Parity of 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 $y$ cannot 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 $y$ cannot both be odd.

$\blacksquare$

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

So $\gcd \left\{{x, y}\right\} \ne 1$ and therefore $\left({x, y, z}\right)$ is not primitive.

$\blacksquare$