What can I say of x given these conditions? I know that $x^2-x$ is integer and also $x^n-x$ is integer for some $ n > 2$. Can I say that $x$ is integer? How can I show it? ($x \in \mathbb{R}$)
 A: False, e.g. a root of $\,x^2\!-\!x\!+\!1\mid x^3\!+\!1\mid x^6\!-\!1\mid x^7\!-\!x\,$ has $\,x^2\!-\!x = -1,\ x^7\!-\!x = 0\, $ but $\,x\not\in\Bbb Z$
A: Suppose that $x^2 - x = k \in \mathbf{Z}$. In order for $x$ to be real, either $k = 0$ and $x \in \{0,1\}$ is an integer, or $k > 0$. For any $k$ and $n$, long division shows that
$$x^n - x = F_{n,k}(x)(x^2 - x - k) + A_{n,k} x + B_{n,k} = A_{n,k} x + B_{n,k}.$$
In particular, for the LHS to be an integer, either $x$ must be an integer, or $A_{n,k}$ must equal zero. (This uses the fact that $x^2 - x = k$ implies that either $x$ is an integer or $x$ is irrational by Gauss' Lemma.) 
I claim that $A_{n,k} > 0$ for $n > 2$ and $k > 0$, and so, if $x$ is real, the LHS is only an integer when $x$ is an integer. 
Lemma: If $k > 0$, then $A_{n,k} > 0$ and $B_{n,k} > 0$ for all $n \ge 3$.
Proof: By induction on $n$. For $n = 3$ and any $k > 0$, we have
$$x^3 - x = (x + 1)(x^2 - x - k) + k x + k.$$
So $A_{n,k} = k > 0$, and $B_{n,k} = k > 0$.
Assume the result is true for $n$. Note that
$$x^{n+1} - x = x(x^n - x) + x^2 - x = x(x^n - x) + (x^2 - x - k) + k.$$
Hence
$$\begin{aligned}
x^{n+1} - x  = & \ x F_{n,k}(x)(x^2 - x - k) + A_{n,k} x^2 + B_{n,k} x + (x^2 - x - k) + k \\
         = & \ (x F_{n,k}(x) + 1)(x^2 - x - k) + A_{n,k} x^2 + B_{n,k} x + k \\
         = & \ (x F_{n,k}(x) + 1)(x^2 - x - k) + A_{n,k} (x^2 - x - k) +  A_{n,k}(x + k) +  B_{n,k}x + k \\
         = & \ (x F_{n,k}(x) + 1 + A_{n,k})(x^2 - x - k) + (A_{n,k} + B_{n,k})x +  A_{n,k} \cdot k + k \end{aligned}$$
And thus (by induction on $n$, and using that $k > 0$)
$$A_{n+1,k} = A_{n,k} + B_{n,k} > 0,$$
$$B_{n+1,k} = A_{n,k} \cdot k + k > 0.$$
