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If $x_1$, $x_2$, $x_3,\ldots$ is a sequence such that $$x_n=\frac{x_{n-2}\space x_{n-1}}{2x_{n-2}-x_{n-1}},$$ where $x_i \in \mathbb R$ and $x_i \ne0 $ for all $i\in \mathbb N$ and $n=3$, $4$, $5,\ldots$

How can I establish necessary and sufficient conditions on $x_1$ and $x_2$ for $x_n$ to be an integer for infinitely many values of $n$?

I have been stuck on this problem for quite sometime now. I can't seem to find a pattern so that i could "make" $x_n$ an integer.

Any help is much appreciated!

Thanks in advance!

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Notice that if $x_1=x_2$, then $x_n = x_1$. – user26872 Mar 25 '12 at 5:44
up vote 2 down vote accepted

Note that

$$ \frac{1}{x_n} = \frac{2}{x_{n-1}} - \frac{1}{x_{n-2}} $$

so $\tfrac{1}{x_n}$ satisfies a linear recursion. This can be solved explicitly: Let

$$ a = \frac{2}{x_1} - \frac{1}{x_2},\ b = \frac{1}{x_2} - \frac{1}{x_1} $$

Then $\tfrac{1}{x_n} = a + nb$, or

$$ x_n = \frac{1}{a+nb}. $$

Now if $b \neq 0$ then $x_n \rightarrow 0$ and in particular $x_n$ can only be an integer finitely many times. So $b=0$ and $a = \tfrac{1}{m}$ for some integer $m$.

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Value of $x_n$ can be rewritten as, $$\frac{1}{x_n} = \frac{2}{x_{n-1}} - \frac{1}{x_{n-2}}.$$ $$\frac{1}{x_n} - \frac{1}{x_{n-1}} = \frac{1}{x_{n-1}} - \frac{1}{x_{n-2}}.$$ This means that difference of reciprocal remans same, hence sequence $1\over{x_n}$ is in arithmetic progression. This implies $${1\over{x_n}} = A + (n-1)D,$$ Where $A$ is first term and $D$ is difference of arithmetic progression. $${{x_n}} = {1\over{A + (n-1)D}},$$Only way $x_n$ can be integer for infinitely many values of n, is $D = 0$ and A is fraction whose numerator is 1. In such cases $x_n$ will be constant $1\over{A}$ for all value of n.

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