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>
       
#include <cmath>
       
 
       
// determine the sum of all numbers between "from" and "to" (inclusive both) which match the problem statement

       
unsigned long long processSlice(unsigned int from, unsigned int to)
       
{
       
  std::vector<unsigned long long> sumSquares(to - from + 1, 0);
       
  // like a prime sieve: add square of all divisors
       
  for (unsigned long long i = 1; i <= to; i++)
       
  {
       
    // position of smallest multiple of i >= from
       
    auto pos = (from / i) * i;
       
    if (pos < from)
       
      pos += i;
       
 
       
    // add i^2 to all multiples of i
       
    for (; pos <= to; pos += i)
       
      sumSquares[pos - from] += i*i;
       
  }
       
 
       
  // find all sums that are perfect squares
       
  unsigned int sum = 0;
       
  for (size_t i = 0; i < sumSquares.size(); i++)
       
  {
       
    auto number  = i + from;
       
    auto current = sumSquares[i];
       
 
       
    // compute integer square root
       
    unsigned long long root = sqrt(current);
       
    // iff root^2 = current then it's a perfect square
       
    if (root * root == current)
       
      sum += number;
       
  }
       
 
       
  return sum;
       
}
       
 
       
int main()
       
{
       
  unsigned int limit = 64000000;
       
  std::cin >> limit;
       
 
       
  // how many number should be analyzed at once (=> influences memory consumption)
       
  unsigned int sliceSize = 2000000;
       
 
       
  // total sum
       
  unsigned int sum  = 0;
       
  // start of current slice
       
  unsigned int from = 1;
       
  while (from < limit)
       
  {
       
    // end of current slice
       
    auto to = from + sliceSize - 1;
       
    if (to >= limit)
       
      to = limit;
       
 
       
    // process current slice
       
    sum += processSlice(from, to);
       
    // next slice
       
    from = to + 1;
       
  }
       
 
       
  // print result
       
  std::cout << sum << std::endl;
       
  return 0;
       
}

This solution contains 10 empty lines, 14 comments and 3 preprocessor commands.