# Divisibility in Pythagorean Triples

Subject:Pythagorean Triples.
Date:Tue, 01 Feb 2000 15:02:32 -0800
Organization:Oracle Corporation

First let me thank you for preparing an excellent site.

Now on Pythagorean Triples:

 (m2 - n2) , 2m·n , (m2 + n2 )

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). Pythagorean triples 