Let $m,n$ be positive integers, and $\displaystyle \Phi\_{m,n}~:~ {\mathbb{R}\_+^\*}^m \to \mathbb{R}\_+^\*, \ \ \ (x\_1,x\_2, \ldots , x\_m) \mapsto \sum\_{k=1}^m \sqrt[n]{x\_k}$.


Clearly for $m=1$ if for all positive integer $n$, we have $\Phi\_{1,n}(x) \in \mathbb Q$, then $x=1$.


It seems that the same conclusion holds for $m>1$ (or at least the subset of ${\mathbb{R}\_+^\*}^m$ for which $\Phi\_{m,n}(x) \in \mathbb Q$ is finite).


Is it true (or even obvious and I missed it)?



Sum of n-th roots is rarely rational
The following conclusion is true: If $\Phi\_{m,n}(x)\in\mathbb{Q}$ for all positive integers n, then x1=x2=...=xn=1.


It follows in what I believe is a fairly routine, or at least not too difficult, manner from the following Claim:


Let K be the extension field of ℚ generated by all n-th roots of all xi. Then K is a finite extension of ℚ.


Proof. Let yi be the N!-th root of xi. Then the power sum symmetric functions of the yi are all rational, hence the elementary symmetric functions are all rational, so the y's lie in a field extension of ℚ of degree at most m. Take N as large as you like. Voila!


