Problem 31: PEN, Linear Recurrences, Problem 12

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Joined: Mar 10, 2015

by henderson, Jul 1, 2016, 4:36 PM

$\bf\color{red}Problem \ 31$ The sequence $\{ a_n \}_{n\geq 1}$ is defined by
$a_1=1, a_2=12, a_3=20, a_{n+3}=2a_{n+2}+2a_{n+1}-a_n.$ Prove that $1+4a_na_{n+1}$ is perfect square for all $n \in \mathbb{N}.$
$\bf\color{red}My \ solution$ $\color{green}\textbf{Claim.}$ $\begin{align*} &1+4a_na_{n+1}=(a_{n+2}-a_{n+1}-a_n)^2 \end{align*}$ for all $n \in \mathbb{N}.$
$\color{green}\textbf{Proof.}$ $1+4a_na_{n+1}=(a_{n+2}-a_{n+1}-a_n)^2$ $\iff$
$1+4a_na_{n+1}=a^2_{n+2}+(a^2_{n+1}+a^2_n+2a_{n+1}a_n)-2a_{n+2}(a_{n+1}+a_n)$ $\iff$
$\begin{align*} &a^2_{n+2}+a^2_{n+1}+a^2_n-2a_{n+2}a_{n+1}-2a_{n+1}a_n-2a_na_{n+2}=1.\ (\bigstar) \end{align*}$ We will prove $(\bigstar)$ by induction.

For $n=1,$ the result is obvious.
Now, let us assume that $(\bigstar)$ is true and then prove it for $n+1.$ Or, using mathematical symbols, we just need to prove the following equation:
$a^2_{n+3}+a^2_{n+2}+a^2_{n+1}-2a_{n+3}a_{n+2}-2a_{n+2}a_{n+1}-2a_{n+1}a_{n+3}=1.$ $\iff$
$\begin{align*} &a^2_{n+2}+a^2_{n+1}+a^2_n-2a_{n+2}a_{n+1}-2a_{n+1}a_n-2a_na_{n+2}=\\ &=a^2_{n+3}+a^2_{n+2}+a^2_{n+1}-2a_{n+3}a_{n+2}-2a_{n+2}a_{n+1}-2a_{n+1}a_{n+3}\\ \end{align*}$ $\iff$
$a^2_n-2a_{n+1}a_n-2a_na_{n+2}=a^2_{n+3}-2a_{n+3}a_{n+2}-2a_{n+1}a_{n+3}$ $\iff$
$a_n(a_n-2a_{n+1}-2a_{n+2})=a_{n+3}(a_{n+3}-2a_{n+2}-2a_{n+1}).$ But the given condition implies
$a_n-2a_{n+1}-2a_{n+2}=-a_{n+3}$ and $a_{n+3}-2a_{n+2}-2a_{n+1}=-a_n,$ so
$a^2_n-2a_{n+1}a_n-2a_na_{n+2}=a^2_{n+3}-2a_{n+3}a_{n+2}-2a_{n+1}a_{n+3}$ $\iff$
$a_n\cdot (-a_{n+3})=a_{n+3}\cdot (-a_n),$ which is true.
We are done!