If $x>0$ is such that $x^{n}+\frac{1}{x^n}$ and $x^{n+1}+\frac{1}{x^{n+1}}\in \mathbb{Q} \implies x+\frac{1}{x}\in\mathbb{Q}$? Let $n \in \mathbb{N}$. If $x>0$ is such that $x^{n}+\frac{1}{x^n}$ and $x^{n+1}+\frac{1}{x^{n+1}}\in \mathbb{Q} \implies x+\frac{1}{x}\in\mathbb{Q}$? 
Any thoughts on how to solve the above problem. Working for $n=2$ says that this result is true, but not sure if one can generalize 

Proffering the following argument based on elementary properties of algebraic numbers.
From $x^n+x^{-n}=q_1$ it follows that $x$ satisfies the polynomial equation $x^{2n}-q_1x^n+1=0$. Furthermore, it is obvious that all the zeros of this polynomial are $x^{\epsilon}\zeta_n^k$, where $\epsilon=\pm1, \zeta_n=e^{2\pi i/n}$ and $k=0,1,2,\ldots,n-1$.
Therefore the zeros of the minimal polynomial $m(T)$ of $x$ (over $\Bbb{Q}$) are among those numbers. 
But, from $x^{n+1}+x^{-(n+1)}=q_2$ it similarly follows that the zeros of the minimal polynomial of $x$ are among the numbers $x^{\epsilon}\zeta_{n+1}^\ell, \ell=0,1,2,\ldots,n$.
Therefore the zeros of $m(T)$ are either just $x$, or both $x$ and $x^{-1}$.
In the former case $x$ is rational, and the claim is immediate. In the latter case $x+\dfrac1x$ is rational because it is the coefficient of the linear term of the minimal polynomial $m(T)=(T-x)(T-1/x)\in\Bbb{Q}[x]$.

This argument also proves that if $x^n+x^{-n}$ and
$x^{n+1}+x^{-(n+1)}$ are both integers, then $x+1/x$ must also be an integer. This is because in this case $x$ is an algebraic integer, and hence the coefficients of $m(T)$ are all integers. This old trick then implies that $x^k+x^{-k}\in\Bbb{Z}$ for all $k\in\Bbb{Z}$.
