Proving that if $n \in \mathbb{Z}$ and $n^2 − 6n + 5$ is even, then $n$ must be odd. Prove that if $n \in \mathbb{Z}$ and $n^2 − 6n + 5$ is even, then $n$ must be odd.
$p= n^2 - 6n + 55$ is even, $Q= n$ is odd
Proof: Assume on contrary $n$ is even. Then $n= 2k$ for some $k \in \mathbb{Z}$. Then, $$n^2 -6n + 5= 2k^2-6(2k)+5=2k^2-12k + 5$$
Unsure of where to go from here. 

Hint: Note that $$n^2-6n=n(n-6).$$ Since $n$ and $n-6$ are both _____ or both _____, then we see that $n$ is _____ if and only if $n(n-6)$ is. Since $5$ is odd and $n(n-6)+5$ is even, then $n(n-6)$ must be _____, and so....
