Supercongruences involving products of two binomial coefficients

TL;DR

This paper establishes new supercongruences involving sums of products of binomial coefficients modulo prime powers, expanding the understanding of binomial sum congruences in number theory.

Contribution

It introduces novel supercongruences for binomial coefficient sums modulo powers of primes, including explicit formulas involving Euler polynomials.

Findings

Proved a supercongruence for sums involving binomial coefficients modulo p.

Derived a supercongruence involving Euler polynomials modulo p^3.

Established a binomial sum congruence modulo p^2 for specific parameters.

Abstract

In this paper we deduce some new supercongruences modulo powers of a prime $p>3$. Let $d\in\{0,1,\ldots,(p-1)/2\}$. We show that $$\sum_{k=0}^{(p-1)/2}\frac{\binom{2k}k\binom{2k}{k+d}}{8^k}\equiv 0\ (\mbox{mod}\ p)\ \ \ \mbox{if}\ d\equiv \frac{p+1}2\ (\mbox{mod}\ 2),$$ and $$\sum_{k=0}^{(p-1)/2}\frac{\binom{2k}k\binom{2k}{k+d}}{16^k} \equiv\left(\frac{-1}p\right)+p^2\frac{(-1)^d}4E_{p-3}\left(d+\frac12\right)\pmod{p^3},$$ where $E_{p-3}(x)$ denotes the Euler polynomial of degree $p-3$, and $(-)$ stands for the Legendre symbol. The paper also contains some other results such as $$\sum_{k=0}^{p-1}k^{(1+(\frac{-1}p))/2}\frac{\binom{6k}{3k}\binom{3k}k}{864^k}\equiv0\pmod{p^2}.$$