Necessary condition for x>0 being an integer

Question

I was trying to solve a number theory problem and then I realized that I was needing to verify (prove or disprove) the following ''fact'' about numbers. I would appreciate any help.

Q: Suppose $x >0$ is such that $x^n \in \mathbb{Z}$ for all $n \geq 2$, then $x$ must be an integer? Furthermore, suppose that $n\geq 3$ is an odd integer, does the conclusion holds?

Answer 1

Suppose that $x^2,x^3\in\Bbb Z$. Then $x^2(x-1)=x^3-x^2\in\Bbb Z$, so $x-1$ is rational. The square of a rational number is an integer if and only if the rational number itself is an integer, so $x-1$ is an integer, and therefore so is $x$.

Answer 2

If $x^2=k\in\mathbb{Z}$, then $x=\sqrt{k}.$ If $x$ is not an integer, then neither is $x^3=k\sqrt{k}$.

A similar argument proves the second statement also.