Parity of Smaller Elements of Primitive Pythagorean Triple

Theorem
Let $$\left({x, y, z}\right)$$ be a Pythagorean triple, i.e. integers such that $$x^2 + y^2 = z^2$$.

Then $$x$$ and $$y$$ can not both be odd.

It follows that if $$\left({x, y, z}\right)$$ is a primitive Pythagorean triple, then $$x$$ and $$y$$ are of opposite parity.

Proof

 * Let $$x$$ and $$y$$ both be odd such that $$\exists z \in \Z: x^2 + y^2 = z^2$$.

Then $$x^2 + y^2 \equiv 1 + 1 \equiv 2 \pmod 4$$.

But from Square Modulo 4, $$z^2 \equiv 0 \pmod 4$$ or $$z^2 \equiv 1 \pmod 4$$.

Thus $$x^2 + y^2$$ can not be square.

Hence from this contradiction $$x$$ and $$y$$ can not both be odd.


 * If $$x$$ and $$y$$ are both even, then they have $$2$$ as a common divisor.

So $$x \not \perp y$$ and therefore $$\left({x, y, z}\right)$$ is not primitive.