<s>[INST] 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 $a\equiv X\pmod 3$ 
 [/INST] The numbers $a$ and $b$ cannot both be multiples of $3$, because then $c$ would be a multiple of $3$ and the triple would not be primitive.
And if none of them is a multiple of $3$, then both of them are of the form $3k\pm1$, for some integer $k$, from which it follows that both squares $a^2$ and $b^2$ are of the form $3k+1$. But then $c^2(=a^2+b^2)$ is of the form $3k+2$. You should be able to show that this is impossible.
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