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?

$\endgroup$
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).

$\endgroup$
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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.