2 Promotion of Test Questions
Let the three sides of a triangle be integers $l, m, n$, and $l > m > n$. Given that $p$ is an odd prime, $M$ is a positive integer,
$$
\begin{array}{l}
(p, M)=1, \text{ and } \\
\left\{\frac{p^{l}}{M}\right\}=\left\{\frac{p^{m}}{M}\right\}=\left\{\frac{p^{n}}{M}\right\},
\end{array}
$$

where $\{x\}=x-[x]$, and $[x]$ denotes the greatest integer not exceeding $x$. Find the minimum perimeter of such a triangle.
Given, we have $p^{l} \equiv p^{m} \equiv p^{n}(\bmod M)$.
$$
p^{l} \equiv p^{m} \equiv p^{n}\left(\bmod p_{i}^{a_{i}}\right), i=1,2, \cdots, s .
$$

From equation (1), we get $p^{l-n} \equiv p^{m-n} \equiv 1\left(\bmod p_{i}^{o_{i}}\right)$.
Since $\left(p, p_{i}^{o_{i}}\right)=1$, according to Euler's theorem, we have
$$
p^{\varphi\left(p_{i}^{a_{i}}\right)} \equiv 1\left(\bmod p_{i}^{a_{i}}\right).
$$

Where $\varphi\left(p_{i}^{a_{i}}\right)=p_{i}^{a_{i}-1}\left(1-\frac{1}{p_{i}}\right)$.
Let $k_{i}$ be the smallest positive integer such that $p^{k} \equiv 1\left(\bmod p_{i}^{a_{i}}\right)$ holds, then it must be that $k_{i} \left\lvert\, p_{i}^{a_{i}-1}\left(1-\frac{1}{p_{i}}\right)\right., i=1,2, \cdots, s$.

Let the least common multiple of $k_{1}, k_{2}, \cdots, k_{s}$ be $k^{\prime}=\left[k_{1}, k_{2}, \cdots, k_{s}\right]$, then $k^{\prime}\left|(l-n), k^{\prime}\right|(m-n)$.

Let $l-n=k^{\prime} u, m-n=k^{\prime} v$, and $u>v$. Then the three sides of the triangle are $k^{\prime} u+n, k^{\prime} v+n, n$, and
$$
k^{\prime} u+n<\left(k^{\prime} v+n\right)+n.
$$

Therefore, $k^{\prime}(u-v)<n$.
To minimize the perimeter of the triangle, take $u=2, v=1$, $n=k^{\prime}+1$, at this time the perimeter of the triangle is
$$
\left(k^{\prime} u+n\right)+\left(k^{\prime} v+n\right)+n=k^{\prime}(u+v)+3 n.
$$