Skip to main content

Questions tagged [congruences]

Filter by
Sorted by
Tagged with
5 votes
1 answer
430 views

About Morley congruence

Let $p>3$ be an odd prime and $a$ be a positive integer, is the following congruence true? $$\binom{p^a-1}{\frac{p^a-1}{2}}\equiv(-1)^{\frac{p^a-1}{2}}4^{p^a-1}\pmod{p^3}.$$ When $a=1$, this is ...
yang's user avatar
  • 51
2 votes
1 answer
295 views

Efficiently lifting $a^2+b^2 \equiv c^2 \pmod{n}$ to coprime integers

Let $n$ be integer with unknown factorization. Assume factoring $n$ is inefficient. Let $a,b,c$ satisfy $a^2+b^2 \equiv c^2 \bmod{n}, 0 \le a,b,c \le n-1$. Is it possibly to lift the above ...
joro's user avatar
  • 25.4k
11 votes
2 answers
1k views

Ramanujan's tau function, $691$ congruence, and $\eta(z)^{12}$

Let $q = e^{2\pi i\,z}$. I. 24th power The Ramanujan tau function $\tau(n)$ is given by the expansion of the Dedekind eta function $\eta(z)$'s $\text{24th}$ power. Then $$\begin{aligned}\eta(z)^{...
Tito Piezas III's user avatar
4 votes
2 answers
616 views

Order of a matrix congruent to the identity modulo p

I couldn't find a demonstration of this theorem: Given $A \in SO_2(\mathbb{Z}[{1 \over q_1},\dots,{1 \over q_k}])$ and $p$ prime $\notin \{q_1,\dots,q_k\}$ $\exists n \in \mathbb{N} : A^n=Id$ and $A\...
DarioDF's user avatar
  • 143
3 votes
1 answer
522 views

Solvability of Diophantine equation using congruences where the variables are bounded?

In page 2 of Mordell "Diophantion equations" it is given a necessary condition for the solvability of a Diophantine equation: Integer solutions of the inhomogeneous equation $f(x) = 0$ can exist only ...
asad's user avatar
  • 841
7 votes
2 answers
780 views

Some equalities involving prime powers

Let $p,a,b,x,y$ be positive integers where $p$ is an odd prime; $x$ and $y$ are odd; $p,x$ and $y$ are all coprime. I'm interested in finding examples of such numbers that satisfy this equation: \...
Nick Gill's user avatar
  • 11.2k
38 votes
4 answers
7k views

What did Yu Jianchun discover about Carmichael numbers?

There's a news story going around (see for example [1]; other accounts are even more breathless) about an amateur mathematician, Yu Jianchun, finding an "alternative method to verify Carmichael ...
Charles's user avatar
  • 9,114
1 vote
0 answers
242 views

Computational number theory

Suppose that $p$ is prime and $q$ is an even number divides $p-1$, such that $q<\frac{p-1}{q}$ and $u$ has order $q$ modulo $p$. Let $S$ be the subgroup of $Z^*_p$ consisting of the powers of $u$. ...
shamym shamim's user avatar
2 votes
0 answers
178 views

Monoid prime ideals and prime congruences

I was wondering what the connection is between the notion of "prime congruence" on a monoid, and the notion of "prime ideal" in a monoid. Starting from a prime ideal $P$ in a monoid $M$, one can ...
THC's user avatar
  • 4,547
4 votes
3 answers
544 views

On a theorem of Hensel about congruence of binomial coefficient

In the paper Binomial coefficients modulo prime powers, Andrew Granville stated the following theorem: Let $n, m$ and $r=n-m$ be three given positive integer and $p^k$ is the exact power of $p$ ...
alphaomega's user avatar
5 votes
1 answer
270 views

Up to $2000$, $A145722(n-1) \equiv \sigma(4n-3) \pmod{5}$

A145722 is Expansion of f(q) * f(q^5) / phi(-q^2)^2 in powers of q where f(), phi() are Ramanujan theta functions. Using the pari program and offset 0, up to $2000$...
joro's user avatar
  • 25.4k
5 votes
1 answer
258 views

For a sufficiently large $a$, are there distinct (mod $a$) integers such that all powers up to the $n$-th are "close" modulo $a$?

