Pythagoras always divisible by 3 Lets look at the pythagoras $a^2+b^2=c^2$ for the integers $a,b,c$. Proof that one of components $a,b,c$ is always divisible by 3. How do i prove that?

Assume none of $a,b,c$ is a multiple of $3$.
then
$a \equiv \pm 1$  mod$ (3)$
$b \equiv \pm 1$ mod $(3)$ and
$c\equiv \pm 1$ mod $(3)$
thus
$a^2\equiv 1$ mod $(3),$
$b^2\equiv 1$ mod $(3)$
$a^2+b^2 \equiv 2$ mod$ (3)$
but
$c^2=a^2+b^2 \equiv 1$ mod$ (3)$
and this is a contradiction.
