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 $k\in\Bbb{Z}$.
The idea is to write $x+\frac{1}{x}$ in terms of $x^n+\frac{1}{x^n}$ and $x^{n+1}+\frac{1}{x^{n+1}}$. A domino-sum does the job here.
Put $w_n=x^n+\frac{1}{x^n}$. Then, the fundamental identity is
$$ w_{i}w_{j}=w_{i-j}+w_{i+j} \ (i,j\in{\mathbb Z}) \tag{1} $$
Say that a number $i$ is nice when $w_i\in{\mathbb Q}$. Note that $0$ is nice since $w_0=2$. It follows then from (1) (with $j=i$) that if $i$ is nice then $2i$ is nice also. Using (1) again (with $j=2i$) we see that $3i$ is nice also. More generally, by induction we have that any multiple of a nice integer is nice.
Next, look at those identities : (they all follow from (1))
$$ \begin{array}{lclclclcl} w_{n+1}w_{n} &=& w_1 &+& w_{2n+1} & & & & \\ w_{n+1}w_{3n} &=& & & w_{2n-1} &+& w_{4n+1} & & \\ w_{n+1}w_{5n} &=& & & & & w_{4n-1} &+& w_{6n+1} \\ \end{array}\tag{2} $$
If we look at terms in the same column, the sum is a multiple of $w_1$ : $w_{2n+1}+w_{2n-1}=w_1w_{2n}$, $w_{4n+1}+w_{4n-1}=w_1w_{4n}$ etc. Formally, we have for any $r>0$,
$$ w_{n+1}\bigg(\sum_{j=1}^r w_{(2j-1)n}\bigg)= w_1+\sum_{j=1}^{r-1} w_1w_{2jn} +w_{2rn+1} \tag{3} $$
We can rewrite this as
$$ w_1=\frac{w_{n+1}\bigg(\sum_{j=1}^r w_{(2j-1)n}\bigg)-w_{2rn+1}}{1+\sum_{j=1}^{r-1} w_{2jn}} \tag{4} $$
In the RHS of (4), all the variables are rational numbers, except possibly $w_{2rn+1}$. When $n$ is even, for suitable $r$ (for example $r=\frac{n+1}{2}$), $2rn+1$ will be divisible by $n+1$, so $w_{2rn+1}$ is rational also, which finishes the proof (note that the denominator is nonzero because it is positive). Similarly, when $n$ is even one can use the identity
$$ w_1=\frac{w_{n+1}\bigg(1+\sum_{j=2}^r w_{2jn}\bigg)-w_{2rn+n+1}}{1+\sum_{j=2}^{r} w_{(2j-1)n}} \tag{4'} $$
and take a suitable value of $r$ so that $2rn$ is a multiple of $n+1$. To derive (4'), notice that $d_j=w_{n+1}w_{2jn}-w_1w_{(2j-1)n}$ can be simplified to $d_j=u_{j}-u_{j-1}$ where $u_{j}=w_{2jn+n+1}$. Then $\sum_{j=1}^r d_j = u_{r}-u_{0}$, and (4') follows.
Well, if $x^n\in\mathbb{Q}$, then $\frac{1}{x^n}\in\mathbb{Q}$, so we can ask the question if $x^n\in\mathbb{Q}$ for all $m$ in $n+m$ if $n$ and $n+1$ is rational, and any rational number to the power of a natural number is indeed rational, and no irrational number can be written as both $\sqrt[2]{a/b}$ and $\sqrt[3]{a/b}$, (so for example $\sqrt{2}$ is rational if squared, but not if cubed), so yes, $x+\frac{1}{x}\in\mathbb{Q}$ if $x^n+\frac{1}{x^n}\in\mathbb{Q}$
