Suppose n is an integer. Use a proof by contrapositive to show if n^3 is even, then n is even

Question

So, we assume that n is not even. Then, $n$ is odd, so $n= 2k+1$ for some integer $k$. Then, $(2k+1)^3 = 8x^3+12k^2+6k+1$.

Would it be legal, then, for me to say $(8k^3+12k^2+6k)+1 = 2(4k^3+6k^2+3k)+1$, and then say that since $4k^3+6k^2+3k$ is an integer, $n^3+1$ is even, and therefore $n$ is odd?

Answer 1

Yes, it is totally true.

Any number $2K+1,K\in\mathbb{Z}$ is odd :)