proof: primitive pythagorean triple, a or b has to be divisible by 3

Any square is congruent to 0 or 1 modulo 3

So having, $a^2 + b^2 = c^2$

Let's suppose neither a nor b is divisible by 3, then, the squares must be 1 modulo 3.

So, the expression can be re-written as:

$(3k + 1) + (3k' + 1) = c^2$

and then

$3 (k + k') + 2 = c^2$

That is, $c^2$ is a square congruent 2 module 3, which is absurd.

Edit: maybe I should add that for definition of Pythagorean triple, only one can be divisible by 3, not both.

\begin{array}{|c|c|} \hline n \pmod 3 & n^2 \pmod 3 \\ \hline 0 & 0 \\ 1 & 1 \\ 2 & 1 \\ \hline \end{array}

If neither $a$ nor $b$ is a multiple of $3$, then $a^2 + b^2 \equiv c^2 \pmod 3$ becomes $1 + 1 \equiv c^2 \pmod 3$, which simplifies to $c^2 \equiv 2 \pmod 3$; which has no solution.