Square Modulo 3

Theorem
Let $x \in \Z$ be an integer.

Then one of the following holds:

Corollary 1
Let $x, y \in \Z$ be integers.

Then:
 * $3 \backslash \left({x^2 + y^2}\right) \iff 3 \backslash x \land 3 \backslash y$

where $3 \backslash x$ means $3$ divides $x$.

Corollary 2
Let $x, y, z \in \Z$ be integers.

Then:
 * $x^2 + y^2 = 3 z^2 \iff x = y = z = 0$

Proof
Let $x$ be an integer.

Using Congruence of Powers throughout, we make use of $x \equiv y \ \left({\bmod \, 3}\right) \implies x^2 \equiv y^2 \ \left({\bmod \, 3}\right)$.

There are three cases to consider:


 * $x \equiv 0 \ \left({\bmod \, 3}\right)$: we have $x^2 \equiv 0^2 \ \left({\bmod \, 3}\right) \equiv 0 \ \left({\bmod \, 3}\right)$.


 * $x \equiv 1 \ \left({\bmod \, 3}\right)$: we have $x^2 \equiv 1^2 \ \left({\bmod \, 3}\right) \equiv 1 \ \left({\bmod \, 3}\right)$.


 * $x \equiv 2 \ \left({\bmod \, 3}\right)$: we have $x^2 \equiv 2^2 \ \left({\bmod \, 3}\right) \equiv 1 \ \left({\bmod \, 3}\right)$.

Proof of Corollary 1
Let $3 \backslash x \land 3 \backslash y$.

Then $x \equiv 0 \ \left({\bmod \, 3}\right)$ and $y \equiv 0 \ \left({\bmod \, 3}\right)$.

From the main proof, $x^2 \equiv 0 \ \left({\bmod \, 3}\right)$ and $y^2 \equiv 0 \ \left({\bmod \, 3}\right)$.

Then $\left({x^2 + y^2}\right) \equiv 0 \ \left({\bmod \, 3}\right)$ from Modulo Addition is Well-Defined.

Thus $3 \backslash \left({x^2 + y^2}\right)$.

Now suppose $3 \backslash \left({x^2 + y^2}\right)$.

That is, $\left({x^2 + y^2}\right) \equiv 0 \ \left({\bmod \, 3}\right)$.

From the main result: and
 * $x^2 \equiv 0 \ \left({\bmod \, 3}\right)$ or
 * $x^2 \equiv 1 \ \left({\bmod \, 3}\right)$
 * $y^2 \equiv 0 \ \left({\bmod \, 3}\right)$ or
 * $y^2 \equiv 1 \ \left({\bmod \, 3}\right)$

The only way $\left({x^2 + y^2}\right) \equiv 0 \ \left({\bmod \, 3}\right)$ is for $x^2 \equiv 0 \ \left({\bmod \, 3}\right)$ and $y^2 \equiv 0 \ \left({\bmod \, 3}\right)$.

Proof of Corollary 2
Proof by the Method of Infinite Descent:

Suppose $u, v, w$ are the smallest non-zero integers such that $u^2 + v^2 = 3 w^2$.

Then from Corollary 1 each of $u, v, w$ are multiples of $3$.

So we have $u', v', w'$ such that $3 u'= u, 3 v' = v, 3 w' = w$.

Then:
 * $\left({3 u'}\right)^2 + \left({3 v'}\right)^2 = 3 \left({3 w'}\right)^2$

which leads to:
 * $\left({u'}\right)^2 + \left({v'}\right)^2 = 3 \left({w'}\right)^2$

contradicting the supposition that $u, v, w$ are the smallest non-zero integers such that $u^2 + v^2 = 3 w^2$.

Thus $x^2 + y^2 = 3 z^2 \implies x = y = z = 0$.

If $x = y = z = 0$ then it trivially follows that $x^2 + y^2 = 3 z^2$.