Choose initial values such that sequence always has integer values

Question

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.

Answer 1

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 $x_1$.

Answer 2

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.