Prove that $x$ is an integer if $x^4-x$ and $x^3-x$ are integers. Given that $x^4-x$ and $x^3-x$ are integers.
Prove that $x$ is an integer.
Expanding, adding and subtracting the given expressions I can conclude that the followings are integers:$x(x-1)(x+1)$, $x^3(x-1)$, $x(x-1)(x^2+x+1)$ are integers. 
Could you help me solve the problem?

I will assume that $x$ is a priori living in some integral commutative ring, e.g. let's say $x$ is a complex number. It wouldn't be as easy (or as true) if $x$ were in some non-commutative or non-integral ring.
Write $a = x^4 - x$ and $b = x^3 - x$, which are integers by assumption.
We have 
(1) $x^4 - x - a = 0$,
(2) $x^3 - x - b = 0$. 
Subtracting $x$ times (2) from (1), we get
(3) $x^2 + (b - 1)x - a = 0$.
Subtracting $(x - (b - 1))$ times (3) from (2), we get
(4) $(b^2 - 2b + a)x = a(b - 1) + b$.
Now there are two possibilities.

Firstly, if $b^2 - 2b + a = 0$, then we must also have $a(b - 1) + b = 0$. 
Substituting $a = 2b - b^2$, we get $b^3 - 3b^2 + b = 0$. But $b$ is an integer, so it must follow that $b = 0$.
Hence $x^3 - x = 0$, or $x = 0, \pm 1$, which are all integers.

Secondly, if $b^2 - 2b + a \neq 0$, then we may divide out and get $x = \frac{a(b - 1) + b}{b^2 - 2b + a}$. This means that $x$ is a rational number.
But $x$ is also a root of the polynomial $P(X) = X^3 - X - b\in\mathbb{Z}[X]$, so $x$ must be an integer, since $\mathbb{Z}$ is integrally closed.
Alternatively, a down-to-earth proof may work like this: write $x = \frac{p}{q}$ with $p, q$ integers and $\gcd(p, q) = 1$.
We then have $x^3 - x = \frac{p^3 - pq^2}{q^3}$, which is assumed to be an integer. This means $q\mid p^3 - pq^2$, or $q\mid p^3$.
But $q$ is prime to $p$, so the only possibility is $q = 1$.
Hence $x = p$ is an integer.
