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I have a problem . The exercise says:

Show that the recurrence $$\begin{align*}&X_0=m\;,\\&X_n=X_{n-1}^2-2\;,\qquad\text{for }n>2\;,\end{align*}$$ has a solution $X_n=\left\lceil\alpha^{2^n}\right\rceil$, if $m$ is an integer greater than $2$, where $\alpha+\alpha^{-1}=m$ and $\alpha>1$. For example, if $m = 3$ the solution is $$X_n=\left\lceil\varphi^{2^{n+1}}\right\rceil\;,\qquad\varphi=\frac{1+\sqrt5}2\;,\qquad\alpha=\varphi^2\;.$$

Well I have found that

$$\begin{align*} &X_1=m^2-2\\ &X_2=(m^2-2)^2-2\\ &X_3=\left((m^2-2)^2-2\right)^2-2\\ &\;\vdots \end{align*}$$

THEN

$$\begin{align*} \alpha+\alpha^{-1}&=m\iff \alpha^2+\alpha^{-2}+2=m^2\iff \alpha^2+\alpha^{-2}=X_1\\ \left(\alpha^2+\alpha^{-2}\right)^2&=X_1^2\iff X_2=\alpha^4+\alpha^{-4}\\ &\;\;\vdots \end{align*}$$

I guess that $X_n=\alpha^{2^n}+\alpha^{-2^n}$, where $\alpha>1$ and $n>0$. That clearly holds for $n=0,1,2$.

Let’s say that it holds for $n$ and it must hold for $n+1$. By induction we see that

$$X_{n+1}=X_n^2-2=\alpha^{2^{n+1}}+\alpha^{-2^{n+1}}+2-2=\alpha^{2^{n+1}}+\alpha^{-2^{n+1}}$$

Now how do I prove that $X_n$ is an integer (which it must be) and what must I do to end in the answer that the exercise asks for?

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    Please use mathjax. – 2017-01-11
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    i tried editing but it there are notations wich are not uniquely interpretable. please modify. – 2017-01-11
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    I dont know what mathjax is .where can i find it ? – 2017-01-11
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    @SidiropoulosAndreas: I’ve edited it for you. You’ll find a MathJax tutorial [here](http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference). – 2017-01-11
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    @ Brian M. Scott thank you for your help – 2017-01-11

1 Answers 1

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You’ve shown that $X_n=\alpha^{2^n}+\alpha^{-2^n}$ for each $n\in\Bbb N$. On the other hand, it’s clear from the definition that $X_n$ is an integer for each $n\in\Bbb N$, so $\alpha^{2^n}+\alpha^{-2^n}$ is an integer for each $n\in\Bbb N$. We’re told that $\alpha>1$, so $0<\alpha^{-1}<1$, and therefore $0<\alpha^{-2^n}<1$ for each $n\in\Bbb N$. Thus,

$$X_n-1

where $X_n-1$ and $X_n$ are consecutive integers, and it follows immediately that

$$X_n=\left\lceil\alpha^{2^n}\right\rceil\;.$$

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    I cannot understand how Xn – 2017-01-11
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    @SidiropoulosAndreas: I accidentally left out a $-1$; it may make better sense now. You know that $X_n$ is an integer: $X_0$ is an integer by definition, and the recurrence ensures that if $X_n$ is an integer, so is $X_{n+1}$, so every $X_n$ is an integer. $X_n-1$ is the next smaller integer, so $X_n-1$ and $X_n$ are consecutive integers. $\alpha^{2^n}$ lies between these two integers, so its ceiling is the larger of the two, which is $X_n$. – 2017-01-11
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    ooo now I see.Thank you. – 2017-01-11
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    @SidiropoulosAndreas: You’re welcome. – 2017-01-11