if $x^3-x\in\mathbb{Z}$ and $x^4-x\in\mathbb{Z}$ for some $x\in\mathbb{R}$, then $x\in\mathbb{Z}$. Assume that $x^3-x\in\mathbb{Z}$ and $x^4-x\in\mathbb{Z}$ for some $x\in\mathbb{R}$.
Prove that $x\in\mathbb{Z}$.
my attempt:
Let $a=x^3-x$ and consider polynomial $X^3-X-a$, then $x$ is a root of it and if $x\in\mathbb{Q}$, then obviously $x\in\mathbb{Z}$.
But it doesn't give anything if $x\notin\mathbb{Q}$, so it's surely not the right way.
 A: Let $x^3-x=a$ and $x^4-x=b$.
Let's do a few steps of the Euclidean algorithm for polynomial GCD:
$x^4-x-b=x(x^3-x-a)+(x^2+(a-1)x-b)\\ \implies x^2+(a-1)x-b=0$
$x^3-x-a=x(x^2+(a-1)x-b)+((1-a)x^2+(b-1)x+a)\\ \implies (1-a)x^2+(b-1)x+a=0$
$0=(1-a)(x^2+(a-1)x-b)-((1-a)x^2+(b-1)x+a)$
$\implies x ((1-a) (a-1)-b+1)-(1-a) b-a=0$
and so $x$ is rational, unless $((1-a) (a-1)-b+1)=0$ and $(1-a) b-a=0$.
The last condition happens for integers $a$ and $b$ iff $a=b=0$, in which case $x=0$ or $x=1$.
A: From the ssumption in the question we can find $a,b\in\mathbb{Z}$ such that: $x^3-x-a=0$ and $x^4-x-b=0$, using this we have: 
$$x.x^3=x^2+ax=x+b$$
then $$ x^2+(a-1)x-b=0\,\, (1)$$
so $x^2=(1-a)x+b$ and one can conclude that $((1-a)x+b)^2=x+b$ after expending this we have :
$$ x^2+(\frac{2b}{1-a}-1)x+\frac{b^2-b}{(1-a)^2}\,\, (2)$$
and using (1) and (2) we conclude that:
$$(\frac{2b}{1-a}-a)x+\frac{b^2-b}{(1-a)^2}+b=0 $$
and if $x\in \mathbb{R-Q}$ then $2b=a(1-a)$ and $b^2-b-b(1-a)^2=0$ and this implies $b=0$ or $b=1+(1-a)^2$
case 1: if $b=0$ then $a=0$ or $a=1$ and there is only integer solutions on $x$
case $2$: if $b=1+(1-a)^2$ then $1-a$ is solution to $2-y+3y^2=0$  which implies $23+(6y-1)^2=0$ which is impossible
in all the cases $x\in \mathbb{Q}$ so you can conclude that $x\in \mathbb{Z}$ also.
