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. 
 A: You were actually very close to finishing off the proof effectively; you were working towards proving the contrapositive (as opposed to a proof by contradiction). Your original statement was

Prove that if $n \in \mathbb{Z}$ and $n^2 − 6n + 5$ is even, then $n$ must be odd.

The contrapositive of the statement above is as follows:

Prove that if $n\in\mathbb{Z}$ and $n$ is even, then $n^2-6n+5$ is odd.

Supposing $n$ is even, as you did by letting $n=2k$ for some $k\in\mathbb{Z}$, we have
$$
n^2 -6n + 5= (2k)^2-6(2k)+5=4k^2-12k + 5=2(2k^2-6k+2)+1=2\eta+1, \eta\in\mathbb{Z}.
$$
Since $n^2-6n+5=2\eta+1$ (with $\eta\in\mathbb{Z}$), we can see that if $n$ is an even number, then $n^2-6n+5$ is an odd number (namely $2\eta+1$). This finishes the proof of the contrapositive, thus proving your original claim. 
A: Hint: Consider taking modulo 2 next in your solution. That wouldn't change even/oddness, but you should see the solution then...if $ k\in\mathbb {Z} $, what is $-12k \text { mod }2 $? Continue from here.
A: 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....
A: $n^2 − 6n + 5$ 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.
