This number theory book I have proved that if $(a, b, c)$ are a primitive Pythagorean triple (i.e., $a, b, c$ have no common factors and $a^2 +b^2 = c^2$), then either $a$ or $b$ is odd. The book did that by assuming that both $a$ and $b$ are even (or odd) and derived a contradiction each time. (If $a$ and $b$ are both even, it showed $c$ would be too; if they're both odd, it ends up with an equation where the left-hand side is odd and the other even.)
As an exercise, the book asks to use a similar reasoning to prove that either $a$ or $b$ must be a multiple of 3. The book didn't get to remainder classes or modular algebra yet, so I guess this "similar reasoning" wouldn't involve them. I don't think the book has even mentioned the word "remainder" yet.
Keeping in mind that $a$ is even, $b$ is odd, and $c$ is odd, I tried to either assume that neither $a$ or $b$ are divisible by 3 and tried to derive a contradiction, or that if either wasn't divisible by 3 then the other was. Neither approach led anywhere, except calculations so long and tedious that you end up forgetting why you're even doing them.
The book comes with no answers to problems, and I should note I suck at figuring out divisibility. Any hint would be greatly appreciated.
If neither a nor b is divisible by 3 (which could be the start of a poem) then they are of the form $3n\pm 1$ so their squares are of the form $3n+1$ so the sum of their squares is of the form $3n+2$ and no square is of this form.