How do you prove this problem?

Question

Suppose (a,b,c) is a primitive Pythagorean Triple. Prove that c cannot be a multiple of 3.

Answer 1

Wlog suppose that $a^2+b^2=c^2$. We only need to work$\mod 3$. Since your hypothesis is that $(a,b,c)$ is a primitive pythagorean triple, then $a$ and $b$ are coprime. So by symmetry these are the cases:

i) $a\equiv 0 \pmod 3$ and $b\equiv 1 \pmod 3$

ii) $a\equiv 0 \pmod 3$ and $b\equiv 2 \pmod 3$

iii) $a\equiv 1 \pmod 3$ and $b\equiv 1 \pmod 3$

iv) $a\equiv 1 \pmod 3$ and $b\equiv 2 \pmod 3$

v) $a\equiv 2 \pmod 3$ and $b\equiv 2 \pmod 3$

Take for example case ii). We have $a^2+b^2\equiv 1 \pmod 3$, then $c^2\equiv 1 \pmod 3$ and therefore $c$ isn't multiple of $3$. The other cases are analogous.