Trying to find what proof works best with this question? If  $ (n+1)^2   $ is even then   $ n $    is odd
The work part isn't the issue for me it's finding what is the best and most simple proof to use on this question
My three options are Direct proof, Indirect proof(Contrapositive) or proof by Contradiction
I know n is odd if $N=2k+1$
N is even if $n=2k$
I want to say solving it by Contrapositive is the best method because you flip n to Even which will make it easier
 A: I would approach this by proving the contrapostive, as you suggest.  The contrapostive of the statement, which is equivalent to the statement, is:

If $n$ is even, then $(n+1)^2$ is odd.

Assume $n$ is even.  So $n= 2k$, for some integer $k$.  Then $(n+1)^2 = n^2+ 2n + 1 =  (2k)^2 + 2(2k) + 1 =  4k^2 + 4k + 1= 2(2k^2+ 2k) +1 = 2m+1$,  which is an odd integer, (noting that we can represent $m = 2k^2+2k$, another integer, given that $k$ is an integer.)
A: If $(n+1)^2$ is even, then $n+1$ is even, and so $n$ is odd. Don't overthink it.
A: I'm reading this 2 years later and ... I don't really like it.  I'd like a redo:
Direct proof: Requires $(n+1)^2\implies .... \implies n$ is odd.  Our obvious second step and second to last steps are:  $(n+1)^2\implies (n+1)^2 = 2k$ for some $k\in \mathbb Z\implies ..... \implies n+1$ is even so $n=(n+1)-1$ is odd.
The missing step is that $\frac {(n+1)^2}2 = k\implies n+1$ is even but why? Well, we'd need Euclid's lemma. As $2$ is prime then $2|(n+1)^2 \implies 2|n+1$ so $n+1$ is even.
That's it.  It's not hard but it does require more thought and explanation than required.
Contrapositive:  Requires proving $n$ not odd $\implies.... \implies (n+1)^2$ is not even.
The obvious second and second to last steps are: $n$ not odd $\implies n$ is even $\implies .... \implies (n+1)^2$ is odd $\implies (n+1)^2$ is not even.
And we need to fill in the blanks.  I don't know about you but it strikes me the obvious relating of $n$ to $(n+1)^2$ is to go from $n$ to $n+1$ to $(n+1)^2$ and it is obvious what to do
$n$ not odd $\implies$
$n$ is even $\implies$
$n+1$ is odd $\implies$
$(n+1)^2$ is odd $\implies$
$(n+1)^2$ is not even.
That seems the most straight forward and simple way.
Proof by contradiction:  Requires assuming $(n+1)^2$ is even and $n$ is even leads to a contradiction.
Not hard.  $n$ is even means $n+1$ is odd and $(n+1)^2$ is odd contradiction $(n+1)^2$ is even.
But we never actually used $(n+1)^2$ is even except as a contrast to $(n+1)^2$ is odd.  This is essentially the contrapositive proof but with as assuming $(n+1)^2$ is even for no real purpose.
Given that... I'd say the contrapositive route is the most natural.
(Actually, I'd just do it and not even think about it....
(Okay, how do I "just do it"?... Well, these are basically toggled either it is odd/ or it is even and as such it falls naturally to me that this is more or less establishing a truth table.  As "parity of $k$ implies parity of $f(k)$" is far more direct and easy to establish than "parity of $f(k)$ implies parity of $x$" I'd say that if we were to do a proof of "parity of $f(k)$ implies parity of $x$ proof, I'd choose the contrapositive of "opposite party of $x$ implies opposite party of $f(x)$" as being far more simple and direct.)
=== old answer below =====
Direct requires proving that if $(n+1)^2$ is even then $n+1$ is even.  And the only real way I see to do that is be contrapositive.  If $n+1=2k$ is even then $(n+1)^2 = 4k^2$ is even but if $(n+1) =2k + 1$ is odd then $(n+1)^2=4k^2 + 4k + 1= 2(2k^2 + 2k) + 1$ is odd.  Or by claiming by Euclid's Lemma that $2$ is prime so if $2|(n+1)^2$ then $2|n+1$ and $n+1$ is even.  Then once we know $n+1$ is even then $n+1 = 2k$ so $n = 2k-1$ is odd.
That's not hard but it's more work.
COntrapositive is fairly.... er, dirct.....  If $n=2k$ is even then $n+1 = 2k+1$ is odd, and $(n+1)^2 = 4k^2+4k + 1$ is odd.  So contrapositively if $(n+1)^2$ is not odd then $n$ is not even.  Done.
Proof by contradiction sets up unnecessary conditions:
Suppose $(n+1)^2$ even and $n$ is even.  Then $n+1$ is odd and $(n+1)^2$ is odd so that's a contradcition.
Fine, but why did we need to suppose $(n+1)^2$ was even.
I'd go with contrapositive.
But they are all relatively easy and straightforward.
