For any primitive pythagorean triple $(a,b,c)$ either $a$ or $b$ must be a multiple of $3$ I'm reading "Friendly Introduction to Number Theory". Now I'm working on Primitive Pythagorean Triples Exercises 2.1 (a) on P18. 

We showed that in any primitive Pythagorean triple $(a, b, c)$, either $a$ or $b$ is even. Use the same sort of argument to show that either $a$ or $b$ must be a multiple of 3.

(1) $a^2 + b^2 = c^2$ with a odd, b even, a,b,c having no common factors
(2) $a^2 = c^2 - b^2 = (c-b)(c+b)$
(3) $c + b = s^2$ and $c - b = t^2$
(4) $c = \frac{(s^2 + t^2)}{2}$ and $b = \frac{(s^2 - t^2)}{2}$
(5) $a = \sqrt{(c-b)(c+b)} = st$
(6) $a = st$, $b = \frac{(s^2 - t^2)}{2}$, $c = \frac{(s^2 + t^2)}{2}$
https://www.math.brown.edu/~jhs/frintch1ch6.pdf
I have no idea how I start doing this. Can you give me a hint? I think I need to show that both the following (1) and (2) are satisfied.
$X \neq 0$
(1) $a\equiv 0\pmod 3$ and $b\equiv X\pmod 3$
(2) $b\equiv 0\pmod 3$ and $a\equiv X\pmod 3$ 

In mod 3, $0^2=0$, $1^2=1$ and $2^2=(-1)^2=1$. Thus, if neither $a$ nor $b$ are 0 mod 3, then $a^2+b^2$ is 2 mod 3, but no number squares to 2 mod 3.
The "principal" condition doesn't really play a part in the proof.
There are also proofs that at least one number must be divisible by 4, and that at least one number must be divisible by 5.
