Divisibility in Pythagorean Triples

First let me thank you for preparing an excellent site.

Now on Pythagorean Triples:

where n and m are integers such that gcd(m, n) = 1 produce Pythagorean Triples. Fine.

Is there another formula ? Not to my limited knowledge.

I have an interesting observation on the Pythagorean Triples generated by the above formula.

1. The first triplet generated is (3, 4, 5).

2. Every triplet (p, q, r) generated will have one number amongst them divisible by 3. And one amongst them (not necessarily a different one ) divisible by 4. And one amongst them again (not necessarily a different one) is divisible by 5.

   The proof is as follows:

   Let m² - n², 2mn and m² + n² be the numbers in the triplet.

   1. One of the numbers in Pythagorean Triples is divisible by 3.

      If either 3|m (3 divides m) or 3|n, we have nothing more to prove.

      The case to prove is m = ±1 (mod 3) and n = ±1 (mod 3). It is immediate that m² = 1 (mod 3) and n² = 1 (mod 3). Hence m² - n² = 0 (mod 3).

   2. One of the numbers in Pythagorean Triples is divisible by 4.

      If either m or n is even then 2mn = 0 (mod 4).

      The case to prove is when both m and n are odd numbers. Then m² - n² = 0 (mod 4) because the square of an odd number is of form (4t +1).

   3. One of the numbers in Pythagorean Triples is divisible by 5.

      If m or n is divisible by 5 there is nothing to prove.

      The case to consider is m = ±1 (mod 5) or m = ±2 (mod 5). And the same is true for n. It then folows that both m² and m² may only be 1 or 4 modulo 5. If they are equal modulo 5, then m² - n² = 0 (mod 5). Otherwise, m² + n² = 0 (mod 5).

• Proof of the formula
• Norm of Gaussian integers
• Gaussian integers
• Divisibilty in Pythagorean Triples
• The Trinary Tree(s) underlying Primitive Pythagorean Triples
• Pythagorean Triples and Perfect Numbers