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I'm thinking about the following problem: Given $n$ pairs of integers $(x_i, y_i)$, when is there a polynomial $f \in \Bbb Z[x]$ with $f(x_i) = y_i$ for each $i$? We can always do this with a rational polynomial, and the following procedure gives a way to determine if it is possible with integer polynomials or not:

Suppose that such an $f = f_0$ exists. By vertical/horizontal shifts, we may assume that $(x_n, y_n) = (0,0)$. Then $f_0(0) = 0$ so $f_0 = xf_1$ for some $f_1 \in \Bbb Z[x]$ that must satisfy $f_1(x_i - x_n) = \frac{y_i-y_n}{x_i-x_n}$ for $i = 1, \ldots, n-1$. Now induct.

All of the steps are reversible, so this shows that if such an integer polynomial exists, we can take it to be of degree $n-1$.

Instead of prescribing the $y_i$ at the beginning, let's ask ``What $n$-tuples of integers are of the form $(f(x_1), \ldots, f(x_n))$ for some $f \in \Bbb Z[x]$ and some fixed $x_i$?" Equivalently, what is the image of the natural map $\Bbb Z[x] \to \Bbb Z^n$ given by evaluation at those $n$ points? Since I'm really interested in the failure of this map to be surjective, the object I care about is the cokernel. To set some notation, if $S$ is a finite set of integers (let's say it's ordered) with $|S| = n$, let $V_S$ be the cokernel of the map $\Bbb Z[x] \to \Bbb Z^n$ given by evaluating at the elements of $S$.

The observation above shows that WLOG we can replace $\Bbb Z[x]$ by the group $\Bbb Z[x]_n$ of polynomials of degree at most $n-1$, and the map $\Bbb Z[x]_n \to \Bbb Z^n$ is given by the Vandermonde matrix with parameters the elements of $S$.

This immediately tells us that the size of $V_S$ is the determinant of the Vandermonde matrix. Also, the procedure above realizes $V_S$ as an extension \begin{equation*} 0 \to V_{(S-a_n) \setminus \{0\}} \to V_S \to \bigoplus_{i=1}^{n-1} \Bbb Z/(a_i-a_n)\Bbb Z \to 0 \end{equation*} which has been helpful for small $n$.

Is anything else known about the structure of $V_S$, and in particular, the map $\Bbb Z^n \to V_S$? Maybe we can determine the corresponding element of an Ext group (coming from the above SES) in some inductive way? I'm also interested in multivariable polynomials, but the problem seems less tractable there.

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    Interesting question! Clearly, $x_i - x_j \mid y_i - y_j$ is necessary for the existence of a required $f \in \mathbb{Z}\left[x\right]$, but it is not sufficient, as witnessed by the counterexample $n = 3$, $(x_1,y_1) = (0,0)$, $(x_2,y_2) = (2,8)$ and $(x_3,y_3) = (4,4)$. – 2017-01-14
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    @darijgrinberg Yep, the issue with that is that if there were such an $f$, then it would be divisible by $x$, and the quotient would be another integer polynomial with $g(x_i) = \frac{y_i}{x_i}$ for $i = 2,3$. That is, $g(2)=4$ and $g(4)=1$, which can't exist since $1-4$ is not divisible by $4-2$. If you just look at the condition $x_i-x_k | y_i - y_k$ for a fixed $k$, then satisfying that condition means you map to $0$ in the SES in my question (I took $k = n$) -- but that doesn't mean you are $0$ in $V_S$. – 2017-01-14

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