Question 204: Find all positive integers $\mathrm{n}$ such that $\mathrm{n}=\mathrm{p}_{1}^{2}+\mathrm{p}_{2}^{2}+\mathrm{p}_{3}^{2}+\mathrm{p}_{4}^{2}$, where $\mathrm{p}_{1}, \mathrm{p}_{2}, \mathrm{p}_{3}, \mathrm{p}_{4}$ are the 4 smallest distinct positive divisors of $\mathrm{n}$.
Question 204, Answer: If $n$ is odd, then $p_{1}, p_{2}, p_{3}, p_{4}$ are all odd, thus $n=p_{1}^{2}+p_{2}^{2}+$ $p_{3}^{2}+p_{4}^{2}$ is even, which is a contradiction. Therefore, $n$ is even.

When $n$ is even, then $p_{1}=1, p_{2}=2$, thus $n=p_{3}^{2}+p_{4}^{2}+5$. If $4 \mid n$, then $p_{3}^{2}+p_{4}^{2}+$ $5 \equiv 0(\bmod 4)$, which is impossible, so $n \equiv 2(\bmod 4)$. This implies that $p_{3}, p_{4}$ are one odd and one even, and thus $p_{4}=2 p_{3}$. This means that the smallest 4 positive divisors of $n$ can only be $1, 2, p_{3}, 2 p_{3}$. At this point, $n=5+5 p_{3}^{2}$, so $p_{3}=5, p_{4}=10$. Therefore, $n=1^{2}+2^{2}+5^{2}+10^{2}=130$.