Proof: $n^2 - 2$ is not divisible by 4 I tried to prove that $n^2 - 2$ is not divisible by 4 via proof by contradiction.  Does this look right?
Suppose $n^2 - 2$ is divisible by $4$.  Then:
$n^2 - 2 = 4g$, $g \in \mathbb{Z}$.
$n^2 = 4g + 2$.
Consider the case where $n$ is even.
$(2x)^2 = 4g + 2$, $x \in \mathbb{Z}$.  
$4x^2 = 4g + 2$.  
$4s = 4g + 2$, $s = x^2, s \in \mathbb{Z}$ as integers are closed under multiplication.  
$2s = 2g + 1$
$2s$ is even, and $2g + 1$ is odd (by definition of even/odd numbers).  An even number cannot equal an odd number, so we have a contradiction.
Consider the case where $n$ is odd.
$(2x + 1)^2 = 4g + 2$, $x \in \mathbb{Z}$
$4x^2 + 4x + 1 = 4g + 2$
$4x^2 + 4x = 4g + 1$
$4(x^2 + x) = 4g + 1$
$4j = 4g + 1$, $j = x^2 + x, j \in \mathbb{Z}$ as integers are closed under addition
$2d = 2e + 1$, $d = 2j, e = 2g; d, e \in \mathbb{Z}$ as integers are closed under multiplication
$2d$ is even, and $2e + 1$ is odd (by definition of even/odd numbers).  An even number cannot equal an odd number, so we have a contradiction.
As both cases have a contradiction, the original supposition is false, and $n^2 - 2$ is not divisible by $4$.
 A: Your proof is perfect.
We can shorten your proof by for example going from $4x^2=4g+2$ (in case 1) to saying "The left-hand side has remainder $0$ after division by $4$, yet the right side has remainder $2$; this is impossible" (basically, looking at the expression $\mod 4$ instead of dividing by $2$ and looking $\mod 2$). 
Also, the second case was trivially impossible, since $n^2=4g+2$ has no solutions if $n$ is odd (since then $n^2$ is odd, but $4g+2$ is even).
Depending on the context (what you know, what you can use, etc), steps like $x^2+x=j$ with the remark that integers are closed under addition and multiplication are mostly considered so trivial that it's not worth mentioning. I repeat however that this is completely dependent on context, and if you want to make sure your audience is aware of these facts and/or steps, you should mention them. More detailed explanations with steps rarely hurt the proof.

The proof can be done a lot quicker however (without contradiction) by looking $\mod 4$. It is quite easy to prove that squares are either $0$ or $1\mod 4$, so $n^2-2$ is either $-2$ or $-1\mod 4$, and thus, $n^2-2$ cannot be divisible by $4$.
A: For odd $n$, $n^2-2$ is odd.
For even $n$, $n^2$ is divisible by $4$, so that $n^2-2$ is not.

By contradiction:
Let $n$ be odd. Then $n^2-2$ is both odd and a multiple of four.
Let $n$ be even. Then $n^2-2$ and $n^2$ are both multiples of $4$, so that $2$ is a multiple of $4$.
A: Assume $4 \mid n^2-2$, for $n \in \mathbb{z}$. This means we can rewrite $n^2-2$ as $4k$, which implies $n^2=2(2k+1)$. This is clearly impossible; this implies $4 \nmid n^2$ but $2 \mid n^2$. So our initial assumption is wrong.
A: Anohter Idea of proof goes as follows:
First of all we need to know that any square of integer is either of the form 4k or 4k+1.
Suppose, to the contrary, that 4 divides $n^2-2$, then:
$$4k = n^2-2$$
$$4k+2 = n^2$$
A contradiction. Thus $n^2-2$ cannot be divisible by 4.
