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?

 

[CellCoree]

To show that a Pythagorean triple is primitive, we need to show that the greatest common divisor (gcd) of the three numbers, x, y, and z, is equal to 1. This means that there are no common factors between the three numbers, making it the simplest form of the triple.

In the case of the Pythagorean triple (a^2-b^2,2ab,a^2+b^2), we can use the fact that the greatest common divisor of two numbers is equal to the greatest common divisor of their difference. This means that gcd(a^2-b^2, 2ab) = gcd(a-b, b) = gcd(a, b).

Since a and b are relatively prime (i.e. they have no common factors), gcd(a, b) = 1. Therefore, gcd(a^2-b^2, 2ab) = 1. We can also see that gcd(a^2+b^2, 2ab) = gcd(a+b, b) = gcd(a, b) = 1.

Finally, we can conclude that gcd(a^2-b^2,2ab,a^2+b^2) = gcd(gcd(a^2-b^2, 2ab), a^2+b^2) = gcd(1, 1) = 1. This shows that the Pythagorean triple (a^2-b^2,2ab,a^2+b^2) is primitive.