Prove that $n^2 - 2n + 7$ is even then $n + 1$ is even I have to use Proof by contradiction to show what if $n^2 - 2n + 7$ is even then $n + 1$ is even. 
Assume $n^2 - 2n + 7$ is even then $n + 1$ is odd. By definition of odd integers, we have $n = 2k+1$. 
What I have done so far:
\begin{align}
& n + 1 = (2k+1)^2 - 2(2k+1) + 7 \\
\implies & (2k+1) = (4k^2+4k+1) - 4k-2+7-1 \\
\implies & 2k+1 = 4k^2+1-2+7-1 \\
\implies & 2k = 4k^2 + 4 \\
\implies & 2(2k^2-k+2) 
\end{align}
Now, this is even but I wanted to prove that this is odd(the contradiction). Can you some help me figure out my mistake? 
Thank you.

HINT
Note that


*

*$n^2-2n+7=n^2-2n+1+6=(n-1)^2+6$ is even $\iff$ $n-1$ is even


and


*

*$n+1=(n-1)+2$

