Show that $4$ does not divide $x^3-2$ Show that $4$ does not divide $x^3-2$ is what I need to prove. 
I think I should put $4k$ is $x^3-2$ and then contradict it somehow.
Alternatively is to factor it out as $x^3$ is $x(x+2)(x-2)$ but I am not sure of that.
Do you know how to show this? 
 A: If $k$ is odd $x^3$ is odd and $x^3-2$ also, so $4$ can't divide it. If $x$ is even it is $2k$ and so $x^3-2=(2k)^3-2=8k^3-2=4(2k^3)-2$ and so it is not a multiple of $4$
A: $$x^3\equiv2\pmod4\implies x^3\equiv2\pmod2\equiv0$$
as $x^3-x=x(x-1)(x+1)\equiv0\pmod2, x^3\equiv x\pmod2$
$\implies x$ is even $\implies x^3\equiv0\pmod4$
A: *

*Using modular arithmetic:


If $x \equiv 0 \mod 4$, then $x^3 \equiv 0 \mod 4$ and $x^3-2 \equiv 2 \mod 4$.
If $x \equiv 1 \mod 4$, then $x^3 \equiv 1 \mod 4$ and $x^3-2 \equiv 3 \mod 4$.
If $x \equiv 2 \mod 4$, then $x^3 \equiv 8 \equiv 0 \mod 4$ and $x^3-2 \equiv 2 \mod 4$.
If $x \equiv 3 \mod 4$, then $x^3 \equiv 27 \equiv 3 \mod 4$ and $x^3-2 \equiv 1 \mod 4$.



*Using a divisibility argument:


If $x$ is odd, then $x^3$ is odd and $x^3-2$ is odd, so it is not divisible by 4.
If $x$ is even, then $x^3$ is a multiple of 8 and thus of 4. So $x^3-2$ is not a multiple of 4. 
A: In $\mathbf Z/4\mathbf Z$, the squares are $0$ and $1$, hence the cubes are $0,1,-1$, not $2$.
A: First remember the obvious fact that $4 = 2 \times 2$, then everything else falls into place.
If $x = 2y$ (meaning that $x$ is even), then $x^3 = (2y)^3 = 8y^3$. Then $$\frac{x^3 - 2}{4} = \frac{8y^3 - 2}{4} = 2y^3 - \frac{1}{2},$$ which is clearly not an integer.
If $x$ is odd, I'm not telling you anything new with that $x^3$ is odd also, as well as $x^3 - 2$, and 4 does not divide odd numbers.
