How to enumerate x^2 + y^2 = z^2 - 1 (with additional constraints)

Question

Lets N be a number (10<=N<=10^5).

I have to break it into 3 numbers (x,y,z) such that it validates the following conditions.

Answer 1

Here is a simple improvement in Python (converting to the faster equivalent in C-based code is left as an exercise for the reader). To get accurate timing for the computation, I removed printing the solutions themselves (after validating them in a previous run).

• Use an outer loop for one free variable (I chose z), constrained only by its relation to N.
• Use an inner loop (I chose y) constrained by the outer loop index.
• The third variable is directly computed per requirement 2.

Timing results:

-------------------- 10 
 1 solutions found in 2.3365020751953125e-05  sec.
-------------------- 100 
 6 solutions found in 0.00040078163146972656  sec.
-------------------- 1000 
 55 solutions found in 0.030081748962402344  sec.
-------------------- 10000 
 543 solutions found in 2.2078349590301514  sec.
-------------------- 100000 
 5512 solutions found in 214.93411707878113  sec.

That's 3:35 for the large case, plus your time to collate and/or print the results.

If you need faster code (this is still pretty brute-force), look into Diophantine equations and parameterizations to generate (y, x) pairs, given the target value of z^2 - 1.

import math
import time

def break3(N):
    """
    10 <= N <= 10^5
    return x, y, z triples such that:
        x <= y <= z
        x^2 + y^2 = z^2 - 1        
        x + y + z <= N
    """

    """
    Observations:
    z <= x + y
    z < N/2
    """

    count = 0
    z_limit = N // 2
    for z in range(3, z_limit):

        # Since y >= x, there's a lower bound on y
        target = z*z - 1
        ymin = int(math.sqrt(target/2))
        for y in range(ymin, z):
            # Given y and z, compute x.
            # That's a solution iff x is integer.
            x_target = target - y*y
            x = int(math.sqrt(x_target))
            if x*x == x_target and x+y+z <= N:
                # print("solution", x, y, z)
                count += 1

    return count


test = [10, 100, 1000, 10**4, 10**5]
border = "-"*20

for case in test: 
    print(border, case)
    start = time.time()
    print(break3(case), "solutions found in", time.time() - start, "sec.")