Year: 2015
Type: preprint
Abstract: Let $p$ be an odd prime, and let $m$ be an integer with $p\nmid m$. In this paper show that $$\sum_{k=0}^{p-1}\frac{\binom{2k}k\binom ak\binom{-1-a}k}{m^k} \equiv 0\pmod p \quad\hbox{implies}\quad\sum_{k=0}^{p-1}\frac{\binom{2k}k\binom ak \binom{-1-a}k}{m^k}\equiv 0\pmod {p^2}.$$
Source: arXiv (Cornell University)
Author: Sun, Zhi-Hong
Institution: Huaiyin Normal University
Language: English
Subfield: Algebra and Number Theory
Field: Mathematics
Domain: Physical Sciences
Open Access status: green
OpenAlex ID: W1845310800