Primes

Note that:

1) Since $p^2+11$ has six divisors, it has to be of the form $k_1^2k_2$ , where $k_1$ , $k_2$ are primes.

2) $3$ is a solution, because $3^2+11=20$ , which has six divisors ( $1, 2, 4, 5, 10, 20)$

3) $2$ is not a solution (easy to check), which implies $p$ is odd, and when we add $11$ , the number becomes even. Thus, we can conclude that either $k_1$ or $k_2$ is $2$ .

Assume $p\not=3$ . Then, $p\equiv1\pmod{3}$ or $p\equiv2\pmod{3}$ . Either way, $p^2\equiv1\pmod{3}$ , and when summed to $11$ , the result will be divisible by $3$ . We can conclude that either $k_1$ or $k_2$ is $3$ .

Since the pairs $(k_1,k_2)=(2,3)$ and $(k_1,k_2)=(3,2)$ don't yield any solutions in primes, we conclude that either $k_1$ or $k_2$ has to be $3$ and $2$ at the same time, which is absurd.

Thus, the only solution is $p=3$ , and so, the sum of the solutions is also $3$ .