Related papers: On $p$-adic congruences involving $\sqrt d$

The purpose of this note is to obtain some congruences modulo a power of a prime $p$ involving the truncated hypergeometric series $$\sum_{k=1}^{p-1} {(x)_k(1-x)_k\over (1)_k^2}\cdot{1\over k^a}$$ for $a=1$ and $a=2$. In the last section,…

Let $p$ be an odd prime, and let $a$ be a rational $p$-adic integer with $a\not\equiv 0\pmod p$. In this paper, using WZ method we establish the congruences for $\sum_{k=0}^{p-1} \binom ak^2(-1)^k(1-\frac 2ak)$ modulo $p^2$ and…

Given a prime p and a positive integer m satisfying a certain inequality, the converse of Wolstenholme's Theorem is shown to hold for the product mp^k where k is any positive integer, generalizing a result by Helou and Terjanian.

Let $p>3$ be a prime, and let $m$ be an integer with $p\nmid m$. In the paper we prove some supercongruences concerning $$\align &\sum_{k=0}^{p-1}\frac{\binom{2k}k\binom{3k}k}{54^k},\…

For any odd prime number $p$, let $(\cdot|p)$ be the Legendre symbol, and let $n_1(p)<n_2(p)<\cdots$ be the sequence of positive nonresidues modulo $p$, i.e., $(n_k|p)=-1$ for each $k$. In 1957, Burgess showed that the upper bound…

Every odd prime number p can be written in exactly (p + 1)/2 ways as a sum ab+cd of two ordered products ab and cd such that min(a, b) > max(c, d). An easy corollary is a proof of Fermat's Theorem expressing primes in 1 + 4N as sums of two…

Let $m$ and $n>0$ be integers. Suppose that $p$ is a prime dividing $m-4$ but not dividing $m$. We show that $\nu_p(\sum_{k=0}^{n-1}\frac{\binom{2k}k}{m^k})$ and $\nu_p(\sum_{k=0}^{n-1}\binom{n-1}{k}(-1)^k\frac{\binom{2k}k}{m^k})$ are at…

It is significant to study congruences involving multiple harmonic sums. Let $p$ be an odd prime, in recent years, the following curious congruence $$\sum_{\substack{i+j+k=p \\ i, j, k>0}} \frac{1}{i j k} \equiv-2 B_{p-3}\pmod p$$ has been…

By a very simple argument, we prove that if $l,m,n$ are nonnegative integers then $$\sum_{k=0}^l(-1)^{m-k}\binom{l}{k}\binom{m-k}{n}\binom{2k}{k-2l+m} =\sum_{k=0}^l\binom{l}{k}\binom{2k}{n}\binom{n-l}{m+n-3k-l}. On the basis of this…

The numbers $R_n$ and $W_n$ are defined as \begin{align*} R_n=\sum_{k=0}^{n}{n+k\choose 2k}{2k\choose k}\frac{1}{2k-1},\ \text{and}\ W_n=\sum_{k=0}^{n}{n+k\choose 2k}{2k\choose k}\frac{3}{2k-3}. \end{align*} We prove that, for any positive…

For various positive integers $k$, the sums of $k$th powers of the first $n$ positive integers, $S_k(n+1)=1^k+2^k+...+n^k$, have got to be some of the most popular sums in all of mathematics. In this note we prove that for each $k\ge 2 $$…

For positive integers $1\leq i\leq k$, we consider the arithmetic properties of quotients of Wronskians in certain normalizations of the Andrews-Gordon $q$-series $$ \prod_{1\leq n\not \equiv 0,\pm i\pmod{2k+1}}\frac{1}{1-q^n}. $$ This…

For a positive integer $n$ let $\mathfrak{P}_n=\prod_{s_p(n)\ge p} p,$ where $p$ runs over all primes and $s_p(n)$ is the sum of the base $p$ digits of $n$. For all $n$ we prove that $\mathfrak{P}_n$ is divisible by all "small" primes with…

In this paper, we prove that the equation $x^2-(p^{2k+2}+1)y^2=-p^{2l+1}$, $l \in \{0,1,\dots,k\}, k \geq 0$, where $p$ is an odd prime number, is not solvable in positive integers $x$ and $y$. By combining that result with other known…

Let E be a real quadratic field with discriminant d and let p be an odd prime not dividing d. For \rho=1 or -1, we determine $\prod_{0<c<d, (d/c)=\rho} binomial coeff.{p-1}{\lfloor pc/d\rfloor}$ modulo p^2 in terms of Lucas numbers, the…

This article examines the nontrivial solutions of the congruence \[ (p-1)\cdots(p-r) \equiv -1 \pmod p. \] We discuss heuristics for the proportion of primes $p$ that have exactly $N$ solutions to this congruence. We supply numerical…

The usual product $m\cdot n$ on $\mathbb{Z}$ can be viewed as the sum of $n$ terms of an arithmetic progression whose first term is $a_{1}=m-n+1$ and whose difference is $d=2$. Generalizing this idea, we define new similar product mappings,…

We show that if {1, b, c, d} is a D(-1) diophantine quadruple with b<c<d and c=1+s^2, then the cases s=p^k, s=2p^k, c=p and c=2p^k do not occur, where p is an odd prime and k is a positive integer. For the integer d=1+x^2, we show that it…

Let $p$ be an odd prime, and let $\sum_{n=0}^{\infty} a_{n}X^{n}\in\mathbb{F}_p[[X]]$ be the reduction modulo $p$ of the Artin-Hasse exponential. We obtain a polynomial expression for $a_{kp}$ in terms of those $a_{rp}$ with $r<k$, for even…

Given a prime number $p$, the study of divisibility properties of a sequence $c(n)$ has two contending approaches: $p$-adic valuations and superconcongruences. The former searches for the highest power of $p$ dividing $c(n)$, for each $n$;…