9. Let $p$ be a prime number, and the number of distinct positive divisors of $p^{2}+71$ does not exceed 10. Find $p$.

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9. Solution: When $p=2$, $p^{2}+71=75=3 \times 5^{2}$, which has $(1+1)(2+1)=6$ positive divisors;

When $p=3$, $p^{2}+71=80=2^{4} \times 5$, which has $(4+1)(1+1)=10$ positive divisors. Therefore, $p=2$ and $p=3$ satisfy the condition.
When $p>3$, $p^{2}+71=(p-1)(p+1)+72$.
Since $p$ is an odd prime, $(p-1)$ and $(p+1)$ are two consecutive even numbers, so one of them must be a multiple of 2 and the other a multiple of 4, and one of them must be a multiple of 3, making $(p-1)(p+1)$ a multiple of 24.
Let $p^{2}+71=24 \times m=2^{3} \times 3 \times m$, where $m \geqslant 4$.
If $m$ has a prime factor other than $2$ and $3$, then the number of positive divisors of $p^{2}+71$ is $\geqslant(3+1)(1+1)(1+1)>10$;
If $m$ contains the prime factor 3, then the number of positive divisors of $p^{2}+71$ is $\geqslant(3+1)(2+1)>10$;
If $m$ contains only the prime factor 2, then the number of positive divisors of $p^{2}+71$ is $\geqslant(5+1)(1+1)>10$.
Therefore, $p>3$ does not satisfy the condition. In conclusion, the prime numbers $p$ that satisfy the condition are 2 or 3.