My code

… was written in C++11 and can be compiled with G++, Clang++, Visual C++. You can download it, too. Or just jump to my GitHub repository.

       
#include <iostream>
       
#include <vector>
       
 
       
// odd prime numbers are marked as "true" in a bitvector

       
std::vector<bool> sieve;
       
 
       
// return true, if x is a prime number

       
bool isPrime(unsigned int x)
       
{
       
  // handle even numbers
       
  if ((x & 1) == 0)
       
    return x == 2;
       
 
       
  // lookup for odd numbers
       
  return sieve[x >> 1];
       
}
       
 
       
// find all prime numbers from 2 to size

       
void fillSieve(unsigned int size)
       
{
       
  // store only odd numbers
       
  const unsigned int half = size >> 1;
       
 
       
  // allocate memory
       
  sieve.resize(half, true);
       
  // 1 is not a prime number
       
  sieve[0] = false;
       
 
       
  // process all relevant prime factors
       
  for (unsigned int i = 1; 2 * i*i < half; i++)
       
    // do we have a prime factor ?
       
    if (sieve[i])
       
    {
       
      // mark all its multiples as false
       
      unsigned int current = 3 * i + 1;
       
      while (current < half)
       
      {
       
        sieve[current] = false;
       
        current += 2 * i + 1;
       
      }
       
    }
       
}
       
 
       
int main()
       
{
       
  unsigned int limit = 100000000;
       
  std::cin >> limit;
       
 
       
  fillSieve(limit + limit/2);
       
 
       
  // 1 is the only odd number where i+1 is prime
       
  unsigned long long sum = 1;
       
 
       
  // scan only even numbers >= 2 of the form 2+4k
       
  for (unsigned int n = 2; n <= limit; n += 4)
       
  {
       
    // quick check for divisor = 1
       
    if (!isPrime(1 + n))
       
      continue;
       
    // n is always divisible by 2, perform a second quick check for divisor = 2
       
    if (!isPrime(2 + n/2))
       
      continue;
       
 
       
    // check all potential divisors 3 ... sqrt(n)
       
    bool valid = true;
       
    for (unsigned int divisor = 3; divisor*divisor <= n; divisor++)
       
    {
       
      // is it a divisor ?
       
      if (n % divisor != 0)
       
        continue;
       
 
       
      // d + n/d
       
      if (!isPrime(divisor + n / divisor))
       
      {
       
        valid = false;
       
        break;
       
      }
       
 
       
      // other divisor is n/d:
       
      // n/d + n/(n/d)
       
      // = n/d + d => same as above
       
    }
       
 
       
    // found another value
       
    if (valid)
       
      sum += n;
       
  }
       
 
       
  // display result
       
  std::cout << sum << std::endl;
       
  return 0;
       
}

This solution contains 15 empty lines, 23 comments and 2 preprocessor commands.