proof: primitive pythagorean triple, a or b has to be divisible by 3 I'm reading "A friendly introduction to number theory" and I'm stuck in this exercise, I'm mentioning this because what I need is a basic answer, all I know about primitive pythagorean triplets is they satisfy $a^2 + b^2 = c^2$ and a, b and c has no common factors.
Now.. my approach (probably kind of silly) was to "classify" the odd numbers (not divisible by 3) as $6k+1$, $6k+2$ and $6k+5$, and the even numbers with $6k+2$ and $6k+4$, then, trying different combinations of that, I could probe all the cases when I assume c is not 3 divisible, but I still have to probe that c cannot be 3 divisible and I don't know how to do it.
Anyway, probably there is a better simpler solution.
(Sorry, if this is a stupid question, I'm trying to teach myself number theory without much math background)

\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.
