How can we prove that a Pythagorean triple is primitive?

[Poirot1]

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

 

[soroban]

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.)

 

[Opalg]

Re: primitive pythagorean triple

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.

 

[Poirot1]

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?