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 $x^2 + y^2 = 3 z^2$.
$\blacksquare$
Sources
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $2$: Some Properties of $\Z$: Exercise $2.9$