45th Mongolian Mathematical Olympiad

We will show that by induction method. For $n=1$ is trivial. Now suppose that $x_1,x_2,...,x_n,x_{n+1}$ are integers. From the given recurrence we get $x_n-x_{n+1}-x_{n+2}=2\sqrt{x_{n+1}\cdot x_{n+2}+3}$. Hence $2\sqrt{x_{n+1}\cdot x_{n+2}+3}$ is integer. This is showing us $x_{n+3}=x_{n+2}+x_{n+1}+2\sqrt{x_{n+1}\cdot x_{n+2}+3}$ is integer. Proof is completed.