2. Find all prime numbers $p$ for which the number $p^{2}+11$ has fewer than 11 positive divisors.
I/2. If $p=2$, then $p^{2}+11=15=3 \cdot 5$ and this number has 4 divisors (1, 3, 5, and 15). If $p=3$, then $p^{2}+11=17$ and it has 2 divisors. For $p>3$, the prime number $p$ is odd, so $p^{2}$ leaves a remainder of 1 when divided by 4, and the number $p^{2}+11$ is divisible by 4. The prime number $p>3$ is not divisible by 3, so $p^{2}$ leaves a remainder of 1 when divided by 3, and the number $p^{2}+11$ is also divisible by 3. Thus, $p^{2}+11=2^{2} \cdot 3 \cdot a$. If $p \geq 11$, then $p^{2}+11 \geq 132$ and $a>11$, so the divisors of the number $p^{2}+11$ include at least the numbers $1,2,3$, $4,6,12, a, 2 a, 3 a, 4 a, 6 a$ and $12 a$, meaning there are at least 11 different divisors (since $a>11$, it can indeed be $a=12$). We must also calculate specifically for $p=5$, where we get $p^{2}+11=36=2^{2} \cdot 3^{2}$, which has 6 divisors, and for $p=7$, where we get $p^{2}+11=60=2^{2} \cdot 3 \cdot 5$, which has 12 divisors. Only for the prime numbers 2, 3, and 5 does the number $p^{2}+11$ have fewer than 11 positive divisors.