Sequence and series question. Condition for $x_n$ to be an integer 
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!
 A: 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 $m$.
A: 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.
