Find $\beta$'s for which there is a sequence of integers with $x_{n+2} = \sqrt{\beta x_{n+1}-x_n}$ 
I would like to find all $\beta \in \mathbf{R}_{>0}$ such that there exist a sequence $(x_i)_{i \in \mathbf{N}}$ such that : 
  $$x_{n+2} = \sqrt{\beta x_{n+1}-x_n}, x_n > 0,\; \forall\, n.$$

I really don't know how to proceed yet here is what I've noticed so far :


*

*$\beta = 1$ work because the sequence $x_0 = 1$, $x_1 = 2$, $x_2 = 1$ works. 

*if $\beta \in \mathbf{N}$ then we must have $\beta \mid (x_{n+2}^2+x_n)$.


EDIT : As Luis noticed the first point is incorrect.
 A: Your point (2), that "if" $\beta \in \mathbb{N}$ it must divide $(x_{n+2}^2 + x_n)$, is the strongest part. We have that $\beta = \frac{(x_{n+2}^2 + x_n)}{x_{n+1}}$ so clearly $\beta$ is rational. Write $\beta = p/q$, where $p$ and $q$ are coprime. Since the numerator of $\frac{(x_{n+2}^2 + x_n)}{x_{n+1}}$ is integer, we know that $x_{n+1}$ must be a multiple of $q$. Since this is true for every $n$ (except possibly $n=0$), we have that every integer in the sequence is a multiple of $q$.
So, let's take two elements in the sequence, $x_n$ and $x_{n+1}$. Pull out all the factors of $q$ possible, so $x_n = aq^b$ and $x_{n+1} = cq^d$, where $q$ is not a factor of $a$ or $c$. Then
$$x_{n+2}^2 = \frac{p}{q}cq^d - aq^b = cpq^{d-1} - aq^b$$
Since the left hand side has at least 2 factors of $q$, we must have at least 2 factors of $q$ on the right as well. If $d=1$, then the right hand side is the sum of a non-multiple of $q$, that is $cp$, and a multiple of $q$; a contradiction. If $d=2$, then we must have $b=1$: otherwise we'd be adding a multiple of only $q$ and a multiple of $q^2$ to get a multiple of $q^2$. So either $d\ge 3$ or ($d=2$ and $b=1$).
But we just showed that every term must have at least two factors of $q$: we need $d\ge 2$. Since this applies to each term just as well, we must also have $b\ge2$. This immediately eliminated the ($d=2$ and $b=1$ possibility), so $d\ge 3$.
But now, induction! If the third term has $k$ factors of $q$, then the left side has $2k$ factors of $q$, so either ($d-1 \ge 2k$ and $b \ge 2k$ and so they add something with $2k$ factors), or ($k <= b = d-1 < 2k$ so that they have $b$ factors already and can add more factors through $cp-a$). This is the same logic as before, just more general. So, again, either $d-1 \ge k$ or $d-1 \ge 2k$: in other words, $d > k$ either way, so the divisor in $x_{n+1}$ must be greater than in $x_{n+2}$. This means that as we go forward in the sequence, the amount of factors of $q$ must strictly decrease. Since at some point we must drop below having any factors of $q$ at all, and they will no longer be rational. So, punchline:
$\beta$ must be an integer.
From here, it's straightforward. If $\beta = 1$, we must have $x_{n+1} > x_n$ so that the square root is positive. But $x_{n+2} = \sqrt{x_{n+1} - x_n} \le \sqrt{x_{n+1}} \le x_{n+1}$, a contradiction, so no solutions there. And we know that if $\beta > 1$, there is the solution with fixed point $x_n = \beta - 1$.
Are there other solutions for $\beta > 1$? It is clear that the sequence must be decreasing if the last term (we'll say $x_{n+1}$) is anywhere above $\beta$, since
$$x_{n+1} > \beta$$
$$\implies x_{n+2}/x_{n+1} = \frac{\sqrt{\beta x_{n+1} - x_n}}{x_{n+1}} \le \frac{\sqrt{\beta x_{n+1}}}{x_{n+1}} = \sqrt{\beta / x_{n+1}} < \sqrt{\beta / \beta} = 1$$
$$ \implies x_{n+2} < x_{n+1}$$
This means that each sequence must eventually get down to at most $\beta$, and then it will never go above that level. Can the sequence oscillate in the range $[1,\beta]$? Once we're at that 'trapped' stage, we know that $x_n \le \beta$. So, assume $x_{n+1} < \beta - 1$:
$$x_{n+2}/x_{n+1} = \frac{\sqrt{\beta x_{n+1} - x_n}}{x_{n+1}} \ge \frac{\sqrt{\beta x_{n+1} - \beta}}{x_{n+1}} = \frac{\sqrt{\beta}\sqrt{x_{n+1} - 1}}{x_{n+1}} > \frac{\sqrt{\beta}\sqrt{x_{n+1} - 1}}{x_{n+1} - 1} $$
$$  = \frac{\sqrt{\beta}}{\sqrt{x_{n+1} - 1}} > \frac{\sqrt{\beta}}{\sqrt{\beta - 2}} > 1$$
so that the sequence must be increasing towards $\beta-1$. This essentially confirms what Sangchul conjectured about the exponential attraction towards $\beta-1$.
So, this shows that every sequence must eventually permanently dip below $\le \beta$, and then must monotonically increase up towards $\beta-1$. This means it must eventually reach a sequence of only $\beta$ and $\beta-1$. Either it's the all-$\beta-1$ sequence, which we already found; or it's an all-$\beta$ sequence, which obviously doesn't work algebraically; or it's an alternating-$(b-1)$-and-$\beta$ sequence, which also doesn't work. So, eventually every sequence of integers must become the all-$beta-1$ sequence. Since we can also calculate thee sequence in reverse, computing $x_n$ from $x_{n+1}$ and $x_{n+2}$, any sequence that ends in that constant sequence must have always been the constant sequence. So, that's it!
