Saudi Arabian Mathematical Competitions

The sequence $x_{n} = \sqrt{f(n+1)} - \sqrt{f(n)}$, $n \geq 1$, contains only integers. We have
$$\lim_{n \rightarrow \infty} x_{n} = \lim_{n \rightarrow \infty} \frac{f(n+1) - f(n)}{\sqrt{f(n+1)} + \sqrt{f(n)}} = \frac{2a}{2\sqrt{a}} = \sqrt{a}$$
It follows that $\sqrt{a}$ is an integer, hence
$$y_{n} = \sqrt{f(n)} - n \sqrt{a} \in \mathbb{Z}$$
for every $n \geq 1$. We have
$$\lim_{n \rightarrow \infty} y_{n} = \frac{b n + c}{\sqrt{f(n)} + n \sqrt{a}} = \frac{b}{2\sqrt{a}}$$
hence $\frac{b}{2\sqrt{a}} \in \mathbb{Z}$, and $y_{n} = \frac{b}{2\sqrt{a}}$ for any $n \geq n_{0}$. It follows $f(n) = (\sqrt{a} n + d)^{2}$ for any $n \geq n_{0}$, and we get $f = g^{2}$, where $g = \sqrt{a} X + d$.