m \paren {F_{-k + 2} - F_{-k + 1} }$ $\ds$ $=$ $\ds F_{m - 1} F_{-k - 1} + F_m F_{-k}$ Definition of Fibonacci Number for Negative Index

So $\map P k \land \map P {k - 1} \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.

Therefore:

$\forall n \in \Z_{\le 0}: F_{m + n} = F_{m - 1} F_n + F_m F_{n + 1}$