Let c be an integer which is not divisible by 3. Then the equation $x^3 − x = c$ has no integer solutions. I am trying to show the following. Let $c$ be an integer which is not divisible by $3$. Then the equation $x^3 − x = c$ has no integer solutions.
My approach is to notice that if 3 does not divide c, then $c \equiv 1$ (mod 3) or $c \equiv 2$ (mod 3). 
Case #1: We have $x^3 - x = (x+1)(x-1)(x) = c = 3k + 1$ for some $k \in \mathbb{Z}$. I'm unsure where to go from here though.
Case #2: We have $x^3 - x = (x+1)(x-1)(x) = c = 3k + 2$ for some $k \in \mathbb{Z}$. I'm also unsure where to go from here though.
Any hints would be appreciated.
 A: Note that no matter which integer $x$ is, $x^3-x=(x-1)x(x+1)$ is divisible by $3$. So basically, study the left-hand side, not the right-hand side of the equation $x^3-x=c$.
A: Look at the equation modulo 3.  There are only 3 residue classes to check.  The remainder is either $0$, $1$, or $2$.
In all casses, you get $0$ (mod 3).
$x=0$:  $\ \ 0^3-0\equiv 0$ (mod 3)
$x=1$:  $\ \ 1^3-1\equiv 0$ (mod 3)
$x=2$: $\ \ 2^3-2=6\equiv 0$ (mod 3)
Thus there can be no integer solutions because the remainders on the left side and the right side do not agree. This is of course because $3 \nmid c$ implies the right side is not $0$ (mod 3),
A: $x^3-x=(x-1)(x)(x+1)$ is a product of three consecutive integers, so it is divisible by $3$, whereas $c$ isn't.   
More generally, by Fermat's little theorem $x^p-x=c$ with $p\nmid c$ has no integer solutions ($p$ is prime).
A: In three consecutive integer numbers one is divisible by three. Then their product ...what property has the product?
A: Note that $2 \equiv -1 \pmod{3}$, thus we can choose $\{-1,0,1\}$ as a system of representatives for the equivalence classes modulo $3$. Since $0^3 = 0$ and $1^3 = 1$ (even in $\Bbb{Z}$), this immediately implies that
$$
x^3 \equiv x \pmod{3}
$$
for every $x \in \Bbb{Z}$, i.e. that $3 \mid x^3 - x$.
