MHB How can we prove that a Pythagorean triple is primitive?

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To prove that a Pythagorean triple (x, y, z) = (a^2 - b^2, 2ab, a^2 + b^2) is primitive, it is essential to establish that gcd(x, y, z) = 1. This can be shown by ensuring that the integers a and b are coprime (gcd(a, b) = 1) and of opposite parity, meaning one is even and the other is odd. A contradiction argument can also be employed, assuming a common factor d greater than 1, leading to a prime factorization. The discussion emphasizes the importance of these conditions in confirming the primitiveness of the triple. Understanding these principles is crucial for mathematical proofs involving Pythagorean triples.
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I know that (a^2-b^2,2ab,a^2+b^2) is pythagorean triple. How to show it is primitive? i.e
gcd(x,y,z)=1
 
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Re: primitive pythagorean triple

Hello, Poirot!

I know that (x,y,z) = (a2 - b2, 2ab, a2 + b2) is a Pythagorean triple.

How to show it is primitive? .i.e. gcd(x,y,z) = 1
I'm not sure how we "show" it, but here is a fact.

For a primitive Pythagorean triple, a and b must be of opposite parity.
. . (One must be even, the other must be odd.)
 
Re: primitive pythagorean triple

Poirot said:
I know that (a^2-b^2,2ab,a^2+b^2) is pythagorean triple. How to show it is primitive? i.e
gcd(x,y,z)=1
The conditions for the triple to be primitive are that gcd(a,b)=1 and a, b are of opposite parity. See Pythagorean triple - Wikipedia, the free encyclopedia.
 
Re: primitive pythagorean triple

I'm pretty sure a contradiction argument is expedient.

assume x,y,z have a common factor d not equal to 1.
Since everything can be factorized into primes can I assume d is prime?
 

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