China Mathematical Competition

Since $M_0(x_0, y_0)$ is an intersection point of $y^2 = nx - 1$ and $y = x$, we get $x_0 = y_0 = \frac{n \pm \sqrt{n^2 - 4}}{2}$. Then obviously $x_0 + \frac{1}{x_0} = n$.

Let $(x_0^m, y_0^m)$ be an intersection point of $y^2 = kx - 1$ and $y = x$. Then we get
$$k = x_0^m + \frac{1}{x_0^m}.$$
We denote $k_m = x_0^m + \frac{1}{x_0^m}$. Then,
$$k_{m+1} = k_m \left(x_0 + \frac{1}{x_0}\right) - k_{m-1} = n k_m - k_{m-1} \quad (m \ge 2). \quad \textcircled{1}$$
Since $k_1 = n$ is an integer,
$$k_2 = x_0^2 + \frac{1}{x_0^2} = \left(x_0 + \frac{1}{x_0}\right)^2 - 2 = n^2 - 2$$
is also an integer. Then, by the principle of mathematical induction and ①, we conclude that for any positive integer $m$, $k_m = x_0^m + \frac{1}{x_0^m}$ is a positive integer too. Let $k = x_0^m + \frac{1}{x_0^m}$. So $(x_0^m, y_0^m)$ is an intersection point of $y^2 = kx - 1$ and $y = x$.