Zhendarov R.G.

Find all such prime numbers $p$ that the number $p^{2}+11$ has exactly six distinct divisors (including one and the number itself).
Note that $p^{2}+11=(p-1)(p+1)+12$. If $p \geq 5$ and is prime, then the numbers $p-1$ and $p+1$ are even, and one of them is divisible by 3. Therefore, the product $(p-1)(p+1)$ is divisible by 12, hence $p^{2}+11$ is also divisible by 12, which means it has at least seven divisors (6 divisors of the number 12 and the number $p^{2}+11>12$ itself). It remains to check $p=2$ and $p=3$. For $p=2$, the number $p^{2}+11=2^{2}+11=15$ has four divisors $(1,3,5,15)$; for $p=3$, the number $p^{2}+$ $11=3^{2}+11=20$ has six divisors $(1,2,4,5,10,20)$.

## Answer

$p=3$.