Square Modulo 3/Corollary 1

Corollary Square Modulo 3
Let $x, y \in \Z$ be integers.

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

where $3 \mathop \backslash x$ denotes that $3$ divides $x$.

Sufficient Condition
Let $3 \mathop \backslash x \land 3 \mathop \backslash y$.

Then by definition of divisibility:
 * $x \equiv 0 \pmod 3$

and
 * $y \equiv 0 \pmod 3$

From Square Modulo 3:
 * $x^2 \equiv 0 \pmod 3$

and
 * $y^2 \equiv 0 \pmod 3$

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

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

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

Then by definition of divisibility:
 * $\left({x^2 + y^2}\right) \pmod 3$

From Square Modulo 3:
 * $x^2 \equiv 0 \pmod 3$ or $x^2 \equiv 1 \pmod 3$

and
 * $y^2 \equiv 0 \pmod 3$ or $y^2 \equiv 1 \pmod 3$

Thus the only way $\left({x^2 + y^2}\right) \equiv 0 \pmod 3$ is for:
 * $x^2 \equiv 0 \pmod 3$

and:
 * $y^2 \equiv 0 \pmod 3$