Prove that if $n^2 -2n +2$ is odd then $n$ is odd 
Prove that if $n^2 -2n +2$ is odd then $n$ is odd

I was wondering if you would prove this by using proof by contrapostive. I tried using proof by contrapostive, but I end up with the wrong answer.
 A: $$n^2-2n+2=(n-3)(n+1)+5$$ will be odd $$(n-3)(n+1)$$ is even
iff $n-3$  is even as $n+1,n-3$  have same parity
Alternatively, 
$n^2-2n+2=(n-1)^2+1$ will be odd iff $(n-1)^2$ is even
$\iff n-1$ is even
A: Yes, you can. Assume that $n$ is even. Then $n=2k$ for some integer $k$. Now $n^2 - 2n + 2 = 4k^2 - 4k + 2 = 2(2k^2 - 2k + 1)$, which is $2$ times an integer, meaning that it is an even number.
A: If $n$ is even then $n-1$ is odd, so $(n-1)^2$ is odd, so $n^2-2n+1$ is odd, so $(n^2-2n+1)+1$ is even, so $ n^2-2n+2$ is even.
A: Prove: $n \text{ even} \implies n^2-2n+2\text{ even}$
If $n$ is even, then $\exists k\in \mathbb Z$ such as $n = 2k$, so
$$n^2 - 2n + 2 = (2k)^2 - 2k + 2 = 4k^2 - 2k + 2 = 2 (\cdots)$$
where $\dots = 2k^2 -k +1$, so $n^2 - 2n + 2$ is even.
A: Also observe that 
\begin{align}
n^{2}-2n+2 &= n^{2}-2n+1+1 \\
           &= (n-1)^2 + 1
\end{align} 
Now, if $n$ is odd then it is of the form $n=2j \pm 1$ for some $j \in \mathbb{N}$. Let us take $$n=2j+1$$
Then
\begin{align}
(2j+1-1)^{2}-1 &= (2j)^{2}-1 \\
               &= 4j^{2} -1
\end{align}
Which is clearly odd. If, on the other hand, $n=2j$ (ie, $n$ is even) then
\begin{align}
(2j-1)^{2}-1 &= 4j^{2}-2j-2j-1-1 \\
             &= 4j^{2}-4j-2 \\
             &= 2(2j^{2}-2j-1)
\end{align}
Rendering this even.
