Integral zeros of the Newton polynomial

Question

I'm trying to understand the following result;

Statement: A newton polynomial of the form $$a_1 {x\choose c_1}+a_2{x\choose c_2}+a_3{x\choose c_3}+⋯+a_s{x\choose c_s},$$ where $0 ≤c_1<c_2<c_3<⋯<c_s$ and $a_i$ are non-zero real numbers has $\textbf{at most}$ $s−1$ distinct roots in $[c_1,\infty) \cap \mathbb{Z}$

Algebraic intuition: I tried to induct on s and argue that adding a new term ${x\choose c_{s+1}}$, which is a monotonically increasing function would make the graph of the polynomial hit the $x$-axis at most once depending on the signs of the coefficients $a_i$, but I can't formalize this idea since there are lots of possibilities to consider.

Geometric intuition: the statement above essentially claims that the vector $(a_1,a_2,⋯,a_s) \in \mathbb{R}^s$ has at most $s-1$ perpendicular vectors in the set $\{({x\choose c_1},{x\choose c_2},⋯,{x\choose c_s}):x \in [c_1,\infty) \cap \mathbb{Z}\}$. I can't formalize this further, either.

Any help would be appreciated. Thanks, in advance.

$\textbf{Edit}$: $c_i$'s are integers.

Answer 1

Given a vector $(a_1, a_2, \cdots a_s) \in \mathbb{R}^s$, there is a unique hyperplane, say $H$, of dimension $s-1$ perpendicular to it. For $s=1$, the polynomial $\binom{x}{c_1}$ is non-zero and monotonic in $[c,\infty)$ and hence has no roots. Assume $s \geq 2$ and let $\tilde{r_i}:= (\binom{r_i}{c_1}, \binom{r_i}{c_2}, \cdots, \binom{r_i}{c_s})$. Order $\tilde{r_i}$ according to the ordering on $r_i$. If the polynomial has two distinct roots in the said domain, then $r_i \geq c_2$, for $i=1$, $2$, and a simple computation with binomial coefficients would show that the vectors $\tilde{r_1}$ and $ \tilde{r_2}$ are linearly independent. Since they lie in $H$, this would prove the case for $s=2$. From here, we proceed iteratively. A third root, if it exists would satisfy $r_3 \geq c_3$, and the vectors $\tilde{r_i}$, for $1 \leq i \leq 3$, would be the columns of the $s \times s$ matrix as in Corollary $2$ of the following paper by Gessel-Veinnot (https://www.sciencedirect.com/science/article/pii/0001870885901215). The determinant of this matrix is non-zero and hence the columns are linearly independent. Since the dimension of $H$ is $s-1$, the number of such root are bounded above by $s-1$.