Find the sum of primes p such that p²+11 has exactly six different positive divisors (including 1 and the number itself).

Answer 1

if the prime factors of p²+1, excluding 1 and itself, are of the form {a²,b | a≠b}, then the possible factors are:
{1,a,b,a², ab, a²b} total exactly 6 factors.

The first few such sums are:
1²+11=12=2²*3 [but 1 is not a prime]
3²+11=20=2²*5 [cannot be the sum of primes]
8²+11=75=3*5² factors are {1,3,5,15,25,75}, and 8=3+5, both primes.
9²+11=92=2²*23; factors are {1,2,4,23,46,92}, and 9=2+2+5
14²+11=207=3²*23; factors are {1,3,9,23,69,207}, and 14=2+5+7
15²+11=236=2²*59; factors are {1,2,4,59,118,236}, and 15=3+5+7
...
I guess we can go on and on.