Choose initial values such that sequence always has integer values We are given a recurrence relation defined by $$x_{n+2}=\frac{x_{n+1}x_n}{2x_n-x_{n+1}}.$$
Place necessary and sufficient values on $x_0$ and $x_1$ such that $x_n$ is an integer for all positive integer values of $n$.
There are two obvious conditions to begin with, that $2x_0-x_1\not=0$, and that $2x_0-x_1 \mid x_1x_0$.
But I know they aren't good enough, since choosing $x_0=-10$, $x_1=5$, we get $x_2=2$, but $x_3=\frac{10}{8}$. 
I'm obviously missing something(s), but I don't know what to do to find other conditions. I've tried finding $x_n$ in terms of $n$, without success. 
 A: By induction, 
$$ x_{n} = \dfrac{x_0 x_1}{n x_0 - (n-1) x_1} = \dfrac{x_0 x_1}{x_0 + (n-1) (x_0 - x_1)} $$
If $x_0 \ne x_1$, the only way this can be an integer for all $n$ is that $x_0 x_1 = 0$, i.e. $x_0 = 0$ or $x_1 = 0$.  
If $x_0 = x_1 \ne 0$, it is always an integer, namely $x_1$.
A: I have found the following condition:
For all positive integer $k$, there is a $r$
and an element $(x_0,x_1)$ of the set $M_k = \{(x_0,x_1) \in \mathbb{Z}\times \mathbb{Z} :(k+1)x_0 = k x_1\}$,
such that $x_r$ is rational.
Proof:
Induction gives for such an element $(x_0,x_1)$
$$x_r = \frac{(k+1)x_0}{k-r+1}.$$
Base Case:
follows trivially from $(k+1)x_0 = k x_1$.
Induction Step:
$$
x_{r+1}
= \frac{x_r x_{r-1}}{2x_{r-1} - x_r}
= \frac{\frac{(k+1)x_0}{(k-r+1)} \frac{(k+1)x_0}{(k-(r-1)+1)}}{2\frac{(k+1)x_0}{(k-(r-1)+1)} - \frac{(k+1)x_0}{(k-r+1)}}
= \frac{\frac{1}{(k-r+1)} \frac{1}{(k-(r-1)+1)}}{\frac{2}{(k-(r-1)+1)} - \frac{1}{(k-r+1)}}
(k+1)x_0\\
= \frac{\frac{1}{(k-r+1)} \frac{1}{(k-(r-1)+1)}}{\frac{2(k-r+1)-(k-(r-1)+1)}{(k-r+1)(k-(r-1)+1)}}
(k+1)x_0
= \frac{(k+1)x_0}{2(k-r+1)-(k-(r-1)+1)}
= \frac{(k+1)x_0}{k-(r+1)+1}.
$$
Now choose $x_0,k,r$ appropriately to make $x_r$ rational.
