Prove by contradiction that if $a,b,c\in \mathbb{Z}$, such that $a^2 +b^2 =c^2$, then at least one of the numbers $a,b$ is odd.

Question

Here is what I have done so far.

We assume that both $a$ and $b$ are even. Then $\exists \kappa,\lambda\in\mathbb{N}_0$ such that $a=2\kappa$ and $b=2\lambda$. Then,

$a^2+b^2=c^2\Leftrightarrow (2\kappa)^2+(2\lambda)^2=c^2\Leftrightarrow c=2\sqrt{{\kappa}^2+{\lambda}^2} $

I am not sure if that helps though, and if it leads to a contradiction. So, any ideas how I should procceed?

Answer 1

This is not true in general. As noted by Arthur, you have $6^2+8^2=10^2$. However, this becomes true if you assume that $a,b,c$ have no common divisor.

Indeed, your equation $(2\kappa)^2+(2\lambda)^2=c^2$ would mean that $c^2$ is a multiple of $4$, hence $c$ would be even. As $a$ and $b$ are assumed to be even, this would contradict our assumption.

Answer 2

The problem here is that the problem isn't really a problem, as it can be reduced to a case when it doesn't apply and a case when it is trivial.

It is possible (but not necessary) for both a and b to be even when a and b share common factors, like in the above example of (a, b, c) = (6, 8, 10). Specifically, if and only if c is even, both a and b must be even, as this is the only way for c^2 to have the correct modulo 4.

When there are no common factors, then exactly one of a and b must be odd, but this is trivial. The statement of 'no common factors' automatically excludes pairs of even numbers sharing the factor 2, so it's a trivial correlation.