Generalization of Integer-Powered Sums Problem

Question

I am trying to solve a problem that involves the sum of the $n$-th roots of positive reals. Specifically, the task is to determine all sets of positive reals $a_1, a_2, a_3$ such that $\sqrt[n]{a_1}+\sqrt[n]{a_2}+\sqrt[n]{a_3}$ is an integer for every integer $n \geq 1$.

From prior discussions and attempts at solutions, it appears that the only solution to this problem is when all numbers $a_1, \ldots, a_3$ are equal to 1. I have considered connections to Newton's sums and other algebraic properties, but these have not provided a full theoretical justification for why the sums are integers only under this specific condition. I found a similar problem here but it did not offer the insights needed for this generalization: $x_1^n + \cdots + x_{\ell} ^n$ an integer for all $n \in \mathbb{N}$.

I would greatly appreciate any explanations, proofs, or references to relevant mathematical literature that can confirm if the solution of all ones is indeed the only set of values that meets these criteria, and explore why no other sets of positive numbers can satisfy this condition.

Answer 1

Write $f(n) = \sqrt[n]{a_1} + \sqrt[n]{a_2} + \sqrt[n]{a_3}$ (there is nothing special about the number $3$ here and it could be replaced by any other positive integer in the following argument). Here is the idea: suppose we had something like $a_1 = 1, a_2 = 2, a_3 = 3$, so we wanted $1 + \sqrt[n]{2} + \sqrt[n]{3}$ to be an integer for all $n \ge 1$. Then it's not hard to see that as $n \to \infty$ all three terms approach a limit of $1$, which means the sum approaches a limit of $3$. But for a large finite value of $n$ it will be slightly larger than $3$, and in particular strictly between $3$ and $4$, and so can't be an integer. This argument generalizes to the case that $a_i \ge 1$ for $i = 1, 2, 3$; we'll always have $f(n) \ge 3$ and for large $n$ it will be slightly larger than $3$ unless $a_1 = a_2 = a_3 = 1$.

But in general we could have, for example, $a_1 <1 , a_2 = 1, a_3 > 1$, in which case it's less clear what happens and we need to do a bit more work. The key technical input is the observation that for large $n$ and fixed $a$ we have an asymptotic expansion

$$\sqrt[n]{a} = \exp \left( \frac{\ln a}{n} \right) = \sum_{k \ge 0} \frac{(\ln a)^k}{k! \, n^k}$$

in powers of $\frac{1}{n}$. This means $f(n)$ has an asymptotic expansion

$$\begin{eqnarray*} f(n) &=& \sum_{i=1}^3 \left( \sum_{k \ge 0} \frac{(\ln a_i)^k}{k! \, n^k} \right) \\ &=& \sum_{k \ge 0} \left( \frac{\sum_{i=1}^3 (\ln a_i)^k}{k! \, n^k} \right). \\ \end{eqnarray*}$$

It follows that in order for $f(n)$ to be an integer for large $n$ it must be exactly equal to $3$ (the constant $k = 0$ term of the above asymptotic expansion), which requires that every term after the constant term in this asymptotic expansion vanishes, meaning that $\sum_{i=1}^3 (\ln a_i)^k =0$ for all $k \ge 1$. But the logarithms $\ln a_i$ are real, so already for $k = 2$ the condition that $\sum_{i=1}^3 (\ln a_i)^2 = 0$ implies that $\ln a_i = 0$ for all $i$, as desired.

So already the quadratic term of the asymptotic expansion can't vanish unless $a_1 = a_2 = a_3 = 1$. However, it is possible for the linear term to vanish, since the coefficient of this term is $\ln a_1 + \ln a_2 + \ln a_3 = \ln a_1 a_2 a_3$ which vanishes if, for example, $a_1 = \frac{1}{2}, a_2 = 1, a_3 = 2$.

Answer 2

It has been established that if the sum is always an integer, then for $n$ sufficiently large it is $3$.

By suitably changing the names of things*, we may assume we have $$ a+b+c=a^2+b^2+c^2=a^3+b^3+c^3=3 $$ Now $$ ab+ac+bc={(a+b+c)^2-(a^2+b^2+c^2)\over2}=3 $$ and $$ abc={(a+b+c)^3-3(a+b+c)(a^2+b^2+c^2)+2(a^3+b^3+c^3)\over6}=1 $$ so $$ (x-a)(x-b)(x-c)=x^3-3x^2+3x-1=(x-1)^3 $$ so $a=b=c=1$.

*Namely, by taking $a=a_1^{1/(6k)}$, $b=a_2^{1/(6k)}$, $c=a_3^{1/(6k)}$, where $k$ is taken large enough to make $.7<a_i^{1/(2k)}<1.3$ for $i=1,2,3$, which can be done since $r>0$ implies $\lim_{k\to\infty}r^{1/k}=1$.