Let $a$, $b$, and $c$ be integers satisfying $a^2 + b^2 = c^2$. Prove: $abc$ must be even. I'm pretty sure that this can be proved by reductio ad absurdum, and have a proof for that. However, I'm not sure how to prove this using any other method of proof. It's my first time taking a course in logic and I'm not sure where to go with this.
Hint: Think about even and odd numbers. Can all 3 variables be odd?
Suppose $a$, $b$, and $c$ are odd. Then $a=2p-1$, $b=2q-1$, and $c=2r-1$. Inserting these into our formula, $a^2+b^2=c^2$, we get $(2p-1)^2+(2q-1)^2=(2r-1)^2$. Expanding, $4p^2-2p+1+4q^2-2q+1=4r^2-2r+1$, or $4p^2-2p+4q^2-2q+1=4r^2-2r$.

When written as $2(2p^2-p+2q^2-q+1)-1$, we see that the LHS is odd. Likewise, we see that the RHS is even. Because we can never have a value that is both even and odd, we get a contradiction. We can then say one of $a$, $b$, or $c$ must be even.

Let's say that $a$ is even; then $a=2n$, and $abc=2nbc$. We see that $abc$ is even. The other two cases are similar.

Therefore, we can say that $abc$ is even, in general.