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)
Best Answer
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 $2$ modulo $3$, which is absurd.
Edit: maybe I should add that for definition of Pythagorean triple, only one can be divisible by 3, not both.