Direct proof problem

Question

Prove: If $n\in \mathbb{Z}$ and $7n-3$ is odd, then $n$ is even.

If I assume that $n$ is odd, I can get the required contradiction, but I want to use a direct proof. Is there anyway to do this?

Answer 1

If $7n-3$ is odd, then $7n$ is even, thus $n$ is even. Simple as that.

Answer 2

Suppose that $7n-3$ is odd. Then $7n-3 =2k+1$ where $k \in \mathbb{Z}$. Subtracting $6n-3$ from both sides of the equation yields:

$n=2k+1-6n+3 = 2(k-3n+2) = 2m$, where $m =k-3n+2 \in \mathbb{Z}$. Thus, $n$ is even.