Elements of Primitive Pythagorean Triple are Pairwise Coprime

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Let $\tuple {x, y, z}$ be a primitive Pythagorean triple.


$x \perp y$
$y \perp z$
$x \perp z$

That is, all elements of $\tuple {x, y, z}$ are pairwise coprime.


We have that $x \perp y$ by definition.

Suppose there is a prime divisor $p$ of both $x$ and $z$.

That is:

$\exists p \in \mathbb P: p \divides x, p \divides z$

Then from Prime Divides Power:

$p \divides x^2, p \divides z^2$

Then from Common Divisor Divides Integer Combination:

$p \divides \paren {z^2 - x^2} = y^2$

So from Prime Divides Power again:

$p \divides y$


$x \not \perp y$

This contradicts our assertion that $\tuple {x, y, z}$ is a primitive Pythagorean triple.

Hence $x \perp z$.

The same argument shows that $y \perp z$.