Problem 33: PEN, Recursive Sequences, Problem 6

by henderson, Jul 2, 2016, 6:45 PM

$\bf\color{red}Problem \ 33$ The sequence $\{ a_n \}_{n\geq 1}$ is defined by
$a_1=1, a_{n+1}=2a_n+\sqrt{3a^2_n+1}.$ Show that $a_n$ is an integer for every $n.$
(Serbia 1998) $\bf\color{red}My \ solution$ $\color{green}\textbf{Claim.}$ $\begin{align*} &4a_{n+1}-a_n=a_{n+2} \end{align*}$ for all $n\in \mathbb{N}.$
$\color{green}\textbf{Proof}.$ Before starting the proof of the claim, we need to prove some facts.

$\color{blue}\textbf{Fact 1}.$ $a_n$ is positive for all $n\in \mathbb{N}.$
$\color{blue}\textbf{Proof of fact 1}.$ We will use induction. For $n=1,$ $a_1=1>0.$ Let's assume that $a_n$ is positive. Then $a_{n+1}=2a_n+\sqrt{3a^2_n+1}>0.$ So, the induction completed.

$\color{red}\textbf{Fact 2}.$ The sequence is increasing.
$\color{red}\textbf{Proof of fact 2}.$ Since $a_n$ is positive for every $n\in \mathbb{N},$ we have
$a_{n+1}=2a_n+\sqrt{3a^2_n+1}=a_n+(a_n+\sqrt{3a^2_n+1})> a_n,$ as desired.

Now, we can turn into the proof of the claim. Starting from the given condition
$\begin{align*} &a_{n+1}=2a_n+\sqrt{3a^2_n+1}\ (\bigstar) \end{align*}$ $\iff$
$a_{n+1}-2a_n=\sqrt{3a^2_n+1}$ $\iff$
$a^2_{n+1}+4a^2_n-4a_{n+1}a_n=3a^2_n+1$ $\iff$
$a^2_{n+1}+a^2_n-4a_{n+1}a_n=1.$ Since $(\bigstar)$ holds for all $n\in \mathbb{N},$
$a^2_{n+1}+a^2_n-4a_{n+1}a_n=1$ holds for all $n\in \mathbb{N},$ too.
Then $a^2_{n+2}+a^2_{n+1}-4a_{n+2}a_{n+1}=1$ is also true. $\Longrightarrow$
$a^2_{n+1}+a^2_n-4a_{n+1}a_n=a^2_{n+2}+a^2_{n+1}-4a_{n+2}a_{n+1}$ $\Longrightarrow$
$a^2_n-4a_{n+1}a_n=a^2_{n+2}-4a_{n+2}a_{n+1}$ $\Longrightarrow$
$\begin{align*} &(a_{n+2}-a_n)(a_{n+2}+a_n)=4a_{n+1}(a_{n+2}-a_n).\ (\clubsuit) \end{align*}$ In the $\color{red}\textbf{fact 2},$ we proved that the sequence is increasing. So,
$a_{n+2}-a_n>a_{n+1}-a_n>0,$ which means that we are able to cancel $a_{n+2}-a_n$ in $(\clubsuit).$ Canceling $a_{n+2}-a_n,$ we obtain
$a_{n+2}+a_n=4a_{n+1}$ $\Longrightarrow$
$4a_{n+1}-a_n=a_{n+2},$ so the claim has been proved.

Setting $n=1$ in
$a_{n+1}=2a_n+\sqrt{3a^2_n+1}$ and using $a_1=1,$
we get that $a_2=4.$
Then, the first $2$ terms of the sequence are integers, and there is a linear relationship between $3$ consecutive terms, so all the terms are integers, as desired. $\quad \square$

This post has been edited 4 times. Last edited by henderson, Sep 15, 2016, 11:26 AM

Sequence

the PEN Project

Recursive Sequences

Comment

1 Comment

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Similar problem:
Define the sequences $(a_n)_{n\geq1}$ by $a_1=1$ and $a_{n+1}=2a_n+\sqrt{3a_n^2-2}$ for all integers $n\geq1.$ Prove that $a_n$ is an integer for all $n.$
$\bf{Solution:}$
We have $(a_{n+1}-2a_n)^2=3a_n^2-2,$ $n\geq1.$
Then $a_{n+1}^2+a_n^2-4a_{n+1}a_n=-2$
And $a_{n+2}^2+a_{n+1}^2-4a_{n+2}a_{n+1}=-2,$ $n\geq 1.$
Subtracting these relations yields
$a_{n+2}^2-a_n^2-4a_{n+1}(a_{n+2}-a_n)=0$
And $(a_{n+2}-a_n)(a_{n+2}+a_n-4a_{n+1})=0,$ $n\geq 1.$
Since the sequence $(a_n)_{n\geq1}$ is increasing,it follows that $a_{n+2}=4a_{n+1}-a_n,$ $n\geq1.$
Taking into account that $a_1=1,a_2=3$ and inducting on $n$ we reach the conclusion.

by Ferid.---., Jul 9, 2016, 12:00 PM

Report

Submit

Math Path

Problem 33: PEN, Recursive Sequences, Problem 6

by henderson, Jul 2, 2016, 6:45 PM

1 Comment