Proof explanation: when to use a particular methodology of proof If $3n+2$ is even, then $n$ is even. 


For example the question above and I have had other questions in my textbook. I am currently confused on when to use a particular proof , such as proof by contradiction or contrapositive or direct proof etc. 

Does it depend on each question ? Just need some advice regarding this problem.
$\newcommand\odd{\mathit{odd}}\newcommand\even{\mathit{even}}$You prove it by proving it.  Any technique will do.

Ask yourself simply "Do I believe $3n+ 2$ is even then $n$ must by even?"  If you do ask yourself "Why do I believe it".  If you don't convince you self it does.

I'd say I basically have two options.  One strategy is to assume we've already proven:

$\odd*\odd = \odd$

$\odd*\even = \even$

$\even*\even = \even$

$\odd \pm \odd = \even$ 

$\odd \pm \even = \odd$ 

$\even \pm \even = \even$.

Then I'd figure I just have to fit $3n + 2=\even$, i.e, $\odd*{???} + \even = \even$

$\odd*{???} = \even - \even = \even$

So $\odd*{???} = \even$.  I go through what I know $\odd*\odd = \odd$ and $\odd*\even = \even$ so it has to be that $???$ is even.

This was a case of going through options and seeing what does and doesn't work; so that indicates a proof by contradiction:

Proof:  Suppose $n$ is odd.  Then $3n$ is also odd as we have proven that $\odd*\odd = \odd$.  Then $3n + 2$ is odd as we have already prove that $\odd + \even$ is odd.  This is a contradiction.

Or if you don't feel you can rely upon previous result and must replicate them... then replicate them.

Proof: Suppose $n = \odd$ then $n = 2m + 1$ for some $m$.  So $3n + 2 = 3(2m+1) + 2 = 6m + 5 = 6m + 4 + 1 = 2(3m+2) + 1$ is an odd number.  This is a contradiction.



Or I can figure I need to prove things directly.  If I know $3n + 2 = \even = 2k$ for some $k$ will I find that 

$3n + 2 = 2k$

$3n = 2k -2 = 2(k-1)$

$n = \frac {2(k-1)}3$  Can I say that $3\nmid 2$ so $3|k-1$? so $n = 2l$ for some $l= \frac {k-1}3$?  I can but it's a little more convoluted then I like.

Pf:  $3n + 2 = 2k$ so $3n = 2k - 2 = 2(k-1)$ Then $3|2(k-1)$ but $3|k-1$ so $k- 1 = 3l....$ .... and let's not go down this route.  It's too convoluted and the way above with a proof by contradiction was simpler.

.... Third option I don't know whether $n$ is even or odd by it must be one or the other.

$n = 2m + k$ where $k = 0$ or $1.

And $3n + 2 = 2j$ for some $j$.

So  can I do $3n + 2 = 3(2m + k) + 2 = 2j$?

$6m + 3k + 2 = 2j$

So $3k = 2j - 6m -2 = 2(j-3m -1)=\even$

If $k=1$ then $3k = 3$ is even.  If $k = 0$ then $3k = 0$ is even.  As these are the only two options, it must be that $k = 0$ and$n = 2m + k = 2m$ is even.

Choose whatever works best for you.
There are only two possible cases for $n\in\mathbb{Z}$:

Case $(1)$:

$n\equiv 0\bmod 2 \quad\text{(n even)}\implies n=2k \quad\text{for some}\quad k\in\mathbb{Z}\implies 3n+2=6k+2=2(3k+1)\implies 3n+2\equiv 0\bmod 2$

Case $(2)$:

$n\equiv 1\bmod 2\quad\text{(n odd)}\implies n=2k+1\quad\text{for some}\quad k\in\mathbb{Z}\implies 3n+2=3(2k+1)+2=6k+4+1=2(3k+2)+1\implies 3n+2\equiv1\bmod 2$

Therefore, by exhaustion of all the possible cases, $3n+2\equiv 0\bmod 2\implies n\equiv 0\bmod 2$ (this is known as proof by exhaustion).