<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] If $n + 1$ is even, then  $n$ must be odd. Let $n = 2k + 1$ since it is in the form of an odd number. Now substitute into the quadratic expression: $$\begin{align} n^2 - 2n + 7 &= (2k+1)^2 - 2(2k+1) + 7 \\ &= (2k)^2 + 1 + 4k - 4k - 2 + 7 \\ &= 4k^2 + 6 \\ &= 2(2k^2 + 3).\end{align}$$ Since the quadratic is an even number, this also completes the proof. For more sensibility, since $n + 1$ is even, we can let $n + 1 = 2r$ and then let $n = 2r - 1$, but the same is implied anyway.
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