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, that $$\exists p \in \mathbb{P}: p \backslash x, p \backslash z$$.

Then $$p \backslash x^2, p \backslash z^2$$ from Prime Divides Power.

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

So from Prime Divides Power again, $$p \backslash y$$ 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 $$y \perp z$$.