Fibonacci Number of Even Index by Golden Mean Modulo 1

Theorem

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

Then:

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

$F_n$ denotes the $n$th Fibonacci number

$\phi$ is the golden mean: $\phi = \dfrac {1 + \sqrt 5} 2$

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

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

We have:

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

$=$

$\ds -1 + F_{2 n + 1} - \hat \phi^{2 n} + \phi^{-2 n}$

$\ds $

$=$

$\ds -1 + F_{2 n + 1} - \paren {-1}^{2 n} \phi^{-2 n} + \phi^{-2 n}$

$\ds $

$=$

$\ds -1 + F_{2 n + 1}$

which is an integer.

Hence the result.

$\blacksquare$