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

保留了源文本的换行和格式。
7. $p=2$ or 3

Analysis: If $p=2$, then $p^{2}+71=75=3 \times 5^{2}$, the number of divisors is 6;
If $p=3$, then $p^{2}+71=80=2^{4} \times 5$, the number of divisors is 10;
If $p \geq 5$, then $p$ is a prime of the form $6k \pm 1$, i.e., $p^{2}+71=36k^{2} \pm 12k+72 \equiv 0(\bmod 24)$, hence the prime factorization of $p^{2}+71$ must include 2 and 3, and the exponent of 2 is no less than 3. If $p^{2}+71=2^{3} \times 3 \times p_{1}$, since the number of divisors is no more than 10, the distinct prime factors of $p^{2}+71$ can only be 2 and 3; thus $p^{2}+71=2^{3} \times 3, 2^{4} \times 3$, as $p^{2}+71 \geq 96$, there is no solution in this case. In summary, $p=2$ or 3.