Elements of Primitive Pythagorean Triples Modulo 4

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

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

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 Elements of Primitive Pythagorean Triple, there can be no such elements $$x$$ of the form $$x = 2 \pmod 4$$.

Hence the result.

Proof of Corollary
From Solutions of Pythagorean Equation, any Pythagorean triple is such that all elements are a multiple of the elements in some primitive Pythagorean triple.

As the latter must have an even element which is a multiple of $$4$$, such an element when multiplied by some integer will still be a multiple of $$4$$.

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