10. Let $p$ be a given odd prime, and let the positive integer $k$ be such that $\sqrt{k-p k}$ is also a positive integer, then $k=$
$$
\frac{(p+1)^{2}}{4}
$$
10.【Analysis and Solution】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 $(m-2 n)(m+2 n)=p^{2}$.
$\because p$ is a prime number, and $p \geqslant 3, \therefore\left\{\begin{array}{l}m-2 n=1 \\ m+2 n=p^{2}\end{array}\right.$, solving gives $\left\{\begin{array}{l}m=\frac{p^{2}+1}{2} \\ n=\frac{p^{2}-1}{4}\end{array}\right.$.
$\therefore k=\frac{p \pm m}{2}=\frac{2 p \pm\left(p^{2}+1\right)}{4}$, hence $k=\frac{(p+1)^{2}}{4}$. (negative value discarded)