You have done the optimization of removing multiples of 2 (by incrementing by 2).

There is another slight improvement that removes multiples of 2 and 3 from the test.

bool IsPrime (int n) {
   if (n == 2)     return true;
   if (n == 3)     return true;
   if (n < 5)      return false;
   if (n % 2 == 0) return false;
   if (n % 3 == 0) return false;

   int inc = 2;
   int dec = 1;
   for (int i = 5; i <= int(sqrt(n)) + 1; i += inc)
   {
       if (n % i == 0)
               return false;
       inc = inc + (2 * dec);  // makes inc go 2/4/2/4/2/4/.....
       dec = dec * -1;
   }
   return true;
}

Another optimization for IsPrime() is that you only need to test using primes. ie no pointing in testing with 6 (its not prime) as all values that are divisible by 6 are also divisible by 3 (is prime). So if your loop only tested by using other primes:

// Assumes you are calling this function with monotimically inreassing
// values of n so that the `foundPrimes` vector will be constructed in order.
bool IsPrime (int n) {

   static std::vector<int>   foundPrimes = { 2, 3, 5, 7, 11 };

   for(std::size_t loop = 0; loop <  foundPrimes.size(); ++loop)
   {
       if (n == foundPrimes[loop])     { return true;}
       if (n % foundPrimes[loop] == 0) { return false;}
   }
   start = foundPrimes.last() + 2;

   for (int i = start; i <= int(sqrt(n)) + 1; i += 2)
   {
       if (n % i == 0)
               return false;
   }
   foundPrimes.push_back(n);
   return true;
}