Nice! :smile: the idea is prove that $\sqrt{a_{n+2}} = 4\sqrt{a_{n+1}} - \sqrt{a_n}$ and this is easy!

e.lopes wrote: Nice! :smile: the idea is prove that $\sqrt{a_{n+2}} = 4\sqrt{a_{n+1}} - \sqrt{a_n}$ and this is easy! Maybe it should be ${a_{n+2}} = 4{a_{n+1}} - {a_n}$

And maybe you should all post solutions instead of ideas.

hawk tiger: i don't no where i find this \sqrt, thanks for your correction peter: sorry peter, i will post complete solutions now :razz:

Peter wrote: The sequence $\{a_{n}\}_{n \ge 1}$ is defined by \[a_{1}=1, \; a_{n+1}=2a_{n}+\sqrt{3a_{n}^{2}+1}.\] Show that $a_{n}$ is an integer for every $n$. Well, as nobady has written a complete solution I will post bellow a solution for a generalization of the problem. Problem:The sequence $\{a_{n}\}_{n \ge 1}$ is defined by $a_{1}=1, \; a_{n+1}=ma_{n}+\sqrt{(m^{2}-1)a_{n}^{2}+1}$,where $m\in{N}$. Show that $a_{n}$ is an integer for every $n$. Solution: Let us rewrite the condition in the following way and square both sides: $a_{n+1}-ma_{n}=\sqrt{(m^{2}-1)a_{n}^{2}+1}$, so $a_{n+1}^{2}+m^{2}a_{n}^{2}-2ma_{n}a_{n+1}=(m^{2}-1)a_{n}^{2}+1$,thus $a_{n+1}^{2}+a_{n}^{2}=2ma_{n+1}a_{n}+1$ and $a_{n}^{2}+a_{n-1}^{2}=2ma_{n}a_{n-1}+1$. Now subtracting the last two equations we get that $a_{n+1}^{2}-a_{n-1}^{2}=2ma_{n}(a_{n+1}-a_{n-1})$, therefore \[a_{n+1}=2ma_{n}-a_{n-1}\] And we can easily see that $a_{1}=1$ and $a_{2}=2m$, consequently all the members of the sequence are integers, as the problem calemed.

Peter wrote: The sequence $\{a_{n}\}_{n \ge 1}$ is defined by \[a_{1}=1, \; a_{n+1}=2a_{n}+\sqrt{3a_{n}^{2}+1}.\]Show that $a_{n}$ is an integer for every $n$. $$\bf\color{red}My \ solution$$$\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. $\square$

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.