10. Let $p$ be a given odd prime. If the positive integer $k$ makes $\sqrt{k^{2}-p k}$ also a positive integer, then $k=$ $\qquad$
Let $k^{2}-p k=n^{2} \Rightarrow k^{2}-p k-n^{2}=0 \Rightarrow k=\frac{p \pm \sqrt{p^{2}+4 n^{2}}}{2}$. Since $k \in \mathbf{N}^{*}$, then $p^{2}+4 n^{2}$ is a perfect square. Let $p^{2}+4 n^{2}=m^{2} \Rightarrow m^{2}-4 n^{2}=p^{2}$ $\Rightarrow(m+2 n)(m-2 n)=p^{2} \Rightarrow\left\{\begin{array}{l}m+2 n=p^{2}, \\ m-2 n=1\end{array} \Rightarrow\left\{\begin{array}{l}m=\frac{p^{2}+1}{2}, \\ n=\frac{p^{2}-1}{4} .\end{array}\right.\right.$
$=\frac{p \pm\left(\frac{p^{2}+1}{2}\right)}{2}=\frac{2 p \pm\left(p^{2}+1\right)}{4}=\frac{(p+1)^{2}}{4}($ negative value discarded $)$.