Square Modulo 3/Corollary 2

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

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

Proof
Proof by the Method of Infinite Descent:

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

Then from Square Modulo 3: 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 positive 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 $x^2 + y^2 = 3 z^2$.