All Elements of Primitive Pythagorean Triple are 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 $y \perp z$.

$\blacksquare$