Need some help with an even/odd proof by contraposition. Let me preface this by saying I am horrible at math, and I apologize for the dumb question.
So, I'm trying to prove that "for all integers , if (^2) + 2 is even, then  is even.", and it has to be by contraposition. This is what I have so far:

By contrapositive, this statement is the same as: for all integers n, if n is odd, then (n^2) + 2 is odd
By definition of odd, n = 2k+1 for any integer k
Thus by substitution, ((2k+1)^2) + 2
= 4k^2 + 4k + 3
=4(k^2 + k) + 3 by basic algebra

Obviously I'm trying to get it to take the 2n+1 form for an odd integer, but right now it's stuck at 4n+3. How do I get it into the proper form to complete the proof? Am I missing something obvious?

I'll continue from the asker's post, which was a good start.  We recognize $2n$ is even for any integer n, and $2n+1$ is odd for any integer $n$.
Then we want to prove the contrapositive of
$(A): $ "For all integers $n$, if $(n^2) + 2$ is even, then $n$ is even."  
The contrapositive of $(A)$ is given by 
$(B)$: "For all integers $n$, if $n$ is odd, then $(n^2)+2$ is odd."  The two statements are logically equivalent.
"Thus by substitution, assume $n$ is an odd integer.  Then $n=2k+1,$ where $k$ is some integer. Then:
$$n^2 + 2 =((2k+1)^2) + 2 = 4k^2 + 4k + 3 =  4k(k + 1) + 3..."$$
From there: 
$$4k^2 + 4k + 3 = 4k^2 + 4k + 2 + 1 = 2\cdot\underbrace{\left(2k^2+ 2k + 1\right)}_{\large j} + 1= 2j+1$$
(where $j$ is the integer given by $j=2k^2+2k+1, \quad k\in \mathbb Z$).
So we summarize:

For all $ n\in \mathbb Z$, if $n$ is odd, then $n^2 + 2$ must be odd.  

Since we have proved the equivalent contraposive, $(B)$, of the original statement we want to prove, we can indeed assert:  

For all $n \in \mathbb Z,\; \text{ if }\,\;n^2+2$ is even, then $n$ must be even."


Side note 

Given two statements $p,q$, $\quad (p\to q) \equiv (\lnot q \to \lnot p)$.  

We use $p:= "n^2 + 2$ is even", and $q:= "n$ is even." So, $\lnot p$ is "$n^2 +2$ is odd"  and $\lnot q$ is "$n$ is odd$.
We first proved $\lnot q \to \lnot p$, to prove its equivalent,  $\,p\to q$
