<s>[INST] if $(a,b,c)$ is a Pythagorean triple then $a$ or $b$ or $c$ can be divided by 3 Prove that if $(a,b,c)$ is a Pythagorean triple then $a$ or $b$ or $c$ can be divided by 3
 [/INST] $$
(3k)^2 = 3k' \; , (3k+1)^2 = 3k'+1  \; and \; (3k+2)^2 = 3k'+4=3k''+1
$$
(so there's no square of the form $3k + 2$)
If $a = 3k$ or $b=3k$ we are done, assume the opposite:
$a\mod3 \neq b \mod 3 \neq 0$
$$
(3m+1)^2+(3n+2)^2 = 3p+1+4=3p'+2 = c^2
$$
(not possible)
$a\mod3 = b \mod 3 \neq 0 $
$$
(3m+1)^2+(3n+1)^2 = 3p+1+1=3p'+2 = c^2
$$
$$
(3m+2)^2+(3n+2)^2 = 3p+4+4=3p'+2 = c^2
$$
(not possible)
So at least one must be divisible by 3
</s>