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Assume that $\alpha_1,\ldots,\alpha_n$ are algebraic numbers. Assuming that

$\sum_{i=1}^n \alpha_i^k \in \mathbb{Z}$

for all $k\in\mathbb{N}$. Does this imply that the $\alpha_i$ are actually algebraic integers? I know that if these $\alpha_i$ are the conjugates of some algebraic number $\alpha$, then the relation implies that $\textrm{Tr}(\alpha^k)\in\mathbb{Z}$ for all $k\in\mathbb{N}$ (trace taken over e.g. $\mathbb{Q}(\alpha)/\mathbb{Q}$). This implies that $\alpha$ is an algebraic integer, so in this special case it's true.

Does anyone know about the general case?

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vess claimed to have proven something slightly better here: artofproblemsolving.com/Forum/viewtopic.php?p=178197#p178197 –  darij grinberg Sep 3 '11 at 22:55
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2 Answers

This is a consequence of American Mathematical Monthly problem E2993 (1983, p. 287), proposed by Michael Larsen. Solutions by A. A. Jagers and me can be found in American Mathematical Monthly 93 (1986), 483-484, http://www.jstor.org/stable/2323483.

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