Pythagorean Triples: Think Backwards

pyTriples.java

// If you have ever tried computing all the possible pythagorean triples (a^2 + b^2 = c^2) without duplicates,
// you might have tried creating 3 nested loops with the first one being a, the second b, and the third c.
// This method kind of works, however, you end up with duplicates and unbelievable slow just getting into the hundreds (for c).
// A more efficient way to solve the problem would be to start with c being the first loop
// and the other two variables in the second and third loop.

// Such a method might look like so:
for (int i = 0; i < 1000; i++) {
    for (int j = 0; j < i; j++) {
        for (int k = 0; k < j; k++) {
            if (Math.pow(i, 2) == Math.pow(j, 2) + Math.pow(k, 2)) {
                System.out.println(k + "^2 + " + j + "^2 = " + i + "^2");
                // OR
                System.out.println(i + "^2 = " + j + "^2 + " + k + "^2");
            }
        }
    }
}

// At the moment, the most optimized version of the problem I have found that computes all the possible combinations looks like so:
for (int i = 5; i < 1000; i++) {
    for (int j = 4; j < i; j++) {
        if (j % 2 != 0) {
            for (int k = 8; k < j; k += 2) {
                if (Math.pow(i, 2) == Math.pow(j, 2) + Math.pow(k, 2)) {
                    System.out.println(i + "^2 = " + j + "^2 + " + k + "^2");
                }
            }
        } else {
            for (int k = 3; k < j; k++) {
                if (Math.pow(i, 2) == Math.pow(j, 2) + Math.pow(k, 2)) {
                    System.out.println(i + "^2 = " + j + "^2 + " + k + "^2");
                }
            }
        }
    }
}