<s>[INST] 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.
 [/INST] 
Prove that $n^2-2n+7$ is even then $n+1$ is even.

If $n^2-2n+7$ is even, then $n^2-2n$ is odd.
For all even numbers, this does not work since $(even)^2-2(even)$ always results in an even number.
Therefore, the number $n$ must be an odd number. Since $n$ is odd, $n+1$ is even.
</s>