Fix integers $n\geqslant1$ and $k\geqslant 0$. For an
integer $i$, the $k$-fold derivative of $x^i$ can be denoted by
$i^{\underline{k}}x^{i-k}$ where $i^{\underline{k}}$ means
$i(i-1)\cdots(i-k+1)$ if $k>0$, and 1 if $k=0$. Let $\alpha$ be an
integer satisfying $\alpha^n\equiv1\pmod n$ and
$f(x)=\sum_{i=0}^{n-1}x^i$. We proved that the following rather weak
conditions on $(k,n,\alpha)$ are equivalent to the $k$-fold derivative
$f^{(k)}(\alpha)\equiv0\pmod n$:
(a) $k+1\not\in\{4,p\}$ where $p$ is prime, or
(b) $k+1=4$ and $4\nmid n$, or
(c) $k+1$ is a prime, call it $p$, and either $p\nmid n$ or
$\alpha\not\equiv1\pmod p$.
Assume $k\leqslant n$, otherwise $f^{(k)}(x)$ is the zero polynomial, and the conclusion is vacuously true.
Questions: Is this easily deduced from known results? (Our proof is 3 pages long!) Have you seen this before? Is it of interest? (We needed the case $k=1$ in a paper, that's how we came up with it.)
