Determine all real numbers $a$ such that there exists an infinite sequence of strictly positive real numbers $x_{0}, x_{1}, x_{2}, x_{3}, \ldots$ satisfying for all $n$ the equality

$$
x_{n+2}=\sqrt{a x_{n+1}-x_{n}} .
$$
If $a>1$, we can take the constant sequence equal to $a-1$. Suppose $a \leqslant 1$. Since the square root is well defined, we necessarily have $a x_{n+1}-x_{n}>0$, so already $a \geqslant 0$, and furthermore $x_{n+1} \geqslant a x_{n+1}>x_{n}$. Consequently, the sequence is strictly increasing.
Since $x_{n+1} \leqslant x_{n+2} \leqslant \sqrt{x_{n+1}-x_{n}}<\sqrt{x_{n+1}}$, we deduce that $x_{n}<1$ for all $n$. Moreover, $x_{n+1}^{2} \leqslant x_{n+1}-x_{n}$, so $x_{n+1} \geqslant x_{n+1}^{2}+x_{n} \geqslant x_{n} x_{0}+x_{n}=\left(1+x_{0}\right) x_{n}$, which shows that $x_{n} \geqslant \left(1+x_{0}\right)^{n} x_{0}$ for all $n$. This contradicts that $x_{n} \leqslant 1$.