Saudi Arabia Mathematical Competitions

The sequence $x_n=\sqrt{f(n+1)}-\sqrt{f(n)}$, $n\ge 1$, contains only integers. We have $$\underset{n\to \infty }{\lim }{x}_n=\underset{n\to \infty }{\lim }\frac{f(n+1)-f(n)}{\sqrt{f(n+1)}+\sqrt{f(n)}}=\frac{2a}{2\sqrt{a}}=\sqrt{a}$$ It follows that $\sqrt{a}$ is an integer, hence $$y_n=\sqrt{f(n)}-n\sqrt{a}\in \mathbb{Z}$$ for every $n\ge 1$. We have $$\underset{n\to \infty }{\lim }{y}_n=\frac{bn+c}{\sqrt{f(n)}+n\sqrt{a}}=\frac{b}{2\sqrt{a}}$$ hence $\frac{b}{2\sqrt{a}}\in \mathbb{Z}$, and $y_n=\frac{b}{2\sqrt{a}}$ for any $n\ge n_0$. It follows $f(n)=(\sqrt{a}n+d{)}^2$ for any $n\ge n_0$, and we get $f=g^2$, where $g=\sqrt{a}X+d$.