$a_1^k+a_2^k+\dots+a_n^k$ integer implies all $a_i$ are integers?

Question

Let $n$ be a positive integer, and let $a_1,\ldots,a_n$ be rational numbers. Suppose that $a_1^k+a_2^k+\dots+a_n^k$ is an integer for all positive integers $k$. Is it true that $a_1,a_2,\dots,a_n$ must be integers?

From the conditions, we can say something about the elementary symmetric polynomials. For example, $\sum_{i<j}a_ia_j=\dfrac12\left(\left(\sum a_i\right)^2-\sum a_i^2\right)$ is half an integer. It is still hard to imply anything about the $a_i$'s themselves.

Answer 1

I think that you can prove more. Suppose that $n\geq 1$, and that $v_k=b_1a_1^k+\cdots+b_n a_n^k\in \mathbb{Z}$ for all $k$, with $b_j$ non zero rational numbers and $a_j$ distinct rational numbers. Then all $a_k$ are integers. Some hints:

1. To prove the case $n=1$, write $a_1=u_1/v_1$ , with $u_1$ and $v_1$ with no prime commun factors. If $p$ is a prime number such that $p$ divide $v_1$, then for large $k$, we get a contradiction. Hence $v_1$ has no prime divisors, and $v_1=1$.

2. Use induction on $n$. If the result is true for $n-1$, with $n\geq 2$, fix a $j$, $1\leq j\leq n+1$, take $d_j$ such that $d_ja_j\in \mathbb{Z}$, and use the induction hypothesis for $w_k=d_j(v_{k+1}-a_jv_k)$ (the term in $a_j^k$ disappears).

Answer 2

Let $p$ be a prime and consider the system in $p$-adic rationals $\mathbb Q_p$.
Wlog. none of the $a_i$ is zero, then $a_i=p^{e_i}b_i$ with $e_i\in\mathbb Z$ and $b_i\in\mathbb Z_p^\times$. Then $b_i^{p-1}\in 1+p\mathbb Z_p$ and $b_i^{(p-1)p^r}\to 1$ as $r\to\infty$. Let $e=\min \{\,e_i\mid 1\le i\le n\}$ and $m=|\{\,i\mid e_i=e\,\}|$. Then for $k=(p-1)p^r$ with $r$ big enough (namely such that $k>m$) we have $a_1^k+\ldots +a_n^k\approx mp^{ke}$. This is an integer only if $e\ge0$. We conclude that none of the original $a_i$ has "$p$ in the denominator". As $p$ was an arbitrary prime, all $a_i$ are integers.