Primitive Pythagorean triple divisible by 3

Question

Prove that for any primitive Pythagorean triple (a, b, c), exactly one of a and b must be a multiple of 3, and c cannot be a multiple of 3.

My attempt:

Let a and b be relatively prime positive integers.

If $a\equiv \pm1 \pmod{3}$ and $b\equiv \pm1 \pmod{3}$,

$c^2=a^2+b^2\equiv 1+1\equiv 2 \pmod{3}$

This is impossible as the only quadratic residues modulo 3 are 0 and 1.

So far, so good.

If one of a, b is $\equiv 0 \pmod{3}$ and the other is $\equiv \pm1 \pmod{3}$,

$c^2=a^2+b^2\equiv 0+1\equiv 1 \pmod{3}$

This is the part I don't understand. Just because $c^2\equiv 1\pmod{3}$ doesn't mean that $c^2$ must be a perfect square. For example, $a=12$ and $b=13$ satisfy the above conditions but $c^2=a^2+b^2=313$, which isn't a perfect square.

Answer 1

Your proof is complete. You are asked to show that one of $a,b,c$ is divisible by $3$. In the first part you show that $a$ and $b$ can't both be non-divisible by $3$. In the second part, you assume that one of $a,b$ is divisible by $3$ and show that $c^2\equiv 1$ (mod $3$) which implies that $c$ is not divisible by $3$ and hence exactly one of $a,b,c$ is divisible by $3$.

Answer 2

Your first step was a good start: It proves that at least one of $a$ and $b$ must be a multiple of $3$. We will have to eliminate the case of both $a$ and $b$ being multiples of $3$, but that is easy: If $a=3p$ and $b=3q$ then $$c^2 = a^2+b^2 = (3p)^2 + (3q)^2 = 9(p^2+q^2) = 9m$$ so $c$ is also a multiple of $3$, in which case the triplet is not primitive, as $3$ divides all three of them.

So exactly one of $a$ and $b$ must be a multiple of $3$.

Your second step has shown that since exactly one of $a$ and $b$ is a multiple of $3$, $c \equiv 1 \pmod{3}$. You don't care that not every number of the form $3k+1$ is a square; all you care about is that the number that happens to be $c^2$ must be expressible as $3k+1$. And this completes the proof that $c$ is not a multiple of $3$.