Given a set $S$ of distinct natural numbers, we know that a subset $T$ that is $S$ with at most $k$ number of elements missing. Let $M_k := \big\{m_j\big|d_j = \sum_{i\in T}i^j, j\in \{1,2,...,k\}\,\big\}$. Does $M_k$ uniquely determine the missing at most $k$ natural numbers, or another word, $T$?
Ross Millikan made the connection of this problem to the moment problem by viewing it as the discrete version of it.
We can also formulate this problem as a binary linear equation or inequality problem.
For $k=2$, I have proved the claim. If $(a_1,a_2)$ and $(b_2,b_2)$ are two pairs of missing numbers, $a_1+a_2 = b_1+b_2$ and $a_1^2+a_2^2 = b_1^2+b_2^2$. Equivalently $a_1-b_1 = b_2-a_2$ and $(a_1-b_1)(a_1+b_1) = (b_2-a_2)(b_2+a_2)$. We conclude $a_1=b_2$ and $a_2 = b_1$. So the missing numbers are uniquely determined if the two moments are.
But the above method does not seem to carry over easily for $k>2$. Please help.
Denote the $k$ missing numbers as $x_1, x_2, \ldots , x_k$. Then we know $\sum_{i=1}^{k}{x_i^j}$ for $j=1, 2, \ldots , k$. Using Newton's identities, we may determine $\sum_{1 \leq i_1<i_2< \ldots<i_l \leq k}{x_{i_1}x_{i_2} \ldots x_{i_l}}$ for $l=1, 2, \ldots , k$, so we may use Vieta's formula to get a monic polynomial of degree $k$ with $x_1, \ldots , x_k$ as roots. This uniquely determines what $x_1, x_2, \ldots , x_k$ are.