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 $$\{\begin{array}{l}m-2n=1,\\ m+2n=p^2.\end{array}$$ Solve the equations above and we get $$\{\begin{array}{l}m=\frac{p^2+1}{2},\\ n=\frac{p^2-1}{4}.\end{array}$$

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).