My code

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

The code contains #ifdefs to switch between the original problem and the Hackerrank version.
Enable #ifdef ORIGINAL to produce the result for the original problem (default setting for most problems).

       
#include <iostream>
       
#include <cmath>
       
 
       
//#define ORIGINAL

       
 
       
int main()
       
{
       
  unsigned int tests;
       
  std::cin >> tests;
       
 
       
  while (tests--)
       
  {
       
    unsigned long long minimum = 1000000000000ULL;
       
 
       
    unsigned long long p = 1;
       
    unsigned long long q = 2;
       
    std::cin >> p >> q >> minimum;
       
 
       
    unsigned long long blue = 0;
       
    unsigned long long red  = 0;
       
 
       
    // special code for p/q = 1/2
       
    if (p == 1 && q == 2)
       
    {
       
      blue = 15;
       
      red  =  6;
       
 
       
      // keep going until limit is reached
       
      while (blue + red < minimum)
       
      {
       
        // at first I brute-forced the first solutions and wrote them down
       
        // then I saw the following relationship for p/q = 1/2:
       
        //  red(i+1) = 2 * blue(i) + red(i) - 1;
       
        // blue(i+1) = 2 * red(i+1)
       
        red   = 2 * blue + red - 1; // seems to be true for most p/q
       
        blue += 2 * red;            // but this line is not correct for p/q != 1/2
       
      }
       
#ifdef ORIGINAL
       
      std::cout << blue << std::endl;
       
#else
       
      std::cout << blue << " " << (red + blue) << std::endl;
       
#endif
       
      continue;
       
    }
       
 
       
    // brute-force smallest solution
       
    bool found = false;
       
    for (blue = 2; blue < 100000; blue++)
       
    {
       
      // sum = red + blue
       
      // blue * (blue - 1) / (sum * (sum - 1)) = p / q
       
      // blue * (blue - 1) * q / p = sum * (sum - 1)
       
      unsigned long long b2 = blue * (blue - 1);
       
      b2 *= q;
       
      // right side must be an integer
       
      if (b2 % p != 0)
       
        continue;
       
      unsigned long long sum2 = b2 / p; // sum2 = sum * (sum - 1)
       
 
       
      // (sum-1)^2 < sum2 < sum^2
       
      unsigned long long sum  = std::sqrt(sum2) + 1;
       
      // sqrt may have returned a floating-point number
       
      if (sum * (sum - 1) != sum2)
       
        continue;
       
 
       
      // now we have the correct solution if minimum is small (< 100000)
       
      red = sum - blue;
       
      if (blue + red >= minimum)
       
      {
       
        found = true;
       
        break;
       
      }
       
    }
       
 
       
    // failed ? TODO: this means just that my simple search aborted
       
    if (!found)
       
    {
       
      std::cout << "No solution" << std::endl;
       
      continue;
       
    }
       
 
       
    // show solution
       
    std::cout << blue << " " << (red + blue) << std::endl;
       
  }
       
 
       
  return 0;
       
}

This solution contains 13 empty lines, 17 comments and 5 preprocessor commands.