If $\sum_{f(\alpha)=0}\alpha^k$ is an integer for each $k\geq 0$, is $f$ monic with integer coefficients? Suppose $f(x)$ is a monic polynomial with rational coefficients. Denote its roots $\alpha_1...\alpha_d$, counting multiplicity. The expressions
$$t_k=\sum_{i=1}^{d}\alpha_i^k$$
are rational for all $k\geq0$, by Galois theory. If $f(x)$ has integer coefficients, then the $\alpha_i$ are algebraic integers and thus the $t_k$ are integers for all $k$.

Does the converse hold? When $t_k$ are integers for all $k\geq0$, does this mean that $f(x)$ has integer coefficients?

I am aware of Newton's identities, which relate these $t_k$ to the coefficients of $f$: https://en.wikipedia.org/wiki/Newton%27s_identities. This bounds the denominators of the coefficients of $f(x)$, but it doesn't tell me they are integers. This question has been bugging me for days now and I'm sure there's some really clean argument I'm just not seeing.
 A: This is a clarification of the answer to this question found on Sil's AoPS forum link.We show that $f$ has integer coefficients by showing that the $\alpha_i$ are all algebraic integers, in a sense skipping the middle man of computing coefficients. Let $K=\mathbb{Q}(\alpha_1,...\alpha_n)$. It is standard that $\alpha_i$ are algebraic integers if and only if they lie in the ring of integers of the localization $K_{\mathfrak{p}}$ of $K$ at every prime ideal $\mathfrak{p}$. Namely, the following proposition sufficies for our purposes:
Proposition: If $K/\mathbb{Q}_p$ is a finite extension and $\alpha_1...\alpha_d\in K$ are such that
$$\sum_{i}\alpha_i^k\in \mathcal{O}_K$$
for every $k\geq0$, then $\alpha_i\in \mathcal{O}_K$.
Proof: Let $\varpi$ be a uniformizer for $K$, and let $v:K^{\times}\to\mathbb{Q}$ be the $\varpi$-adic valuation normalized so that $v(p)=e$, $v(\varpi)=1$. (NB: This is different normalization than in the AoPS forum). For the sake of contradiction, suppose that the minimal valuation among the $\alpha_i$ is negative, say $-r$.
Set $\beta_i=\varpi^r \alpha_i$. These values are all in $\mathcal{O}_K$, some of them are in $\mathcal{O}_K^{\times}$, and
$$v\left(\sum_{i}\beta_i^k\right)\geq r\cdot k\geq k.$$
Seeing as removing values of $\beta_i\in \varpi \mathcal{O}_K$won't ruin this last property, we can WLOG assume all of the $\beta_i\in \mathcal{O}_K^{\times}$ to deduce our contradiction.
Set $\gamma_1...\gamma_{p^f}$ be a full set of representatives for $\mathcal{O}_K^{\times}$. Let $n_1...n_{p^f}$ be the count of elements $\beta_i$ in each representative class. It is a straightforward computation (i.e, expand binomial coefficients) to show that
$$\sum_{i}\beta_i^k\equiv \sum_{j}n_j \gamma_j^k \,\,\text{mod}\, \varpi^{v(k)/e}.$$
Divide out so that at least one of the $n_j$ is not a multiple of $p$. Choose $k_0$ so that $\sum_{i}\beta_i^k\equiv \sum_{j}n_j \gamma_j^{k_0}$ is not a multiple of $\varpi$. Choosing larger and larger values of $k$ congruent to $k_0$ modulo $p^f$, the fact that $\sum_{i}\beta_i^k\equiv \sum_{j}n_j \gamma_j^{k}$ is not a multiple of varpi changes, but the lower bound on valuation becomes non trivial. This gives a contradiction, yielding the result.
A: Assume the $t_k$ are all integers. The generating function for the sequence $(t_k)$ is
$$\sum_{k=0}^\infty t_k X^k = \sum_{k=0}^\infty(\alpha_1^kX^k + \ldots+ \alpha_d^k X^k) =  \frac{1}{1-\alpha_1 X} + \ldots + \frac{1}{1-\alpha_d X} =: \frac{p(X)}{q(X)}$$
for some relatively prime polynomials $p(X),\, q(X) \in \mathbb{Z}[X]$. It's not too hard to show that $q(0) = \pm 1$ if and only if $f(x)$ has integer coefficients.
Since $p$ and $q$ are relatively prime, we can use the Euclidean algorithm to find $a(X)$ and $b(X)$ in $\mathbb{Z}[X]$ such that
$$ a(X)p(X) + b(X)q(X) = d$$
for some integer $d$. Then
$$a(X)\cdot \frac{p(X)}{q(X)} + b(X) = \frac{d}{q(X)}.$$
Since $\frac{p(X)}{q(X)}$ has an integral power series expansion, so does $\frac{d}{q(X)}$.
Write
$$d = q(X)\cdot(c_0 + c_1 X + c_2X^2 + \ldots).$$
We can assume any common factors have been divided out of both sides. In particular, for every prime $\ell$ both sides of the equation are nonzero mod $\ell$, i.e. $d=q(0)\,c_0$ is nonzero mod $\ell$.
Since $q(0)$ is not divisible by any prime $\ell$, it must be $\pm 1$. So $f(x)$ has integer coefficients.
