Proof of some conjectural supercongruences of Guo and Schlosser
This paper deduces some some supercongruence relations concerning trancated hypergeometric series, and in particular, establishes some supercongruences conjectured by Guo and Schlosser. More precisely, the following supercongruence relation is proved for truncated hypergeometric series $_5F_4$. Theorem 1. Let $p>3$ be a prime. Let $n$ be a positive integer and $x$ a non-zero integer such that $p\equiv x\pmod n$ with $|x|<n<p$. If $\frac{x}{n}+\frac{1}{2}$ is not non-positive integer, then \[_5F_4\left(\begin{matrix} \frac{x}{n}& \frac{x+2n}{2n}& \frac{1}{2}&\frac{x}{n}&\frac{x}{n}\\ && \frac{x}{2n}&\frac{x}{n}+\frac{1}{2}&1&1\end{matrix}\,|\, 1\right) \equiv (-1)^{\frac{p-x}{n}}+\frac{p+1}{2}\frac{p}{x}\Gamma_p\left(\frac{1}{2}-\frac{x}{n}\right) \Gamma_p\left(\frac{1}{2} +\frac{x}{n}\right) \pmod{p^3}.\] Here $\Gamma_p$ is the $p$-adic Gamma function. There are several corollaries to Theorem 1. Corollary. 1. Let $p\equiv 1\pmod 3$. Then \[\sum_{k=0}^{(p-1)/3} (6k+1)\frac{(\frac{1}{3})_k^4(2k)!}{k!^4(\frac{2}{3})_{2k}}\equiv p\pmod{p^3}.\] 2. Let $p\equiv 2\pmod 3$. Then \[\sum_{k=0}^{(p+1)/2} (6k-1)\frac{(-\frac{1}{3})^4_k(2k)!}{k!^4(-\frac{2}{3})_{2k}}\equiv p\pmod{p^3}.\] Theorem 2. Let $p\geq 5$ be a prime. Then \[\sum_{k=0}^{(p-1)/2} (-1)^k(4k+1)\frac{(\frac{1}{2})_k(-\frac{1}{2})_k^2}{(k+1)^2 k!} \equiv \begin{cases} 0\pmod{p^2}\quad\mbox{if }p\equiv 3\pmod 4\\ \frac{8p}{3}\pmod{p^2}\quad\mbox{if }p\equiv 1\pmod 4.\end{cases}\] Corollaries and Theorem 2 establish the conjectures posed by \textit{V. J. W. Guo} and \textit{M. J. Schlosser} [``Some $q$-supercongruences from transformation formulas for basic hypergeometric series'', Preprint, \url{arxiv:1812.06324}]. Proofs use some hypergeometric series identities.
scientific article; zbMATH DE number 7344950
scientific article