if $(a,b,c)$ is a Pythagorean triple then $a$ or $b$ or $c$ can be divided by 3

Question

Prove that if $(a,b,c)$ is a Pythagorean triple then $a$ or $b$ or $c$ can be divided by 3

Answer 1

Hint: $$x^2\equiv \text{0 or 1}\pmod3$$

Answer 2

We know that $a^2+b^2=c^2$, and that the perfect squares mod $3$ are $0$ and $1$. Therefore $c^2$ is not $2$ mod $3$, and so it's not the case that both $a^2$ and $b^2$ are $1$ mod $3$. Therefore one of them is $0$ mod $3$. Let it be $a^2$. Then $a$ is also $0$ mod $3$ because $1^2=2^2=1\pmod{3}$. So $3|a$.

Answer 3

$$ (3k)^2 = 3k' \; , (3k+1)^2 = 3k'+1 \; and \; (3k+2)^2 = 3k'+4=3k''+1 $$ (so there's no square of the form $3k + 2$)

If $a = 3k$ or $b=3k$ we are done, assume the opposite:

$a\mod3 \neq b \mod 3 \neq 0$ $$ (3m+1)^2+(3n+2)^2 = 3p+1+4=3p'+2 = c^2 $$ (not possible)

$a\mod3 = b \mod 3 \neq 0 $ $$ (3m+1)^2+(3n+1)^2 = 3p+1+1=3p'+2 = c^2 $$ $$ (3m+2)^2+(3n+2)^2 = 3p+4+4=3p'+2 = c^2 $$ (not possible)

So at least one must be divisible by 3