$12 \cdot 117$ Find all positive integers $m, n$ and prime $p$ such that $C_{m}^{2}-1=p^{n}$.
(Canadian Mathematical Olympiad Training Problem, 1992)
\[ \text{Solution:} \] Given the equations can be transformed into
\[
\begin{array}{c}
\frac{m(m-1)}{2}-1=p^{n} \\
\frac{(m+1)(m-2)}{2}=p^{n}
\end{array}
\]
(1) If \( m \) is odd. Since \( p \) is a prime number,
then for some integer \( t, 0 \leqslant t \leqslant n \), we have
\[
\left\{\begin{array}{l}
\frac{m+1}{2}=p^{t} \\
m-2=p^{n-t}
\end{array}\right.
\]

Eliminating \( m \) from (1) and (2) gives
\[
p^{n-t}+3=2 p^{t}
\]

When \( t=0 \), we have
\[
p^{n}+3=2 \text{.}
\]

This is clearly impossible. Therefore,
\[
t>0 \text{.}
\]

When \( n-t>0 \), from (3), the prime number \( p \) is a divisor of 3, so \( p=3 \).
At this point, equation (3) becomes
\[
3^{n-t-1}+1=2 \cdot 3^{t-1} \text{.}
\]

If \( t-1>0 \), then the right-hand side of (4) is a multiple of 3, while the left-hand side is not divisible by 3, which is impossible.
If \( t-1=0 \), then we have
\[
\begin{array}{l}
3^{n-t-1}+1=2 \\
n-t-1=0
\end{array}
\]

This gives \( t=1, n=2, m=p^{n-t}+2=5 \).
When \( n-t=0 \), equation (3) becomes
\[
1+3=2 p^{t} \text{.}
\]

Thus,
\[
p=2, n=t=1, m=3 \text{.}
\]
(2) If \( m \) is even, since \( p \) is a prime number,

then for some integer \( t, 0 \leqslant t \leqslant n \), we have
\[
\left\{\begin{array}{l}
m+1=p^{t} \\
\frac{m-2}{2}=p^{n-t}
\end{array}\right.
\]

Eliminating \( m \) from (5) and (6) gives
\[
p^{t}-3=2 p^{n-t} \text{.}
\]

Clearly, \( t>0 \), otherwise the left-hand side is negative and the right-hand side is positive, making (7) impossible. If \( n-t>0 \), then from (7), \( p \) is a divisor of 3, and since \( p \) is a prime, \( p=3 \). At this point, equation (7) becomes
\[
3^{t-1}-1=2 \cdot 3^{n-t-1} \text{.}
\]

Thus, \( n-t-1=0, t-1=1 \), i.e.,
\[
t=2, n=3, m=8 \text{.}
\]

If \( n-t=0 \), then equation (7) becomes
\[
p^{t}-3=2 \text{.}
\]

Thus,
\[
p=5 \text{.}
\]

At this point, \( t=1, n=t=1, m=4 \).
From the above, the equation has only four solutions:
\[
\left\{\begin{array}{l}
m=3, \\
n=1, \\
p=2,
\end{array} \quad \left\{\begin{array}{l}
m=5, \\
n=2, \\
p=3,
\end{array} \left\{\begin{array}{l}
m=8, \\
n=3, \\
p=3,
\end{array} \left\{\begin{array}{l}
m=4, \\
n=1, \\
p=5.
\end{array}\right.\right.\right.\right.
\]