110) Let $p$ be a given odd prime, and let the positive integer $k$ be such that $\sqrt{k^{2}-p k}$ is also a positive integer. Then $k=$ $\qquad$
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, let it be $m^{2}, m \in \mathbf{N}^{*}$, then $(m-2 n)(m+2 n)=p^{2}$.
Since $p$ is a prime number, and $p \geqslant 3$, we have $\left\{\begin{array}{l}m-2 n=1, \\ m+2 n=p^{2},\end{array}\right.$ solving this 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).