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, $$k_{m+1}=k_m\left(x_0+\frac{1}{x_0}\right)-k_{m-1}=n{k}_m-k_{m-1}(m\ge 2).\stackrel{◯}{1}$$ 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 $y=x$.