Pythagoras always divisible by 3

Question

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?

Answer 1

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.

Answer 2

Had an idea: Since the square of a number is always of the form $3k+1$ (assuming that it is not divisible by $3$ because if it is divisible by $3$ then we have nothing to prove), let $a^2=3m+1$ and let $c^2=3n+1$ for some $m$ and $n$ in $\mathbb{Z}$. Then $b^2=3(n-m)\Rightarrow 3|b^2\Rightarrow 3|b.$

Answer 3

Hint : Remainder of $n^2/3$, is either $1$ or $0$. So one number, which is a square should be divisible by 3.