# $x_1^n + \cdots + x_{\ell} ^n$ an integer for all $n \in \mathbb{N}$

Let $$x_1$$, $$\ldots$$, $$x_{\ell}$$ complex numbers such that $$h_n \colon =\sum_{k=1}^{\ell} x_k^n \in \mathbb{Z}$$ for all $$n \in \mathbb{N}$$. To show that $$e_k \colon = \sum x_{i_1} \cdots x_{i_k}$$ are integers for all $$1 \le k \le \ell$$

Note: The converse is clear from the formulas for the Newton's sums.

An attempt: One sees that $$k ! e_{k}$$ are integers for all $$1\le k \le \ell$$.

Now, since $$e_{\ell} ( x_1^d, \ldots, x_{\ell}^d) = x_1^d \cdots x_{\ell}^d = e_{\ell} ( x_1, \ldots, x_{\ell})^d$$

and

$$h_{n} (x_1^d, \ldots, x_{\ell}^d) = h_{ n d} (x_1, \ldots, x_{\ell})$$

we conclude that

$$\ell ! (x_1 \cdots x_{\ell})^d \in \mathbb{Z}$$ for all $$d \ge 1$$. This implies $$e_{\ell} =x_1\cdots x_{\ell} \in \mathbb{Z}$$.

Now, if $$\ell = 3$$, we can also show that $$e_2 \in \mathbb{Z}$$. Indeed, $$2 e_2 \in \mathbb{Z}$$, and then we have

$$h_4 = e_1^4 - 4 e_1^2 e_2 + 4 e_1 e_3 - 2 e_2^2$$

and we conclude $$2 e_2^2 \in \mathbb{Z}$$, and this implies $$e_2 \in \mathbb{Z}$$.

$$\bf{Added:}$$ For $$\ell = 4$$, one proceeds by showing that $$e_4$$, then $$e_2$$, then $$e_3$$ are integers ( note $$e_1 = h_1$$, so we don't worry about it). Perhaps this works for small $$\ell$$'s. Note that if the statement is true for $$\ell+1$$ it is also true for $$\ell$$ ( take the last number to be $$0$$).

Another possible approach: show that all of the $$x_k$$ are algebraic integers.( Clearly they are algebraic numbers, since we already know $$e_k$$ rational). So one might ask a related question:

Assume that all of the sums $$h_n =\sum x_k^n$$ are algebraic integers. Show that the $$x_k$$'s are algebraic integers.

$$\bf{Added:}$$ Another possible approach, and a generalization:

Let $$x_1$$, $$\ldots$$, $$x_{\ell}$$ distinct complex numbers, $$\alpha_1$$, $$\ldots$$, $$\alpha_{\ell}$$ non-zero numbers, such that

$$\sum_{k=1}^n \alpha_k x_k^n$$ is an algebraic integer for all $$n$$. Then $$x_k$$'s are algebraic integers ( and $$\alpha_k$$ are algebraic numbers).

Maybe looking at it in this way could provide a solution.

$$\bf{Added:}$$ I am happy that the answer is positive, as @Aphelli: has shown so neatly.

$$\bf{Added:}$$ Another similar approach:

Check that $$N x_k$$ are (algebraic) integers for some fixed $$N$$ ( in this case it's because the $$e_k$$'s are rational). But also $$N x_k^d$$ are integral for all $$d$$ and this implies $$x_k$$ integral. This procedure can work in other cases too, not only in the symmetric case. More precisely, suppose we have a polynomial $$P$$ in $$x_1$$, $$\ldots$$, $$x_{\ell}$$ ( integral coefficients) and $$P(x_1^n, \ldots, x_{\ell}^n)$$ is an integer for all $$n$$. Can we conclude that $$x_k$$'s are integers?

Assume that the extension $$\mathbb{Q}[ P[x_1, \ldots, x_{\ell}], P[x^2_1, \ldots, x^2_{\ell}], \ldots] \subset \mathbb{Q}[x_1, \ldots, x_{\ell}]$$

is integral. Then the same method applies.

Note that this will not work for the polynomial $$P[x_1, x_2] = x_1 - x_2$$. The statement clearly does not hold, since we could have $$x_1= x_2$$, otherwise arbitrary.

Actually, your argument is almost enough to conclude. There is a number field $$K$$ containing all the $$x_i$$. Consider any prime ideal $$\mathfrak{p}$$ of $$K$$ and let $$A$$ be the valuation ring at $$\mathfrak{p}$$.
I claim the following: let $$z_1,\ldots,z_{\ell} \in K$$ be such that for all $$n \geq 1$$, $$h_n(z_1,\ldots,z_{\ell}) \in A$$. Then all the $$z_i$$ are in $$A$$.
Proof: let $$S$$ be the set of indices $$i$$ such that $$z_i \notin A$$. Then for all $$n,d \geq 1$$, $$\sum_{i \in S}{(z_i^d)^n} \in A$$. Thus by your argument (that is, Newton sums), $$|S|!\prod_{i \in S}{(z_i)^d} \in A$$. But as $$d$$ goes to infinity, the LHS gets a negative valuation, unless $$S$$ is empty.
In particular, all the $$x_i$$ are in $$A$$. Thus, the $$x_i$$ have non-negative valuation at any prime of $$K$$, so they are algebraic integers (and thus their symmetric polynomials are algebraic integers as well).