* Find all positive integers $m, n$ and prime $p$ such that $\mathrm{C}_{m}^{2}-1=p^{n}$.

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From the given, we have $(m+1)(m-2)=2 p^{n}$. (1) Since $(m+1, m-2)=(m+1,3)=1$ or 3, there are the following cases:
(1) $(m+1, m-2)=1$. In this case, since $m+1>m-2$,
$$
\left\{\begin{array}{l}
m+1=2 p^{n}, \\
m-2=1
\end{array} \text { or } \left\{\begin{array}{l}
m+1=p^{n}, \\
m-2=2
\end{array}\right.\right.
$$

The former gives $m=3, p=2, n=1$. The latter gives $m=4, p=5, n=1$.
(2) $(m+1, m-2)=3$. In this case, $p=3$, and

The former gives $m=5, n=2$. The latter gives $m=8, n=3$.
Therefore, there are four solutions: $(m, n, p)=(3,1,2),(4,1,5),(5,2,3),(8,3,3)$.