<s>[INST] 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? 
 [/INST] *

*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. 
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