Related papers: A primality criterion based on a Lucas' congruence

Let p be any prime, and let a and n be nonnegative integers. Let $r\in Z$ and $f(x)\in Z[x]$. We establish the congruence $$p^{\deg f}\sum_{k=r(mod p^a)}\binom{n}{k}(-1)^k f((k-r)/p^a) =0 (mod p^{\sum_{i=a}^{\infty}[n/p^i]})$$ (motivated by…

In 1876, Edouard Lucas showed that if an integer $b$ exists such that $b^{n-1} \equiv 1 (\mathrm{mod} \ n)$ and $b^{(n-1)/p} \not\equiv 1( \mathrm{mod} \ n)$ for all prime divisors $p$ of $n-1$ , then $n$ is prime, a result known as Lucas's…

In this note we prove that {equation*} {np^s\choose mp^s+r}\equiv (-1)^{r-1}r^{-1}(m+1){n\choose m+1}p^s \pmod{p^{s+1}} {equation*} where $p$ is any prime, $n$, $m$, $s$ and $r$ are nonnegative integers such that $n\ge m$, $s\ge 1$, $1\le…

In 1878 \'E. Lucas proved a remarkable result which provides a simple way to compute the binomial coefficient ${n\choose m}$ modulo a prime $p$ in terms of the binomial coefficients of the base-$p$ digits of $n$ and $m$: {\it If $p$ is a…

In this paper we prove three conjectures on congruences involving central binomial coefficients or Lucas sequences. Let $p$ be an odd prime and let $a$ be a positive integer. We show that if $p\equiv 1\pmod{4}$ or $a>1$ then $$…

We prove that if $q$ is a power of a prime $p$ and $p^k$ divides $a$, with $k\ge 0$, then \[ 1+(q-1)\sum_{0\le b(q-1)<a} \binom{a}{b(q-1)}\equiv 0\pmod{p^{k+1}}. \] The special case of this congruence where $q=p$ was proved by Carlitz in…

Let $p$ be a prime, and let $k,n,m,n_0$ and $m_0$ be nonnegative integers such that $k\ge 1$, and $_0$ and $m_0$ are both less than $p$. K. Davis and W. Webb established that for a prime $p\ge 5$ the following variation of Lucas' Theorem…

In this paper we obtain some congruences involving central binomial coefficients and Lucas sequences. For example, we show that if p>5 is a prime then $\sum_{k=0}^{p-1}F_k*binom(2k,k)/12^k$ is congruent to 0,1,-1 modulo p according as p=1,4…

In this paper we establish some new supercongruences motivated by the well-known fact $\lim_{n\to\infty}(1+1/n)^n=e$. Let $p>3$ be a prime. We prove that $$\sum_{k=0}^{p-1}\binom{-1/(p+1)}k^{p+1}\equiv 0\ \pmod{p^5}\ \ \ \mbox{and}\ \ \…

Let n be a positive odd integer and let p>n+1 be a prime. We mainly derive the following congruence: $$\sum_{0<i_1<...<i_n<p}(i_1/3)(-1)^{i_1}/(i_1...i_n)=0 (mod p).$$

We will prove several congruences modulo a power of a prime such as $$ \sum_{0<k_1<...<k_{n}<p}\leg{p-k_{n}}{3} {(-1)^{k_{n}}\over k_1... k_{n}}\equiv {lll} -{2^{n+1}+2\over 6^{n+1}} p B_{p-n-1}({1\over 3}) &\pmod{p^2} &{if $n$ is odd}…

We show that if p is an odd prime then $$\sum_{k=0}^{p-1}E_kE_{p-1-k}=1 (mod p)$$ and $$\sum_{k=0}^{p-3}E_kE_{p-3-k}=(-1)^{(p-1)/2}2E_{p-3} (mod p),$$ where E_0,E_1,E_2,... are Euler numbers. Moreover, we prove that for any positive integer…

Let $P,Q\in\Bbb Z$, $U_0=0,\ U_1=1$ and $U_{n+1}=PU_n-QU_{n+1}$. In this paper we obtain a general congruence for $U_{kmn^r}/U_k\pmod {n^{r+1}}$, where $k,m,n,r$ are positive integers. As applications we extend Lucas' law of repetition and…

A frequently cited theorem says that for n > 0 and prime p, the sum of the first p n-th powers is congruent to -1 modulo p if p-1 divides n, and to 0 otherwise. We survey the main ingredients in several known proofs. Then we give an…

In this work we consider the congruence $\sum_{j=1}^{n-1} j^{k(n-1)} \equiv -1 \pmod n$ for each $k \in \mathbb{N}$, thus extending Giuga's ideas for $k=1$. In particular, it is proved that a pair $(n,k)\in \mathbb{N}^2$ satisfies this…

A prime number $p$ is said to be a Wolstenholme prime if it satisfies the congruence ${2p-1\choose p-1} \equiv 1 \,\,(\bmod{\,\,p^4})$. For such a prime $p$, we establish the expression for ${2p-1\choose p-1}\,\,(\bmod{\,\,p^8})$ given in…

For a positive integer $n$ let $H_n=\sum_{k=1}^{n}1/n$ be the $n$th harmonic number. In this note we prove that for any prime $p\ge 7$, $$ \sum_{k=1}^{p-1}\frac{H_k}{k^2}\equiv \sum_{k=1}^{p-1}\frac{H_k^2}{k}…

Using an identity arising in the known Banach probability problem on matchboxes, we prove an unexpected congruence for odd prime $p:$ for $1\leq k\leq \frac{p-1}{2},\enskip \sum_{i=1}^{p-2k-1}2^{i-1}\binom{k-1+i}{k}\equiv 0\pmod p.$

We show that for any prime prime $p\not=2$ $$\sum_{k=1}^{p-1} {(-1)^k\over k}{-{1\over 2} \choose k} \equiv -\sum_{k=1}^{(p-1)/2}{1\over k} \pmod{p^3}$$ by expressing the l.h.s. as a combination of alternating multiple harmonic sums.

In 2003, Zhao discovered a curious congruence involving harmonic series and Bernoulli numbers: for any odd prime $p$, $$\sum_{\substack{i,j,k\ge 1\\\gcd(ijk,p)=1\\i+j+k=p}}\frac{1}{ijk}\equiv -2B_{p-3} \pmod{p},$$ where $B_n$ is the $n$-th…