China Mathematical Competition (Hainan)

Set $\sqrt{k^2 - pk} = n$, $n \in \mathbb{N}$. Thus $k^2 - pk - n^2 = 0$, and $k = \frac{p \pm \sqrt{p^2 + 4n^2}}{2}$, which implies that $p^2 + 4n^2$ is a perfect square, say $m^2$, where $m \in \mathbb{N}$. So $(m-2n)(m+2n) = p^2$.

Since $p$ is a prime and $p \ge 3$, we have
$$\begin{cases} m - 2n = 1, \\ m + 2n = p^2. \end{cases}$$
Solve the equations above and we get
$$\begin{cases} m = \frac{p^2 + 1}{2}, \\ n = \frac{p^2 - 1}{4}. \end{cases}$$

Consequently, $k = \frac{p \pm m}{2} = \frac{2p \pm (p^2 + 1)}{4}$. Thus $k = \frac{(p+1)^2}{4}$ (the negative value is omitted).