Show that if $x^2 + y^2 = z^2$, then at leeast one of x and y is a multiple of 3.

Question

Show that if $x^2 + y^2 = z^2$, then at least one of $x$ and $y$ is a multiple of 3.

Attempt:

Given $x,y,z$ is pyhagorean triple, $$ x^2+y^2=z^2. $$ Assume neither $x$ nor $y$ is divisible by $3$. Since $x^2+y^2=z^2$, $x^2+y^2 \equiv z^2 \pmod 3$. But neither $x$ nor $y$ is congruent to $0 \pmod 3$. Therefore, \begin{align} x \equiv \pm 1 &\pmod 3 \\ y \equiv \pm 1 &\pmod 3. \end{align} Hence, \begin{align} x^2 \equiv 1 &\pmod 3 \\ y^2 \equiv 1 &\pmod 3, \end{align} and so $$ x^2+y^2 \equiv 2 \pmod 3. $$ But $2$ is not a square modulo $3$. Therefore, $$ 2 \not\equiv z^2 \pmod 3, $$ which is a contradiction.

Answer 1

Suppose none of $x,y,z$ is divisible by $3$.

then

$$x\equiv \pm 1\mod 3,$$

$$y\equiv \pm 1 \mod 3,$$

and $$z\equiv \pm 1 \mod 3$$

$\implies$

$$x^2+y^2-z^2\equiv 1+1-1\mod 3$$

$$\implies x^2+y^2\neq z^2.$$