<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] As pointed out bt @Bram28 and others you use contradiction by making the appropriate changes. 
However, I suggest that you try to use contrapositive instead of contradiction, whenever possible as it makes fewer assumptions.  
The easiest (and cleanest) way to solve this is proving the contrapositive instead i.e. if $n+1$ is odd, then $n^2-2n+7$ is odd. 
Since $n+1$ is odd, so $n$ is even and we can write $n=2k$ for some interger $k$. 
Note that $n^2-2n+7=4k^2-4k+6+1=2(2k^2-2k+3)+1=2m+1$, where $m=2k^2-2k+3$ is an integer, proving the result. 
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