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. 
 A: We want to prove, if $3n+2$ is even, then $n$ is even.
In this case, it's actually easier to prove the contrapositive, that is, 
if n is odd, then 3n+2 is odd.
Suppose $n$ is odd. This means that $n=2k+1$, for some $k\in\mathbb{N}$.
$3n+2$, therefore, can we written as $3(2k+1)+2=6k+3+2=6k+5=2(3k)+5$. We know for a fact that $2$ times anything, is $even$. We also know that $even+odd=odd$. So we have $2(3k)$ which is even, $+ 5$, which is odd, so therefore it is odd.
$\therefore$ by contrapositive, if $3n+2$ is even, then $n$ is even.

Also, whenever you are asked to prove is something is even, or something is odd, always try to see if the contrapositive is easier. In this case, it would've been harder to show that $3n+2$ is even, and then show that $n$ is even, because then you would have to deal with fractions. It is always easier to show that if $n$ is even/odd, then a expression of $n$ is even/odd.
A: If an integer $N$ is even, there is another integer $m$ for which $N = 2m$. Thus, supposing $3n+2$ is even, there must be an $m$ for which $3n = 2(m-1)$; that is, $3n$ is even. Since the product of two integers can only be even when at least one of them is even, $n$ must be even as well. 
A: If $3n+2$ is even, then $\frac{3n+2}{2} = \frac{3}{2}n + 1$ is an integer, so $\frac{3}{2}n$ is an integer, so $n$ is divisible by 2 (even).
Edit: to your general question, the method of proof definitely varies from problem to problem. I, for one, rarely see a problem and immediately jump to a certain proof strategy. I like to just remind myself of all the definitions involved and then try some things. Learning to recognize when to use a certain method just comes with experience.
A: *

*There is another possibility: to disprove, by finding a counterexample.

*It's easy to be confused, you just have to learn with lots of examples. 

*It depends very, very heavily on each question.

*One way to start, for this question and for many similar questions is: don't just jump into proving the statement. First, test the statement by plugging in examples of $n$ and seeing if the implication always holds. To test this problem, plug in $n=1$, $n=2$, $n=3$, et cetera, and see if the implication holds for all the values of $n$ that you plug in. If it does not hold for all the values of $n$ that you plug in, then you've found a counterexample, and therefore disproved the statement, and you're done. If it does hold for all the values of $n$ that you plug in, then you might want to think about how to prove the statement by observing the patterns and trying to make some general conclusions... but I'll stop here regarding general strategy.


To start you off, if you plug in $n=2$, then you get that $3n+2=8$ is even and $n=2$ is even, so the implication it true for the test case $n=2$. 
But, test a few more values of $n$ before you jump into proving the statement.
A: Here is a simple direct proof:
If $3n+2$ is even, then $3n+2=2m$. But then $n =2m-2n-2=2(m-n-1)$ is even.
A: It seems a modular arithmetic problem, so at least initially similar problems with that kind of statements can be attacked by trying to use a modular arithmetic approach, but as it was said in the comments, there is not a generic rule for this.
In the case of your example, the residue of the expression must be $0$ when divided by $2$, so you can use the following steps to arrive to the conclusion you are looking for (using modular arithmetic rules):

$$(3n+2) \pmod{2} \equiv 0 \Leftrightarrow$$
$$(3 \pmod {2})\cdot (n \pmod 2)) + (2 \pmod{2})  \equiv 0 \Leftrightarrow$$
$$1 \cdot (n \pmod {2}) + 0 \equiv 0  \Leftrightarrow$$
$$n \pmod {2} \equiv {0} \Leftrightarrow$$
$$n=2k \ , k \ \in \Bbb Z$$
So $n$ must be even (including $0$, which is indeed an even number) too if we want the expression to be even.

I would dare to say that one of the nice points or applying modular arithmetic rules is that the steps are quite clean and easy to follow.
A: $\newcommand\odd{\mathit{odd}}\newcommand\even{\mathit{even}}$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.
A: $3n+2$ even implies that substracting it $2$ keeps it even, i.e, $3n$ is even. Therefore $n$ has to be even, since if it were odd, then $3n$ wouldn't be even.
Note that first I used direct proof and then I used proof by contradiction. Quite often, when something looks pretty obvious or pretty hard, contradiction is the way to go.
A: If 3n + 2 is even, then (3n + 2)-2 = 3n is also even.
3n can also be written "2n + n". 
2n is even by definition. 
Therefore since 2n + n is even then n itself must be even.
This general technique may have a name, I don't know! It relies on the known addition/subtraction rules of even numbers, but not the multiplication rules. This makes it a bit easier to follow than some of the above proofs, I think.
A: 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).
