<s>[INST] Show that if $x^2 + y^2 = z^2$, then at leeast one of x and y is a multiple of 3. 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.
 [/INST] Suppose none of $x,y,z$ is divisible by $3$.
then
$$x\equiv \pm 1\mod 3,$$
$$y\equiv \pm 1 \mod 3,$$
and
$$z\equiv \pm 1 \mod 3$$
$\implies$
$$x^2+y^2-z^2\equiv 1+1-1\mod 3$$
$$\implies x^2+y^2\neq z^2.$$
</s>