How to find all possible values of four variables when squared sum to N?

Question

A^2+B^2+C^2+D^2 = N Given an integer N, print out all possible combinations of integer values of ABCD which solve the equation.

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Answer 1

nebffa has a great answer. one suggestion:

 step 3a:  max_b = min(a, floor(square_root(n - a^2))) // since b <= a

max_c and max_d can be improved in the same way too.

Here is another try:

1. generate array S: {0, 1, 2^2, 3^2,.... nr^2} where nr = floor(square_root(N)). 

now the problem is to find 4 numbers from the array that sum(a, b,c,d) = N;

2. according to neffa's post (step 1a & 1b), a (which is the largest among all 4 numbers) is between [nr/2 .. nr]. 

We can loop a from nr down to nr/2 and calculate r = N - S[a]; now the question is to find 3 numbers from S the sum(b,c,d) = r = N -S[a];

here is code:

nr = square_root(N);
S = {0, 1, 2^2, 3^2, 4^2,.... nr^2};
for (a = nr; a >= nr/2; a--)
{
   r = N - S[a];
   // it is now a 3SUM problem
   for(b = a; b >= 0; b--)
   {
      r1 = r - S[b];
      if (r1 < 0) 
          continue;

      if (r1 > N/2) // because (a^2 + b^2) >= (c^2 + d^2)
          break;

      for (c = 0, d = b; c <= d;)
      {
          sum = S[c] + S[d];
          if (sum == r1) 
          {
               print a, b, c, d;
               c++; d--;
          }
          else if (sum < r1)
               c++;
          else
               d--;
      }
   }
}

runtime is O(sqare_root(N)^3).

Here is the test result running java on my VM (time in milliseconds, result# is total num of valid combination, time 1 with printout, time2 without printout):

N            result#  time1     time2
-----------  -------- --------  -----------
  1,000,000   1302       859       281
 10,000,000   6262     16109      7938
100,000,000  30912    442469    344359