26. Find all positive integers $m, n$ and prime $p$ such that $\binom{m}{2}-1=p^{\mathrm{n}}$ holds.
26. There is $\frac{(m+1)(m-2)}{2}=p^{n}$. If $m$ is an odd number, then for some integer $t, 0 \leqslant t \leqslant n$, we have $\frac{m+1}{2}=p^{\mathrm{t}}$ and $m-2=p^{n-t}$, thus $p^{n-t}+3=2 p^{t}$. Clearly, $t>0$.
When $n-t>0$, then it must be that $p=3$, hence $3^{n-t-1}+1=2 \cdot 3^{t-1}$.
From $n-t-1=0$, we can deduce $t-1=0$. Conversely, it is also true. Therefore, $t=1, n=2, m=5$.
When $n-t=0$, then $1+3=2 p^{t}$. Thus, $p=2, n=t=1, m=3$.
If $m$ is even, then for some integer $t, 0 \leqslant t \leqslant n$, we have $m+1=p^{t}$ and $\frac{m-2}{2}=p^{n-t}$, i.e., $p^{t}-3=2 p^{n-t}$. Clearly, $t>0$. Similarly, we get $p=3$, $t=2, n=3, m=8$; and $p=5, n=t-1, m=4$. In total, there are only four solutions:
$$
\begin{aligned}
(m, n, p)= & (3,1,2),(4,1,5), \\
& (5,2,3),(8,3,3) .
\end{aligned}
$$