1. (2004 National High School Competition Question) Let $p$ be a given odd prime, and let $k$ be a positive integer such that $\sqrt{k^{2}-p k}$ is also a positive integer, then $k=$ $\qquad$ .
1. Let $\sqrt{k^{2}-p k}=n, n \in \mathbf{N}^{*}$, then $k^{2}-p k-n^{2}=0, k=\frac{p \pm \sqrt{p^{2}+4 n^{2}}}{2}$, thus $p^{2}+4 n^{2}$ is a perfect square,

set it as $m^{2}, m \in \mathbf{N}^{*}$, then $(m-2 n)(m+2 n)=p^{2}$.
Therefore, $k=\frac{p \pm m}{2}=\frac{2 p \pm\left(p^{2}+1\right)}{4}$, so $k=\frac{(p+1)^{2}}{4}$ (negative value discarded).