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Let $n_1 < \dotsb < n_k$ and $m_1, \dotsc, m_k$ be integers. Can we find conditions for the existence polynomial $p$ with integer coefficients such that $p(n_i) = m_i$ for every $i$?

This isn't always possible: for example, since $p(i) \equiv p(0) \mod{i}$, we cannot have $p(0) = 0$ and $p(2) = 1$.

By this answer, if such a polynomial exists then the Lagrange interpolating polynomial has integer coefficients, so the question may be restated as when does the Lagrange interpolating polynomial have integer coefficients.

Community
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Oliver Miller
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2 Answers2

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The coefficients of the Lagrange polynomial that goes through $(n_1,m_1),\ldots,(n_k,m_k)$, can be found by solving $$\begin{pmatrix}1&n_1&\cdots&n_1^{k-1}\\\vdots\\1&n_k&\cdots&n_k^{k-1}\end{pmatrix}\begin{pmatrix}a_0\\\vdots\\a_{n-1}\end{pmatrix}=\begin{pmatrix}m_1\\\vdots\\m_k\end{pmatrix}$$

Hence the coefficients $a_i$ are integers precisely when the determinant of the Vandermonde matrix above divides its adjugate matrix applied to $(m_1,\ldots,m_k)$.

For example, for $n_1<n_2<n_3$ and $m_1,m_2,m_3$, the condition becomes $$(n_1-n_2)(n_1-n_3)(n_2-n_3)\quad | \quad \begin{pmatrix}n_2(n_2 - n_3) n_3 m_1 + n_1 n_3 (n_3-n_1) m_2 + n_1 (n_1 - n_2) n_2 m_3\\ n_3^2 (m_1 - m_2) + n_1^2 (m_2 - m_3) + n_2^2 (m_3-m_1)\\ n_3 (m_2-m_1) + n_2 (m_1 - m_3) + n_1 (m_3-m_2)\end{pmatrix}$$ (by which is meant that the left side divides each of the terms on the right).

Chrystomath
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I don’t have enough reputation to comment, this is not a complete answer.

I have a necessary and sufficient condition for consecutive ni, which is that the terms of the nth level difference are divisible by n! (We define the kth level difference to be the differences between consecutive terms of the (k-1)th level difference, with the 0th level difference being the sequence itself

Ex:

2nd level differences: 2 2

1st level differences: 2 4 6

The sequence itself: 0 2 6 12

This is proven by observing that any integer polynomial can be written as the sum of integer combinations of the polynomials x(x-1)(x-2)...(x-i)

user105362
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