# Fibonacci Number of Odd Index by Golden Mean Modulo 1

## Theorem

Let $n \in \Z$ be an integer.

Then:

$F_{2 n + 1} \phi \bmod 1 = \phi^{-2 n - 1}$

where:

$F_n$ denotes the $n$th Fibonacci number
$\phi$ is the golden mean: $\phi = \dfrac {1 + \sqrt 5} 2$

## Proof

From definition of$\bmod 1$, the statement above is equivalent to the statement:

$F_{2 n + 1} \phi - \phi^{-2 n - 1}$ is an integer

We have:

 $\ds \phi^2 - \phi \sqrt 5$ $=$ $\ds \paren {\frac {1 + \sqrt 5} 2}^2 - \paren {\frac {1 + \sqrt 5} 2} \sqrt 5$ $\ds$ $=$ $\ds \frac {6 + 2 \sqrt 5} 4 - \frac {5 + \sqrt 5} 2$ $\ds$ $=$ $\ds -1$

Hence:

 $\ds F_{2 n + 1} \phi - \phi^{-2 n - 1}$ $=$ $\ds \frac {\phi^{2 n + 1} - \paren {-1}^{2 n + 1} \phi^{-2 n - 1} } {\sqrt 5} \phi - \phi^{-2 n - 1}$ Euler-Binet Formula $\ds$ $=$ $\ds \frac {\phi^{2 n + 2} + \phi^{-2 n} - \phi^{-2 n - 1} \sqrt 5} {\sqrt 5}$ $\ds$ $=$ $\ds \frac {\phi^{2 n + 2} + \phi^{-2 n - 2} \paren {\phi^2 - \phi \sqrt 5} } {\sqrt 5}$ $\ds$ $=$ $\ds \frac {\phi^{2 n + 2} - \phi^{-2 n - 2} } {\sqrt 5}$ as $\phi^2 - \phi \sqrt 5 = -1$ $\ds$ $=$ $\ds \frac {\phi^{2 n + 2} - \paren {-1}^{-2 n - 2} \phi^{-2 n - 2} } {\sqrt 5}$ $\ds$ $=$ $\ds F_{2 n + 2}$ Euler-Binet Formula

which is an integer.

$\blacksquare$