Let's find those prime numbers $p$ for which the number $p^{2}+11$ has exactly 6 positive divisors.
I. solution. We know that if the prime factorization of a positive integer $n$ is $p_{1}^{\alpha_{1}} \cdot p_{2}^{\alpha_{2}} \cdots \cdots p_{n}^{\alpha_{n}}$, then the number of positive divisors of the number is: $d(n)=\left(\alpha_{1}+1\right)\left(\alpha_{2}+1\right) \ldots\left(\alpha_{n}+1\right)$.

A number will have exactly 6 different divisors if it is either
a) of the form $p_{1}^{5}$, where $p_{1}$ is a prime number, or
b) of the form $p_{1}^{2} \cdot p_{2}$, where $p_{1}, p_{2}$ are different prime numbers.

If $p=2$, then $N=p^{2}+11=15=3 \cdot 5$; it does not satisfy either condition, so $p$ must be an odd prime number.

If $p=3$, then $N=3^{2}+11=20=2^{2} \cdot 5$ satisfies condition b); 20 indeed has 6 different positive divisors: $1,2,4,5,10$ and 20.

If $p$ is odd and not 3, then its square gives a remainder of 1 when divided by 4, since $p^{2}=(2 k+1)^{2}=4 k^{2}+4 k+1$. Therefore, $N=p^{2}+11$ is divisible by 4. If now $N=p_{1}^{5}$, then $p_{1}$ can only be 2, but from $p^{2}+11=32$ we do not get an integer value for $p$. Therefore, $N$ cannot be of the form $p_{1}^{5}$. Thus, $N$ must be of the form $p_{1}^{2} p_{2}$, and due to its divisibility by 4, $p_{1}=2$. But $N$ is also divisible by 3, since $p^{2}$ gives a remainder of 1 when divided by 3 (since $p$ is either of the form $3 l+1$ or $3 l-1$), so $p_{2}=3$. Thus, $N=p^{2}+11=2^{2} \cdot 3=12$, and from this $p=1$, which is not a prime number.

Therefore, the only solution to the problem is $p=3$.

Lajos Gáthy (Fehérgyarmat, Zalka M. Gymnasium, 9th grade)

II. solution. Notice that

$$
p^{2}+11=(p-1)(p+1)+12
$$

If $p>3$, then both $p-1$ and $p+1$ are even and one of them is also divisible by 3, so their product is divisible by 12.

Moreover, $p^{2}+11>12$, so it has at least 7 different divisors, which are $1,2,3,4$, 612 and $p^{2}+11$. Therefore, $p$ cannot be greater than 3.

If $p=2$, then $p^{2}+11=15$, which has only 4 different positive divisors.

If $p=3$, then $p^{2}+11=20$, and as we have already seen, it has 6 different positive divisors, which is the only solution.