2. What is the smallest number of natural divisors that the number $p^{2}+2011$ can have for a prime $p$?
Answer: 6. Solution. First, note that the number $2^{2}+2011=2015=5 \cdot 13 \cdot 31$ has 8 natural divisors, which is more than 6. Let $p$ be odd. Then $p^{2}+2011$ is necessarily divisible by 4, but not by 8. Therefore, $p^{2}+2011=4 m=2^{2} m$, where $m$ is odd. This means the number of divisors is no less than 6, and it will be exactly 6 if $m$ is prime. This is the case, for example, when $p=5$; then $5^{2}+2011=2036=4 \cdot 509$.