Given $n\in\Bbb N$ is there an $a_n\in\Bbb N$ such that for every $a>a_n$ there are two distinct integers $0<b<c<a$ such that $b^i\bmod a,c^i\bmod a\in(\sqrt a,\sqrt a\log a)$ for every $i\...
Turbo's user avatar
  • 13.9k
7 votes
1 answer
860 views

Is $n=6$ the only integer satisfies ${\sigma}_x(n) \equiv 0\bmod{n}$ for every odd integer $x > 0$ and $2 (\bmod n)$ if $x$ is even integer? [closed]

After a few computations in wolfram alpha about the divisor function for some values of $n$ to look the behavior of $\sigma_x(n)\bmod n$ for $\,n=6,\,$ i got this result : $\sigma_x(6)=0 \bmod 6$ for $...
zeraoulia rafik's user avatar
13 votes
1 answer
1k views

On cubic reciprocity for $x^3+y^3+z^3 = 996$?

I. The Diophantine equation, $$x^3+y^3+z^3 = 3w^3\tag1$$ with $x\geq y \geq z$ and $w=1$ has only two known solutions, namely $1,1,1$ and $4,4,-5$. Are there larger ones? As Noam Elkies points out ...
Tito Piezas III's user avatar
0 votes
0 answers
600 views

Systems of linear modular equations with unknowns in the moduli

I am interested in systems of linear modular equations, where the unknowns also appear in the moduli. The general form would be: $A \vec{x}= \vec{b} \;\textrm{mod} \; (C \vec{x}+\vec{d})$ where A ...
Arnaud's user avatar
  • 940
4 votes
0 answers
182 views

Computing the density of a set of multiples

Erdős and his coauthors often wrote about problems relating to the densities of sets of multiples. I have a computational question about the same topic. I have a finite* set $A=a_1<\cdots<a_r$ ...
Charles's user avatar
  • 9,114
10 votes
2 answers
1k views

Is there a fixed integer $x>1$ satisfing ${\sigma}^{k}(x)\equiv 0\pmod{x}$ for all positive integers $k$?

This question related to this question from SE. I'm interested to know if there exists an integer $x>1$ that satisfies $${\sigma}^{k}(x)\equiv 0\pmod{x}$$ for all positive integers $k$. Note. $\...
zeraoulia rafik's user avatar
19 votes
2 answers
1k views

A congruence involving binomial coefficients

The following open problem was shown to me by Maxim Kontsevich. I state it in a different but equivalent form. Let $a(n)$ be the sequence at http://oeis.org/A131868, that is, $$ a(n) =\frac{1}{2n^2}\...
Richard Stanley's user avatar
3 votes
1 answer
585 views

Trying to prove a congruence for Stirling numbers of the second kind

This a repost of a question I asked at Stack Exchange, but I got no answer so far, so I am trying here, even though it may not suit the "research level" requirement. Proposition: When $n$ and $m$ are ...
René Gy's user avatar
  • 505
2 votes
1 answer
377 views

Cardinality of the prime divisor set of a k-power sum

Let $a_{1},\dots,a_{n}$ be positive natural numbers ($n>2$) such that $a_{i}\neq a_{j}$ if $i\neq j$. I want to prove that $$ \left\lvert \left\{ p \text{ prime} \; : \; p \mid \sum_{i=1}^n a_{i}^{...
peppo's user avatar
  • 45
7 votes
1 answer
1k views

Roots of a polynomial in a finite field related to Fermat's Last Theorem

In my class, we proved the following condition: define the polynomial $P_l(x)$ as $$P_l(x) = \sum_{r=1}^{l-1}{\frac{1}{r}x^{l-1-r}}$$ Then if for all $a \in \mathbb{Z}/l\mathbb{Z}-\{0,1\},$ $P_l(x)$...
TZE's user avatar
  • 495
22 votes
1 answer
1k views

Does $\binom{2n}{n} \equiv 2 \pmod p$ ever hold?

Well, the title does not tell the whole story; the complete question is: Are there any primes of the form $p=2n(n-1)+1$, with integer $n\ge 1$, such that $$ \binom{2n}{n} \equiv 2\pmod p ? $$ ...
Seva's user avatar
  • 23k
11 votes
1 answer
625 views

A congruence conjecture regarding $(r-s)^4-1 \equiv 0\!\pmod{4r^2s}$

Is the following conjecture true? Conjecture. If $r > s \ge 1$ are relatively prime integers such that \begin{equation} (r-s)^4-1 \equiv 0\!\pmod{4r^2s}, \tag{1} \end{equation} then $r-s = 1$ ...
Kieren MacMillan's user avatar
0 votes
2 answers
470 views

What are all positive integers n for which the congruence $a^{n+1} \equiv a (mod n)$ holds? [closed]

Fermat's little theorem says that the congruence $a^p \equiv a (mod p)$ if $p$ is a prime number. $a^{n+1} \equiv a (mod n)$ works for all integers $a$ and some positive integers $n$, how can we ...
Josh Learner's user avatar
1 vote
0 answers
63 views

"embedding" various matrix equivalences into the equivalence of particular linear map

Consider the square matrices over a (local) ring $R$, up to conjugation, $A\rightarrow UAU^{-1}$, where $U$ is an invertible matrix over $R$. Such an equivalence embeds into the "left-right" ...
Dmitry Kerner's user avatar
2 votes
1 answer
267 views

Does the congruence $a^p \equiv 1 \pmod{b^p}$ with prime $p \ge 5$ force $b \le p$?

