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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)?

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How does your definition of Phi depend on n? –  Kevin Ventullo May 8 '10 at 21:16
    
@Kevin: Those are n-th roots. –  JBL May 8 '10 at 21:58
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up vote 3 down vote accepted

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!

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