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. 

$n^2 − 6n + 5$ even $\implies$ $n^2+5$ even because $6n$ is even.
$n^2+5$ even implies $n^2$ odd because $5$ is odd.
$n^2$ odd implies $n$ odd.
Or use the contrapositive: $n$ even $\implies$ $n^2 − 6n + 5$ odd:
$n$ even $\implies$ $n=2k$ $\implies$ $n^2 − 6n + 5=4k^2-12k+5=2(2k^2-6k+2)+1$, which is odd.
