Answer to interesting question

I have an answer to this question: https://web2.0calc.com/questions/interesting-question_7

The answer is yes; we proceed by induction:

First, note that we must have $\{x^n\} + \frac1{x^n} = 1$ is encompassed by the more general statement $x^n+\frac1{x^n}\in \mathbb Z$, which we will prove. We have our base case(s) $x+\frac1x\in \mathbb Z$ and $x^0 + \frac1{x^0}=2\in \mathbb Z$. 

Suppose that $x^k+\frac1{x^k} \in \mathbb Z$ and $x^{k-1} + \frac1{x^{k-1}} \in \mathbb Z$. We multiply $x^k+\frac1{x^k}$ by $x+\frac1x$ to get 

$$\left(x^k+\frac1{x^k}\right)\left(x+\frac1x\right)=\left(x^{k+1}+\frac1{x^{k+1}}\right) + \left(x^{k-1}+\frac1{x^{k-1}}\right)$$

Note that the LHS is an integer, so $x^{k+1} + \frac1{x^{k+1}}$ is an integer as well, and we are done.