Fibonacci Number of Odd Index by Golden Mean Modulo 1

From ProofWiki
Jump to navigation Jump to search

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.

$\blacksquare$

Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Fibonacci_Number_of_Odd_Index_by_Golden_Mean_Modulo_1&oldid=474760"