Elements of Primitive Pythagorean Triple are Pairwise Coprime

Theorem
Let $\left({x, y, z}\right)$ be a primitive Pythagorean triple.

Then:
 * $x \perp y$
 * $y \perp z$
 * $x \perp z$

That is, all elements of $\left({x, y, z}\right)$ are pairwise coprime.

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

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

That is:
 * $\exists p \in \mathbb P: p \mathop \backslash x, p \mathop \backslash z$

Then from Prime Divides Power:
 * $p \mathop \backslash x^2, p \mathop \backslash z^2$

Then from Common Divisor Divides Integer Combination:
 * $p \mathop \backslash \left({z^2 - x^2}\right) = y^2$

So from Prime Divides Power again:
 * $p \mathop \backslash y$

and:
 * $x \not \perp y$

This contradicts our assertion that $\left({x, y, z}\right)$ is a primitive Pythagorean triple.

Hence $x \perp z$.

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