Divisibility in Pythagorean Triples
Subject: | Pythagorean Triples. |
---|---|
Date: | Tue, 01 Feb 2000 15:02:32 -0800 |
From: | Syamala Tadigadapa |
Organization: | Oracle Corporation |
First let me thank you for preparing an excellent site.
Now on Pythagorean Triples:
(m^{2} - n^{2}) , 2m·n , (m^{2} + n^{2} ) |
where n and m are integers such that
Is there another formula ? Not to my limited knowledge.
I have an interesting observation on the Pythagorean Triples generated by the above formula.
The first triplet generated is (3, 4, 5).
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 m^{2} - n^{2}, 2mn and m^{2} + n^{2} be the numbers in the triplet.
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
m^{2} = 1 (mod 3) andn^{2} = 1 (mod 3). Hencem^{2} - n^{2} = 0 (mod 3). 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
m^{2} - n^{2} = 0 (mod 4) because the square of an odd number is of form (4t +1).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 m^{2} and m^{2} may only be 1 or 4 modulo 5. If they are equal modulo 5, then
m^{2} - n^{2} = 0 (mod 5). Otherwise,m^{2} + n^{2} = 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
|Front page| |Contents| |Algebra| |Pythagorean Theorem|
Copyright © 1996-2017 Alexander Bogomolny62644255 |