Mongolian Mathematical Olympiad

$x_{n+1} = 23x_n + y_n + 2$, $y_{n+1} = 551x_n + 24y_n + 64$. Multiplying the first equation by $24$ then subtracting these two, we get $y_{n+2} = 24x_{n+1} - x_n + 18$. And we have $x_{n+2} = 23x_{n+1} + y_{n+1} + 2$. Adding last two equations, we have
$$x_{n+2} = 47x_{n+1} - x_n + 1 \quad (*)$$
Thus, $x_{n+1} = 47x_n - x_{n-1} + 18$. Subtracting these two equations, we get
$$x_{n+2} = 48x_{n+1} - 48x_n + x_{n-1} \quad (**).$$
Let $a_n = \sqrt{x_n}$. It suffices to prove that $a_n$ is integer for $n \in \mathbb{N}$. From $x_1 = 1$ and $y_1 = 39$, we get $x_2 = 64$ and $x_3 = 3025$. Then $(*)$ is equivalent to
$$a_{n+2}^2 = 48a_{n+1}^2 - 48a_n^2 + a_{n-1}^2.$$
From this,
$$a_{n+2}^2 - (49a_{n+1}^2 - 14a_{n+1}a_n + a_n^2) = a_{n-1}^2 - (49a_n^2 - 14a_n a_{n+1} + a_{n+1}^2)$$
and
$$a_{n+2}^2 - (7a_{n+1} - a_n)^2 = a_{n-1}^2 - (7a_n - a_{n+1})^2, \quad n \in \mathbb{N}.$$
Furthermore,
$$(a_{n+2} + a_n - 7a_{n+1})(a_{n+2} + 7a_{n+1} - a_n) = (a_{n-1} - 7a_n + a_{n+1})(a_{n-1} + 7a_n - a_{n+1}).$$
Let $b_n = a_{n+2} + a_n - 7a_{n+1}$, then $b_1 = a_3 + a_1 - 7a_2 = \sqrt{3025} + \sqrt{1} - 7\sqrt{64} = 0$. $b_n(2a_{n+2} - b_n) = b_{n-1}(2a_{n-1} - b_{n-1})$ holds for $n \in \mathbb{N}$ $(**)$. Applying for $n = 2$, we get $b_2(2a_1 - b_2) = b_1(2a_1 - b_1) = 0$. Thus, $b_2 = 0$ or $2a_4 - b_2 = 0$. Since $x_n$ is monotonically increasing, $a_n$ is monotonically increasing too. Hence $2a_{n+2} - b_n = a_{n+2} + 7a_{n+1} - a_n > 0$. Therefore, $b_2 = 0$. Applying $(***)$ for $n \in \mathbb{N}$, $b_n = 0$. Hence $a_{n+2} = 7a_{n+1} - a_n$, for $n \in \mathbb{N}$. Since $a_1$ and $a_2$ are integers, $a_n$ is integer for $n \in \mathbb{N}$.