8. Given that $n$ is a positive integer, $P$ is a prime, $p n$ has exactly 8 positive divisors, which are arranged in ascending order as $1=d_{1}<d_{2}<$ $\cdots<d_{8}=p n$, and $d_{17 p-d_{3}}=\left(d_{1}+d_{2}+d_{3}\right)\left(d_{3}+d_{4}+13 p\right)$, then $n=$ $\qquad$ .
Solve (provided by Liang Tianxiang) 2021.
From $d_{17 p-d_{3}} \mid p n$ we have $d_{1}+d_{2}+d_{3} \mid p n$ and $d_{3}+d_{4}+13 p \mid p n$, thus $d_{1}+d_{2}+d_{3} \geq d_{4}, d_{3}+d_{4}+13 p \geq d_{5}$, and $d_{4} d_{5}=p n$ and $\left(d_{1}+d_{2}+d_{3}\right)\left(d_{3}+d_{4}+13 p\right) \leq p n$, so the aforementioned inequalities must all be equal, i.e.,
$$
\left\{\begin{array}{l}
d_{1}+d_{2}+d_{3}=d_{4} \\
d_{3}+d_{4}+13 p=d_{5} \\
17 p-d_{3}=8
\end{array}\right.
$$

From $d_{3}=17 p-8 \geq 9 p>p \Rightarrow p3$, thus $d_{2}=p=3$, from which we deduce $d_{4}=47, d_{5}=129$, hence $n=\frac{d_{4} d_{5}}{p}=2021$. It is easy to verify that $n=2021$ satisfies the conditions of the problem.