1. Let $p$ be a given odd prime, and the positive integer $k$ makes $\sqrt{k^{2}-p k}$ also a positive integer. Then $k=$ $\qquad$
(2004, National High School Mathematics Competition)
(Prompt: Let $\sqrt{k^{2}-p k}=n, n \in \mathbf{N}_{+}$, then $k^{2}-p k-n^{2}$ $=0, k=\frac{p \pm \sqrt{p^{2}+4 n^{2}}}{2}$. Hence $p^{2}+4 n^{2}$ is a perfect square. Let it be $m^{2}, m \in \mathbf{N}_{+}$, it is easy to get $m=\frac{p^{2}+1}{2}, n=\frac{p^{2}-1}{4}$. Therefore, $k=\frac{(p+1)^{2}}{4}$. )