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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.

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!

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  • $\begingroup$ We notice that if $x_n$ was a rational number, then it would be an integer, since it is an algebraic integer. $\endgroup$
    – Watson
    Oct 22, 2016 at 11:40

2 Answers 2

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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.

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    $\begingroup$ A very transparent approach, nice. $\endgroup$
    – Hubble
    Sep 11, 2016 at 16:33
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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.

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  • $\begingroup$ It should be $n^{j/\ell}$, though I fixed it. I copied it from above when I wrote it the first time, but realized that it was a typo. $\endgroup$
    – user357980
    Sep 11, 2016 at 16:35
  • $\begingroup$ Eisenstein's criterion works since the leading coefficient is $1$ and there is a prime $p$ that divides $\ell$ but not twice. That is, all that I need to assume is that $\ell$ has one prime divisor that has multiplicity one. $\endgroup$
    – user357980
    Sep 11, 2016 at 16:37
  • $\begingroup$ In fact you can get away with a weaker condition than squarefree; this proof works as long as you can write $(k + \sqrt[\ell]{n})^m$ as a positive linear combination of at least two elements independent over $\Bbb Q$: so you need that $\{1,n^{1/\ell},\dots,n^{(\ell - 1)/\ell}\}$ contains one non-rational number (which happens precisely when $n$ is not an $\ell$th power). $\endgroup$
    – Stahl
    Sep 11, 2016 at 16:45
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    $\begingroup$ Sorry, I was editting it. Yes, I did make a mistake, but I fixed it now. The reason that it made no difference since I was adding a collection of positive terms so there was going to be no cancellation. Basically, what I did was take the polynomial like $(5 + a)^7$, where $a = 2^{1/3}$ write it as $$a^7 + 5a^6 + 5^2a^5 + 5^3a^4 + 5^4a^3 + 5^5a^2 + 5^6a + 5^7,$$ which equals $$4a + 5(4) + 5^2(2)a^2 + 5^3(2)a + 5^4(2) + 5^5a^2 + 5^6a + 5^7,$$ which equals $$ (2(5^2) + 5^5)a^2 + (4 + 2(5^3) + 5^6)a + (5(4) + 5^4(2) + 5^7)$$ and noted that this could not be rational by independence. $\endgroup$
    – user357980
    Sep 11, 2016 at 16:46
  • $\begingroup$ If you know that one of $n^{1/\ell}, \dots, n^{(\ell-1)/\ell}$ is irrational, then that means that $n^{1/\ell}$ is irrational since they are gotten by successive powers of that term. However, I am not sure if that implies that they are linearlly independent over $\mathbb{Q}$, so I think that I might need it. I weakened it to knowing that $\ell$ has a prime divisor that has multiplicity $1$ to use Eisenstein's Criterion. If you can think of a weaker condition, that'd be cool. $\endgroup$
    – user357980
    Sep 11, 2016 at 16:53

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