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.
 A: Since you are doing a proof by contradiction, your start is:  Assume $n+1$ is odd. But if $n+1$ is odd, then $n$ is even, and so yo should plug in $n=2k$, rather than what you did, which was plugging in $2k+1$
A: 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.
A: 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. 
A: 
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.
A: HINT
Note that


*

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


and


*

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

A: $n^2-2n+7
=n^2+2n+1-(4n+6)
=(n+1)^2-2(2n+3)
$.
If this is even then,
since $2(2n+3)$ is even,
their sum is even,
so $(n+1)^2$ is even
so $n+1$ is even.
