TITLE: If $a^m_{1}+a^m_{2}+\cdots+a^m_{n}\in\mathbb Z$, show $a_{i}\in\mathbb Z$
QUESTION [5 upvotes]: Let $a_{1},a_{2},\dots,a_{n}$ be rational numbers. If for any positive integer $m$
$$a^m_{1}+a^m_{2}+\cdots+a^m_{n}\in\mathbb Z$$
show that $a_{i}\in\mathbb Z$, $i=1,2,\dots,n$.

I tried: let $S(m)=a^m_{1}+a^m_{2}+\cdots+a^m_{n}$, then  use Newton's identities, I can prove $e_{i}\in\mathbb Z$ where $e_{i}$ is the $i$-th elementary symmetric polynomial, but I can't prove every $a_{i}\in\mathbb Z$. Thanks

REPLY [4 votes]: Suppose the result to be false and consider a counterexample with $n$  minimal. Let $b$ be the denominator of any $a_i$ and replace all the $a_i$ by $a_ib$. Since at least one $a_ib$ is an integer it can be ignored and so, by minimality of $n$ all the $a_ib$ are integers.
Similarly, we need only consider $b$ to be a prime $p$ say. Let $c_i=a_ip$, then each $c_i$ is coprime to $p$.
Now suppose $p$ is odd and consider just powers $m=(p-1)p^k$. Then each $c_i^m$ is congruent to 1 modulo $p^{k+1}$ and therefore the total sum of the $c_i^m$ is congruent to $n$ modulo $p^{k+1}$. For sufficiently large $k$ this sum cannot be divisible by $p^m$, a contradiction.
The case $p=2$ is handled the same way.