TITLE: Deg $n$ integral polynomial $P(x)$ with $n+1$ integer solutions to $0\leq P\leq d$
QUESTION [5 upvotes]: Let $d\in\mathbf{N}$ be as follows: there exists a polynomial $P(x)$ with degree $n>1$ and integer coefficients, such that $P$ has $n+1$ integer solutions to
\begin{equation*}
0\leq P(x) \leq d
\end{equation*}
What is the minimum possible value of $d$? What if $P$ has $n+2$ integer solutions?

REPLY [2 votes]: Let $x_0<\dots<x_n$ be the points where small values $y_0,\dots,y_n$ are attained. By Lagrange's interpolation formula, the number
$$
  \sum_{i=0}^n y_i\cdot\left(\prod_{j\neq i} (x_i-x_j)\right)^{-1}
\qquad(*)
$$
is a nonzero integer $p$, as it is the leading coefficient of $P$. This yields an easy lower bound for the maximum of the $y_i$, as the inverse to the sum of a half of the coefficients in $(*)$ (even or odd ones, depending on the sign of $p$).
This bound is minimal when $x_i=i$, and in this case both bounds equal $ n!/2^{n-1}$. Hence this is a total lower bound. It seems to differ from the upper bound mentioned by @Wolfgang in the comment by a factor of $\Theta(\sqrt n)$.
On the other hand, perhaps, this bound may even be (almost) achieved, especially when it is integral (which happens when $n$ is a power of 2)?