Let's find all prime numbers for which $p^{3}+p^{2}+11 p+2$ is also prime.
Every prime number, with the exception of 3, is either of the form $x+1$ or $x+2$, where $x$ is a multiple of 3. Thus, we distinguish three cases:
a) $p=3$. In this case, $p^{3}+p^{2}+11 p+2=71$, which is a prime number.
b) $p=x+1$, and $x$ is divisible by 3. Then

$$
p^{3}+p^{2}+11 p+2=(x+1)^{3}+(x+1)^{2}+11(x+1)+2=x^{3}+4 x^{2}+16 x+15
$$

Since $x$ is divisible by 3, $x^{3}+4 x^{2}+16 x$ is also divisible by 3, and 15 is as well. Thus, the expression in question is divisible by 3. Furthermore, since $x>0$, its value is at least 15, so it cannot be exactly 3, and a number greater than 3 and divisible by 3 is certainly not prime.
c) $p=x+2$ and $x$ is divisible by 3. Then

$$
p^{3}+p^{2}+11 p+2=(x+2)^{3}+(x+2)^{2}+11(x+2)+2=x^{3}+7 x^{2}+27 x+36
$$

This expression is also divisible by 3, and its value is at least 36, so it is certainly not a prime number.

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

Gábor Elek (Budapest, Váci utcai Ált. Isk., 6th grade)