If $a^m_{1}+a^m_{2}+\cdots+a^m_{n}\in\mathbb Z$, show $a_{i}\in\mathbb Z$

Question

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

Answer 1

Easily deduced from the following more general result (I'm still trying to find the source):

Let $a_1,\ldots,a_n\geqslant 1$ be distinct rational numbers. Suppose that there are nonzero real numbers $c_1,\ldots,c_n$ and a sequence $\{b_m\}$ of integers such that $$\lim_{m\to\infty}\left(b_m-\sum_{k=1}^n c_k a_k^m\right)=0.$$ Then $a_1,\ldots,a_n$ are integers.

The proof I know uses induction on $n$. Denote $r_m=b_m-\sum_{k=1}^n c_k a_k^m$, take arbitrary $1\leqslant\ell\leqslant n$, and let $a_\ell=p/q$ with $\gcd(p,q)=1$. Then $$qr_{m+1}-pr_m=(qb_{m+1}-pb_m)-\sum_{k\neq\ell}c_k(qa_k-p)a_k^m$$ tends to $0$ as $m\to\infty$. If $n>1$, this immediately gives the induction step: let $b_m'=qb_{m+1}-pb_m$ and $c_k'=c_k(qa_k-p)$ for $k\neq\ell$ (the latter are nonzero because $a_k$ are distinct); by induction, $a_k$ are integers for $k\neq\ell$, and since $\ell$ is arbitrary, we're done.

Now consider [the induction base] $n(=\ell)=1$. The above reads: $qb_{m+1}-pb_m\to 0$ as $m\to\infty$. Since these are integers, we have $qb_{m+1}-pb_m=0$ for all sufficiently large $m$, and we may also assume that $b_m\neq 0$ for these $m$ (since $a_1\geqslant 1$). Then $q^k b_{m+k}=p^k b_m$ for all $k$, hence $q^k\mid b_m$ for all $k$, which is possible only if $q=1$.

Answer 2

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.