# Elements of Primitive Pythagorean Triples Modulo 4

## Theorem

Let $x \in \Z: x > 2$.

Then $x$ is an element of some primitive Pythagorean triple if and only if $x \not \equiv 2 \pmod 4$.

### Corollary

In every Pythagorean triple, at least one element is a multiple of $4$.

## Proof

Let $m = k + 1, n = k$ where $k \in \Z: k \ge 1$.

From Consecutive Integers are Coprime, $m \perp n$.

Then we have:

$m, n \in \Z$ are positive integers
$m \perp n$, i.e. $m$ and $n$ are coprime
$m$ and $n$ are of opposite parity
$m > n$.

From Solutions of Pythagorean Equation, these conditions are necessary and sufficient for $\left({2 m n, m^2 - n^2, m^2 + n^2}\right)$ to be a primitive Pythagorean triple.

Substituting for $m$ and $n$, we get:

$\left({2 k^2 + 2 k, 2 k + 1, 2 k^2 + 2 k + 1}\right)$ is a primitive Pythagorean triple.

So we see that for all $k \ge 1$, $2 k + 1$ is an element of a primitive Pythagorean triple.

So every odd integer from $3$ upwards is an element of some primitive Pythagorean triple.

That is, any integer $x$ such that $x \equiv 1$ or $x \equiv 3 \pmod 4$.

Now, consider $m = 2 k, n = 1$ where $k \in \Z: k \ge 1$.

These also fit the criteria for $\left({2 m n, m^2 - n^2, m^2 + n^2}\right)$ to be a primitive Pythagorean triple.

Substituting for $m$ and $n$, we get:

$\left({4 k, 4 k^2 - 1, 4 k^2 + 1}\right)$ is a primitive Pythagorean triple.

This means that for all $k \ge 1$, $4 k$ is an element of a primitive Pythagorean triple.

So every multiple of $4$ is an element of some primitive Pythagorean triple.

That is, any integer $x$ such that $x \equiv 0 \pmod 4$.

Now, consider the general primitive Pythagorean triple $\left({2 m n, m^2 - n^2, m^2 + n^2}\right)$.

As $m$ and $n$ are of opposite parity, they can be expressed as $2r$ and $2s + 1$.

So $2 m n = 2 \left({2r}\right)\left({2s + 1}\right) = 4 r \left({2s + 1}\right)$.

So $2 m n$ is divisible by $4$.

As the only even elements of a primitive Pythagorean triple are of this form $2 m n$ from Parity of Smaller Elements of Primitive Pythagorean Triple, there can be no such elements $x$ of the form $x = 2 \pmod 4$.

Hence the result.

$\blacksquare$