Is $(5+\sqrt[3]2)^n$ ever an integer for $n \in \Bbb Z \setminus \{0\}$?

Question

In general, I would like to prove that if $m>2$ is an integer, then $(5+\sqrt[m]2)^n$ is never an integer (unless for $n=0$).

First, I'm interested in the simple case $m=3$ (I already solved it for $m=2$). I actually think that $x_n=(1+\sqrt[3]2)^n$ is never a rational.

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Here is what I tried.

Suppose that $r:=x_n \in \mathbb{Q}$ for some $n>0$. Then $\sqrt[n]{r} = 5+\sqrt[3]{2}$ has degree $3$ over $\Bbb Q$ and its minimal polynomial is $x^3-15 x^2+75 x-127$. It has to divide $X^n-r$ in $\Bbb Q[X]$. Therefore, we could find three integers $0≤r_1<r_2<r_3<n$ such that $$(X-\sqrt[n]{r}\zeta_n^{r_1}) (X-\sqrt[n]{r}\zeta_n^{r_2}) (X-\sqrt[n]{r}\zeta_n^{r_3}) = X^3-15 X^2+75 X-127$$ (and the product of the other linear factors $(X-\sqrt[n]{r}\zeta_n^{j})$ should have rational coefficients). I tried to compare the constant terms... without success. Even if Galois theory was useful for the case $m=2$, I'm not sure how to possibly use it here $(m≥3)$. Computing the norms $N_{\Bbb Q(\sqrt[m]{2})/\Bbb Q}$ could give necessary conditions on $n$ for $x_n$ being a rational number.

I also tried with field extensions and degrees, but I don't know anything about the irreducibility of $X^n-r$, for instance.

Thank you in advance!

Answer 1

Suppose $(5+\sqrt[m]{2})^n$ is rational. Then it is fixed by the Galois group of $x^m-2$. This contains $\sqrt[m]{2}\mapsto \zeta \sqrt[m]{2}$ where $\zeta=e^{2\pi i/m}$ is a primitive $m$th root of 1.

Thus $(5+\sqrt[m]{2})^n= (5+\zeta \sqrt[m]{2})^n$. This is impossible: they don't even have the same absolute value.

Answer 2

In general, $(k + \sqrt[\ell]{n})^m \not \in \mathbb{Z}$ for any $k, l, m, n \in \mathbb{Z}_+$, $\ell \geq 2$, and $n$ square free (which includes the case $n = 2$).

This is seen by using the binomial formula: $$(k + \sqrt[\ell]{n})^m = \sum_{j=0}^{m}\binom{m}{j}k^{m-j}n^{j/\ell}$$

Since $x^\ell - n$ is irreducible over $\mathbb{Q}$ by eisenstein's criterion and $n^{1/\ell}$ is a root of that polynomial, $n^{1/\ell}, n^{2/\ell}, \cdots, n^{(\ell - 1)/\ell}$ are independent over $\mathbb{Q}$, since $n^{1/\ell}$ has a minimal polynomial of degree $\ell$. So, when we rewrite the above binomial expansion in terms of these powers, (if $i\equiv j$ is interpreted modulo $\ell$) $$(k + \sqrt[\ell]{n})^m = \sum_{j=1}^{m}\binom{m}{j}k^{m-j}n^{j/\ell} = \sum_{i=0}^{\ell-1}\left(\sum_{j\equiv i}\binom{m}{j}k^{m-j}n^{(j-i)/\ell}\right)n^{i/\ell}.$$

Note that $j-i$ is a multiple of $\ell$ since $i \equiv j$.

Since this is a positive integer linear combination of elements that are independent over the the rational numbers, if it were rational then all the coefficients would be zero except those of $n^{0/\ell}$, however, we see that this is not true.