If $n^2+10$ is odd then $n$ is odd. I have been having a little trouble with proofs, and would like to ask for a hint for this statement. I've already started it, however I'm unsure what I should be doing next.

Prove that if $n^2 + 10$ is odd then $n$ is odd.

My answer so far: Suppose that $n$ is an odd integer, and we want to prove $n^2 + 10$ is odd. There exists a $k$ that $n=2k+1$. By substituting for $n$, we get... 
$(2k+1)^2 +10$ = $n^2 = 10$
$4k^2 + 4k + 11$ = $n^2 +10$
I am unsure what to do next.
EDIT: Thank you all for your hints and answers! They were all brilliant and my understanding for all kinds of proofs is a heck lot better! So much appreciated! 
 A: For a direct proof, you need to assume that $n^2 + 10$ is odd, and show this means $n$ is odd. In your proof, you assume $n$ is odd, which is what is to be proven.
For a proof by contraposition, we assume $n$ is not odd, i.e., that $n$ is even, and show this means $n^2 + 10$ is even. So assuming $n$ is even, there is an integer $k$ such that $n = 2k$. Then substitute $2k$ into n: $$n^2 + 10 = (2k)^2 + 10 = 4k^2 + 10 = 2(2k + 5)$$ which is clearly divisible by $2$ and is therefore even.
Remember, what the contrapositive of a statement is: To prove $p \rightarrow q$, we can prove the equivalent statement $\lnot q \rightarrow \lnot p$. 
A: Unfortunately, you're working backward. You've (very nearly) proved that if $n$ is odd, then $n^2+10$ is odd.
Now, if you wanted a proof by contrapositive (the simplest way, really), then suppose that $n$ is even, and prove that $n^2+10$ is also even. For a direct proof, assume that $n^2+10$ is odd, and prove that $n^2$ is then odd, and so $n$ is odd (why?). For a proof by contradiction, assume that $n^2+10$ is odd, but $n$ is even, and derive an absurdity.
A: Your assumptions are backwards. Given $n^2+10$ is odd, you want to prove that $n$ is odd. Perhaps the easiest way to do this is by proving the contrapositive, which can be stated as

If $n$ is even, then $n^2 + 10$ is even.

A: What you want to do is prove the contrapositive in proofs of this sort. The tautology is
$$(P \Rightarrow Q) \Leftrightarrow (\neg Q \Rightarrow \neg P).$$
By proving: "If $n$ is even, then $n^2+10$ is even" you will have proven your original theorem.
A: If $n^2+10$ is odd, then $(n^2+10)-11$ is even. Thus, $n^2-1$ is even.
Then $2|(n^2-1)=(n-1)(n+1)$. As $2$ is prime, we get $2|n-1$ OR $2|n+1$. Thus $n-1$ is even or $n+1$ is even, which proves that $n$ is odd.
A: First note that $n^2 +10$ is odd if and only if $n^2$ is odd. Then you could note that if $n$ is even, then $n^2$ is even. From this it follows that if $n^2$ is not even, then $n$ is not even.
A: Below is how I would do it:
If $n^2 + 10$ is odd, then $n^2$ is odd. 
This makes sense because an odd integer $+1$ is even, an even integer $+1$ is odd, etc. This is to say that an odd integer plus an even integer will give you an odd integer. So it's reasonable to say that if $n^2 + 10$ is odd, then $n^2$ is odd.
Now we want to prove that if $n^2$ is odd, then $n$ must be odd. We note that there are only two possibilities: either $n$ is even or $n$ is odd. Suppose $n$ is even. Then $n$ is a multiple of two, i.e. $2c = n$ for some $c \in \mathbb{Z}$. So $(2c)^2 = 4c^2$, which is also a multiple of two. So if $n$ is even, then $n^2$ must be even. But $n^2$ isn't even, so $n$ can't be even either. (This step was called proving the contrapositive - explanation below). So $n$ must be odd.
Proving the Contrapositive
This is a useful technique. If we have some statement that says $P$ implies $Q$, i.e. if P, then Q, then the statement if not Q, then not P is logically equivalent to the first one (can you figure out why?). Sometimes it can be easier to prove  if not Q, then not P than if P, then Q.
A: Suppose n=2k is even. Then n^2+10 = 4k^2+10=2(2k^2+5) is also even. Consequently, since all integers > 0 are either even or odd (but not both), when n^2+10 is odd, so is n.
A: or even simpler, 
if $n^2+10$ is odd, then there exists $k \epsilon \mathbb{N}$ such that
$n^2+10 = 2k+1 \Rightarrow n^2 = 2(k-5)+1$
so, $n^2$ must be odd, which implies $n$ is odd.
