Saudi Arabian Mathematical Competitions, 2015

It is easy to prove by induction that $x_{n}$ is an integer for all $n \geq 1$. Therefore, we have $x_{n+1}=\left\lfloor(3+\sqrt{5}) x_{n}\right\rfloor$ for all $n \geq 1$, and we deduce that
$$\begin{aligned} x_{n+2} & =\left\lfloor(3+\sqrt{5}) x_{n+1}\right\rfloor=6 x_{n+1}+\left\lfloor-(3-\sqrt{5})\left\lfloor(3+\sqrt{5}) x_{n}\right\rfloor\right\rfloor \\ & =6 x_{n+1}+\left\lfloor-4 \frac{\left\lfloor(3+\sqrt{5}) x_{n}\right\rfloor}{(3+\sqrt{5})}\right\rfloor=6 x_{n+1}-4 x_{n} \end{aligned}$$
Indeed, for the last equality we have
$$(3+\sqrt{5}) x_{n}-1<\left\lfloor(3+\sqrt{5}) x_{n}\right\rfloor \leq(3+\sqrt{5}) x_{n},$$
from which we deduce
$$-4 x_{n} \leq-4 \frac{\left\lfloor(3+\sqrt{5}) x_{n}\right\rfloor}{(3+\sqrt{5})}<-4 x_{n}+\frac{4}{3+\sqrt{5}}<-4 x_{n}+1 .$$
Let $y_{n}=x_{n} x_{n+2}-x_{n+1}^{2}$, for all $n \geq 1$. We will prove by induction that $y_{n}=4^{n-1}$, for all $n \geq 1$.
For $n=1$, we have $x_{1}=1, x_{2}=\lfloor(3+\sqrt{5})\rfloor=5$ and $x_{3}=\lfloor 5(3+\sqrt{5})\rfloor=26$, and therefore $y_{1}=4^{0}$. Assume $y_{n}=4^{n-1}$ for some positive integer $n$. We have
$$\begin{aligned} y_{n+1} & =x_{n+1} x_{n+3}-x_{n+2}^{2}=x_{n+1}\left(6 x_{n+2}-4 x_{n+1}\right)-x_{n+2}\left(6 x_{n+1}-4 x_{n}\right) \\ & =4 y_{n}=4^{n} \end{aligned}$$
This ends the induction and solves the problem.