Problem 36: PEN, Linear Recurrences, Problem 5

by henderson, Jul 10, 2016, 9:20 AM

$\color{red}\bf{Problem \ 36}$ The Fibonacci sequence $\{F_n \}$ is defined by
$F_1=1, F_2=1, F_{n+2}=F_{n+1}+F_n.$ Show that $\begin{align*} &F^2_{2n-1}+F^2_{2n+1}+1=3F_{2n-1}F_{2n+1} \end{align*}$ for all $n\geq 1.$
$\color{red}\bf{My \ solution}$ $F^2_{2n-1}+F^2_{2n+1}+1=3F_{2n-1}F_{2n+1}$ $\iff$
$F^2_{2n-1}+(F_{2n-1}+F_{2n})^2+1=3F_{2n-1}(F_{2n-1}+F_{2n})$ $\iff$
$F^2_{2n-1}+F^2_{2n-1}+2F_{2n-1}F_{2n}+F^2_{2n}+1=3F^2_{2n-1}+3F_{2n-1}F_{2n}$ $\iff$
$F^2_{2n}+1=F^2_{2n-1}+F_{2n-1}F_{2n}$ $\iff$
$F^2_{2n}+1=F_{2n-1}(F_{2n-1}+F_{2n})$ $\iff$
$\begin{align*} &F^2_{2n}+1=F_{2n-1}\cdot F_{2n+1}. \ ({\color{blue}{\clubsuit}}) \end{align*}$

We will prove $({\color{blue}{\clubsuit}})$ using induction:
For $n=1,$ it's obvious. Let's assume that $({\color{blue}{\clubsuit}})$ holds and then prove it for $n+1,$ or just the following equation:
$F^2_{2n+2}+1=F_{2n+1}\cdot F_{2n+3}.$

$F^2_{2n+2}+1=F_{2n+1}\cdot F_{2n+3}$ $\iff$
$F^2_{2n+2}+1=F_{2n+1}(F_{2n+1}+F_{2n+2})$ $\iff$
$F^2_{2n+2}+1=F^2_{2n+1}+F_{2n+1}\cdot F_{2n+2}$ $\iff$
$(F_{2n}+F_{2n+1})^2+1=F^2_{2n+1}+F_{2n+1}\cdot F_{2n+2}$ $\iff$
$F^2_{2n}+2F_{2n}F_{2n+1}+F^2_{2n+1}+1=F^2_{2n+1}+F_{2n+1}\cdot F_{2n+2}$ $\iff$
$F^2_{2n}+2F_{2n}F_{2n+1}+1=F_{2n+1}\cdot F_{2n+2}$ $\iff$
$F^2_{2n}+1=F_{2n+1}\cdot F_{2n+2}-2F_{2n}F_{2n+1}$ $\iff$
$F^2_{2n}+1=F_{2n+1}(F_{2n+2}-2F_{2n})$ $\iff$
$F^2_{2n}+1=F_{2n+1}(F_{2n+2}-F_{2n}-F_{2n})$ $\iff$
$F^2_{2n}+1=F_{2n+1}(F_{2n+1}-F_{2n})$ $\iff$
$F^2_{2n}+1=F_{2n+1}\cdot F_{2n-1},$ which is equivalent to $({\color{blue}{\clubsuit}}),$ and we have assumed that it's true. So, the induction is completed. $\quad \quad \quad \quad \quad \square$

$\color{red}\textbf{Remark.}$ Here is a related problem:
Let $x$ and $y$ be positive integers such that $xy$ divides $x^2+y^2+1.$ Show that
$\frac{x^2+y^2+1}{xy}=3.$ $\centerline{\color{red}\bf{PEN, Divisibility Theory, Problem 5}}$

This post has been edited 9 times. Last edited by henderson, Sep 11, 2016, 10:47 AM

Comment

0 Comments

Submit

Math Path

Problem 36: PEN, Linear Recurrences, Problem 5

by henderson, Jul 10, 2016, 9:20 AM

0 Comments