English
Related papers

Related papers: Congruences for $q^{[p/8]}\pmod p$ II

200 papers

We produce congruences modulo a prime $p>3$ for sums $\sum_k\binom{3k}{k}x^k$ over ranges $0\le k<q$ and $0\le k<q/3$, where $q$ is a power of $p$. Here $x$ equals either $c^2/(1-c)^3$, or $4s^2/\bigl(27(s^2-1)\bigr)$, where $c$ and $s$ are…

Number Theory · Mathematics 2022-10-13 Sandro Mattarei , Roberto Tauraso

Let $p>3$ be a prime. In this paper, we obtain the congruences for $$\sum_{k=0}^{p-1}\frac{w(k)\binom{2k}k^3}{(-8)^k},\ \sum_{k=0}^{p-1}\frac{w(k)\binom{2k}k^2\binom{3k}k}{(-192)^k},\…

Number Theory · Mathematics 2022-11-28 Zhi-Hong Sun

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…

Number Theory · Mathematics 2013-10-31 Zhi-Wei Sun

In this paper, we show that if $p\equiv 1\pmod 4$ is prime, then $4F_p$ admits a representation of the form $u^2-pv^2$ for some integers $u$ and $v$, where $F_n$ is the $n$th Fibonacci number. We prove a similar result when $p\equiv -1\pmod…

Number Theory · Mathematics 2015-02-17 Pedro Berrizbeitia , Florian Luca , Alberto Mendoza

Using the $q$-Wilf--Zeilberger method and a $q$-analogue of a "divergent" Ramanujan-type supercongruence, we give several $q$-supercongruences modulo the fourth power of a cyclotomic polynomial. One of them is a $q$-analogue of a…

Number Theory · Mathematics 2020-04-23 Victor J. W. Guo

Let k>1 be an integer and let p be a prime. We show that if $p^a\le k<2p^a$ or $k=p^aq+1$ (with 2q<p) for some a=1,2,..., then the set {\binom{n}{k}: n=0,1,2,...} is dense in the ring Z_p of p-adic integers, i.e., it contains a complete…

Number Theory · Mathematics 2011-01-26 Zhi-Wei Sun , Wei Zhang

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…

Number Theory · Mathematics 2007-07-25 Zhi-Wei Sun , Donald M. Davis

Let $p$ be an odd prime and let $a,b\in\mathbb Z$ with $p\nmid ab$. In this paper we mainly evaluate $$T_p^{(\delta)}(a,b,x):=\det\left[x+\tan\pi\frac{aj^2+bk^2}p\right]_{\delta\le j,k\le (p-1)/2}\ \ (\delta=0,1).$$ For example, in the case…

Number Theory · Mathematics 2023-12-25 Zhi-Wei Sun

In this work we construct a new class of maximal partial spreads in $PG(4,q)$, that we call $q$-added maximal partial spreads. We obtain them by depriving a spread of a hyperplane of some lines and adding $q+1$ lines not of the hyperplane…

Combinatorics · Mathematics 2013-01-24 Sandro Rajola , Maurizio Iurlo

In this paper, the divisibility property of the type 2 $(p, q)$-analogue of the $r$-Whitney numbers of the second kind is established. More precisely, a congruence relation modulo $pq$ for this $(p,q)$-analogue is derived.

Combinatorics · Mathematics 2022-08-10 Roberto B. Corcino , Cristina B. Corcino

Let p1, p2,..., pn be distinct prime numbers, and let Nn be their product. We prove that, for any positive integer L that is divisible by the least common multiple of p1 minus one, p2 minus one, and so on, and for integers a1, a2,..., an…

Number Theory · Mathematics 2025-10-14 Shao-Yuan Huang , Hsiu-Yu Wu

We confirm several conjectures of Sun involving quadratic residues modulo odd primes. For any prime $p\equiv 1\pmod 4$ and integer $a\not\equiv0\pmod p$, we prove that \begin{align*}&(-1)^{|\{1\le k<\frac p4:\ (\frac kp)=-1\}|}\prod_{1\le…

Number Theory · Mathematics 2020-03-13 Fedor Petrov , Zhi-Wei Sun

Let $p$ be a prime. In 1878 \'{E}. Lucas proved that the congruence $$ {p-1\choose k}\equiv (-1)^k\pmod{p}$$ holds for any nonnegative integer $k\in\{0,1,\ldots,p-1\}$. The converse statement was given in Problem 1494 of {\it Mathematics…

Number Theory · Mathematics 2018-04-10 Romeo Mestrovic

Let p be any prime, and $p^(\nu_p(n!))$ the maximal power of $p$ dividing $n!$. It is proved that there exists a positive integer $n_0$, which depends only on $p$, such that $q^(\nu_q(n!)) < p^(\nu_p(n!))$ for all $n \ge n_0$ and all primes…

Number Theory · Mathematics 2026-04-28 Dan Levy

A well-known conjecture asserts that there are infinitely many primes $p$ for which $p - 1$ is a perfect square. We obtain upper and lower bounds of matching order on the number of pairs of distinct primes $p,q \le x$ for which $(p - 1)(q -…

Number Theory · Mathematics 2015-07-23 Tristan Freiberg , Carl Pomerance

Let $p$ be an odd prime. Define the Gaussian power sum \[ G_n(p)=\sum_{a=1}^{p-1}\sum_{b=1}^{p-1}(a+bi)^n\in\mathbb Z[i]. \] We determine $G_p(p)$ modulo high powers of $p$: if $p\equiv 1\pmod 4$ then $$G_p(p)\equiv p^2(1+i)\pmod{p^3},$$…

General Mathematics · Mathematics 2026-02-04 Nikita Kalinin , Faith Shadow Zottor

Let $m>2$ and $q>0$ be integers with $m$ even or $q$ odd. We show the supercongruence $$\sum_{k=0}^{p-1}(-1)^{km}\binom{p/m-q}{k}^m\equiv0\pmod{p^3}.$$ for any prime $p>mq$. This confirms a conjecture of Sun.

Number Theory · Mathematics 2015-04-29 Xiang-Zi Meng , Zhi-Wei Sun

It is well known that when a prime $p$ is congruent to 1 modulo 4, the sum of the quadratic residues equals the sum of the quadratic nonresidues. In this note we give analogous results for the case where $p$ is congruent to 3 modulo 4.

Number Theory · Mathematics 2015-12-04 Christian Aebi , Grant Cairns

We give q-analogues of the following congruences by Z.-W. Sun: \sum_{k=1}^{p-1}\frac{D_k}{k} \equiv -\frac{2^{p-1}-1}{p} \pmod p,\\ \sum_{k=1}^{p-1}\frac{H_k}{k 2^k}\equiv 0 \pmod{p},\quad p\geqslant 5, where p is a prime,…

Number Theory · Mathematics 2013-12-24 Victor J. W. Guo

Let $p$ be a prime congruent to 1 or 3 modulo 8 so that the equation $p=a^2+2b^2$ is solvable in integers. In this paper, we obtain closed-form expressions for $a$ and $b$ in terms of Jacobsthal sums. This is analogous to a classical…

Number Theory · Mathematics 2011-10-27 Ki-Ichiro Hashimoto , Ling Long , Yifan Yang
‹ Prev 1 3 4 5 6 7 10 Next ›