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 $a\equiv X\pmod 3$ 

Using https://en.m.wikipedia.org/wiki/Pythagorean_triple or        https://mathcs.clarku.edu/~djoyce/java/elements/bookX/propX29.html
$a=2mn,b=m^2-n^2$ where $m,n$ are coprime integers and not both are odd
$ab=2mn(m^2-n^2)=2(m^3-m)n-2m(n^3-n)$
Now use 
The product of $n$ consecutive integers is divisible by $n$ factorial as $m^3-m=(m-1)m(m+1)$