I'm considering the congruence in the title, i.e., $$a^p \equiv 1 \pmod{b^p},$$ where $a \ge b \ge 1$ are positive integers and $p$ is an odd prime. For $p=3$, a brute-force computer search found ...
Kieren MacMillan's user avatar
6 votes
2 answers
245 views

Congruence equation for Apery numbers

Does the system of congruence equations \begin{eqnarray} A_{17k}&\equiv& 0 \pmod {17^2}, \nonumber \\ A_{17k+1}&\equiv& 0 \pmod {17^2}, \tag{1} \end{eqnarray} has solutions other ...
Zurab Silagadze's user avatar
9 votes
0 answers
315 views

congruences of level 1 and level p modular forms

I've been carrying out some experiments on the computer and I noticed the following congruence phenomenon: fixing a prime $p$, it seems that any modular form over $SL_2(\mathbb{Z})$ and of weight $k \...
Nadim Rustom's user avatar
-1 votes
1 answer
815 views

how to solve system of congruence with multivariables [closed]

There n variables x1,x2,...,xn represented as X, n equations whose coefficient matrix (n*n) is represented as A, and this system ...
Ecolss's user avatar
  • 1
0 votes
0 answers
407 views

Solutions to a quadratic congruence

Fix an odd prime $p$. Let $\alpha = (\alpha_0,\dots,\alpha_k)$ be a solution to the congruence $\sum_{i=0}^{k} \alpha_i^2 \equiv x \mod p$. Now consider the number $N_\alpha$ of solutions to the ...
Gary's user avatar
  • 1
1 vote
2 answers
185 views

Number of solutions to $mx^2+ny^2 \equiv k\pmod{p}$

I need a reference for the result which gives the number of solutions to the congruence $mx^2+ny^2 \equiv k\pmod{p}$. This result seems to be something that would be discussed in Gauss' ...
Mike Decaro's user avatar
2 votes
1 answer
268 views

perfect shuffle of 2n cards

permutation is given by f(i) = 2i if i<=n and 2(i-n)-1 if i>n where i denotes the position of cards. eg pack of cards (1,2,3,4)-->(3,1,4,2) Basically trying to find the multiplicative order of 2 ...
BenJammin's user avatar
  • 121
22 votes
0 answers
1k views

Given a lattice L with n elements, are there finite groups H < G such that L $\cong$ the lattice of subgroups between H and G?

If there is no restriction on $n$, this is a famous open problem. I'm wondering if any recent work has been done for small $n>6$. I believe the question is answered (positively) for $n=6$ by ...
William DeMeo's user avatar
2 votes
0 answers
488 views

An elementary question in modular arithmetic

Let us fix a positive natural number $N$. When $i$ is a natural number smaller than $N$, coprime with $N$, we let $\mu(i)$ be the unique number in $\{1, \ldots, N-1\}$ that is the multiplicative ...
Mathoverflow's user avatar
6 votes
3 answers
2k views

Finite subgroup of $Gl(n,\mathbb Z)$ and congruences

Suppose we have an invertible matrix q in a finite subgroup $Q$ of $Gl(n,\mathbb Z)$, the group of all invertible integer matrices. Now I want to find all $x\; mod\; \mathbb Z^n$ for which $(q+q^2+q^...
2 votes
2 answers
490 views

Proving Congruence Without Leech Lattice

Let $\sigma_{11}(n)$ denote the sum of the 11th powers of the positive integral divisors of the positive integer n. Let $\tau(n)$ denote Ramanujan's tau function, which is the coefficient of $q^n$ in ...
DavidLHarden's user avatar
  • 3,645
2 votes
4 answers
4k views

Best way to introduce the Chinese Remainder Theorem (to a high school student)

What do you think to be the most effective way to teach the Chinese remainder theorem to a smart high school student, which is supposed to only have a soft idea about how modular arithmetic works, and ...
Maurizio Monge's user avatar
18 votes
2 answers
3k views

Binomial supercongruences: is there any reason for them?

One of the recent questions, in fact the answer to it, reminded me about the binomial sequence $$ a_n=\sum_{k=0}^n{\binom{n}{k}}^2{\binom{n+k}{k}}^2, \qquad n=0,1,2,\dots, $$ of the Apéry ...
Wadim Zudilin's user avatar

1 2
3