New congruences involving products of two binomial coefficients

TL;DR

This paper proves new supercongruences involving sums of products of binomial coefficients modulo prime powers, confirming conjectures and extending known results in number theory.

Contribution

The paper establishes novel congruences for sums of binomial coefficient products modulo prime powers, confirming conjectures of Z.-W. Sun and extending previous results.

Findings

Proved a supercongruence for a sum involving inom{2k}k^2/16^k modulo p^3.

Established new congruences for sums involving inom{2k}kinom{3k}k and others modulo p^2.

Confirmed conjectures posed by Rodriguez-Villeguez in 2003.

Abstract

Let $p>3$ be a prime and let $a$ be a positive integer. We show that if $p\equiv1\pmod 4$ or $a>1$ then $$\sum_{k=0}^{\lfloor\frac34p^a\rfloor}\frac{\binom{2k}k^2}{16^k}\equiv\l(\frac{-1}{p^a}\r)\pmod{p^3}$$ with $(-)$ the Jacobi symbol, which confirms a conjecture of Z.-W. Sun. We also establish the following new congruences: \begin{align*}\sum_{k=0}^{(p-1)/2}\frac{\binom{2k}k\binom{3k}k}{27^k}\equiv&\l(\frac p3\r)\frac{2^p+1}3\pmod{p^2}, \\\sum_{k=0}^{(p-1)/2}\frac{\binom{6k}{3k}\binom{3k}k}{(2k+1)432^k}\equiv&\l(\frac p3\r)\frac{3^p+1}4\pmod{p^2}, \\\sum_{k=0}^{(p-1)/2}\frac{\binom{4k}{2k}\binom{2k}k}{(2k+1)64^k}\equiv&\l(\frac{-1}p\r)2^{p-1}\pmod{p^2}. \end{align*} Note that in 2003 Rodriguez-Villeguez posed conjectures on $$\sum_{k=0}^{p-1}\frac{\binom{2k}k^2}{16^k},\ \sum_{k=0}^{p-1}\frac{\binom{2k}k\binom{3k}k}{27^k},\ \sum_{k=1}^{p-1}\frac{\binom{4k}{2k}\binom{2k}k}{64^k},\ \sum_{k=1}^{p-1}\frac{\binom{6k}{3k}\binom{3k}k}{432^k}$$ modulo $p^2$ which were later proved.