3
$\begingroup$

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?

Improve this question
$\endgroup$
1

2 Answers 2

Reset to default
6
$\begingroup$

Proved by grobber (Alexandru Chirvasitu) on AoPS: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=61&t=157732

EDIT: Also, http://www.artofproblemsolving.com/Forum/viewtopic.php?f=38&t=335001 might contain a solution (posts #4, #7 and #8).

Improve this answer
$\endgroup$
7
$\begingroup$

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.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .