Square Modulo 3/Corollary 2

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Corollary to Square Modulo 3

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:

$\paren {3 u'}^2 + \paren {3 v'}^2 = 3 \paren {3 w'}^2$

which leads to:

$\paren {u'}^2 + \paren {v'}^2 = 3 \paren {w'}^2$

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 $x^2 + y^2 = 3 z^2$.

$\blacksquare$


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