Questions tagged [congruences]

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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
  • 8,984
35 votes
2 answers
2k views

Is the sum of digits of $3^{1000}$ divisible by $7$?

Is the sum of digits of $3^{1000}$ a multiple of $7$? The sum of the digits of $3^{1000}$ can be computed using a computer. It is equal to $2142$, so the answer is positive. Is there a short proof ...
Lezraf's user avatar
  • 453
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
  • 22.8k
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
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
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
18 votes
0 answers
746 views

Two curious series for $1/\pi$

On Jan. 18, 2012 I conjectured that for any prime $p>3$ we have $$R_p^2\equiv\frac1{10}\left(512\left(\frac{10}p\right)-27\left(\frac{-15}p\right)-475\right)\pmod p,$$ where $(\frac{\cdot}p)$ ...
Zhi-Wei Sun's user avatar
  • 14.5k
13 votes
1 answer
3k views

A good reference to the general Chinese Remainder Theorem

I am writing a paper on the topology of the Golomb space and need a good (standard) reference to the following General Chinese Remainder Theorem. For integer numbers $a_1,\dots,a_n$ and positive ...
Taras Banakh's user avatar
  • 40.9k
13 votes
1 answer
570 views

A congruence for a product of binomial coefficients?

For every prime $p\geq 5$ one seems to have the congruence $$(-1)^{(p-1)/2}\prod_{k=0}^{p-1}{p-1\choose k}\equiv 1-p+\frac{3}{2}p^2-\frac{7}{6}p^3\pmod{p^4}\ .$$ (I have checked all primes up to $5000$...
Roland Bacher'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
12 votes
4 answers
3k views

Prove that $\sum_{n = 1}^{p - 1} n^{p - 1} \equiv (p - 1)! + p \pmod {p^2}$ with $p$ being an odd prime

First, I have to admit that I have already asked the same question on MSE several days ago. If I am bending any rules, I apologize for that and moderator can delete or close this question without ...
Saša's user avatar
  • 223
12 votes
1 answer
821 views

Numbers $k$ with $\{\binom nk:\ n\in\mathbb N\}$ dense in $\mathbb Z_p$ for any prime $p\le k$

Let $k$ be a positive integer and let $p$ be a prime. In my 2011 PAMS paper joint with my former student W. Zhang [Proc. Amer. Math. Soc. 139(2011), 1569-1577], we studied when $$S(k)=\left\{\binom nk:...
Zhi-Wei Sun's user avatar
  • 14.5k
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
11 votes
2 answers
361 views

Harmonic congruence

There are a number of interesting congruences for harmonic sums, not the least of which is Wolstenholme's theorem: $H_{p-1}:=\sum_{j=1}^{p-1}\frac1j\equiv 0\mod p^2$. It appears that $\sum_{j=1}^{p-1}...
Lewi_Sol's user avatar
  • 309
11 votes
1 answer
620 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
10 votes
1 answer
618 views

A conjecture on binomial coefficients and roots of unity

Is the following true? Let $p$ be a prime and let $w$ be a $(p-1)$st root of unity (not necessarily primitive). Then $$\binom{w}{n}=\frac{w(w-1)\cdots(w-n+1)}{n!}$$ is $p$-integral; i.e., it can be ...
Ira Gessel's user avatar
  • 16.2k
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
9 votes
2 answers
274 views

For which values of $k$ is it known that there are infinitely many $n$, such that $2^{n+k}\equiv 1\pmod{n}$?

I know that there are no solutions to $2^n\equiv 1\pmod{n}$ for $n>1$ and I can prove that there are infinitely many $n$ such that $2^{n+1}\equiv1\pmod{n}$. My question is: Do we know other ...
Konstantinos Gaitanas's user avatar
9 votes
1 answer
354 views

On the permanent $\text{per}[i^{j-1}]_{1\le i,j\le p-1}$ modulo $p^2$

Let $p$ be an odd prime. It is well-known that $$\det[i^{j-1}]_{1\le i,j\le p-1}=\prod_{1\le i<j\le p-1}(j-i)\not\equiv0\pmod p.$$ I'm curious about the behavior of the permanent $\text{per}[i^{j-...
Zhi-Wei Sun's user avatar
  • 14.5k
9 votes
0 answers
306 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
8 votes
1 answer
279 views

On the determinant $\det[\gcd(i-j,n)]_{1\le i,j\le n}$

In Sept. 2013, I investigated the determinant $$D_n=\det[\gcd(i-j,n)]_{1\le i,j\le n}$$ and computed the values $D_1,\ldots,D_{100}$ (cf. http://oeis.org/A228884). To my surprise, they are all ...
Zhi-Wei Sun's user avatar
  • 14.5k
8 votes
1 answer
310 views

Subgroups of Sp(2g,Z) that map onto all Sp(2g,Z/m)

I stumbled into the following problem. I apologize for being a bit naive. For $g\geq 3$, consider the group $\mathrm{Sp}(2g,\mathbb{Z})$ of symplectic square matrices of order $2g$ with integral ...
Gabriele Mondello's user avatar
8 votes
1 answer
385 views

Cardinality of the image of a polynomial modulo $p^n$

Let $f \in \mathbb{Z}[x]$ be a nonconstant polynomial and let $p$ be a prime number. I'm looking for results about $$N_f(p^k) := \#\{(f(n) \bmod p^k) : n \in \mathbb{Z}\},$$ as $k \to +\infty$, where $...
OhhiMark's user avatar
8 votes
0 answers
317 views

A hypergeometric series for $\sqrt3\pi$ with converging rate $1/9$

Recently, I found a (conjectural) new series for $\sqrt3\pi$: $$\sum_{k=1}^\infty\frac{(8k-3)\binom{4k}{2k}}{k(4k-1)9^k\binom{2k}k^2}=\frac{\sqrt3\pi}{18}.\label{1}\tag{1}$$ The series converges fast ...
Zhi-Wei Sun's user avatar
  • 14.5k
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
  • 475
7 votes
2 answers
369 views

Is there a nonzero solution to this infinite system of congruences?

Is there a triple of nonzero even integers $(a,b,c)$ that satisfies the following infinite system of congruences? $$ a+b+c\equiv 0 \pmod{4} \\ a+3b+3c\equiv 0 \pmod{8} \\ 3a+5b+9c\equiv 0 \pmod{16} \\ ...
M Wright's user avatar
  • 403
7 votes
2 answers
758 views

Could a nice principle be extracted from this lemma of Gauss

I asked the following question in the math SE, with a bounty of 200 pts, without result. question: To prove the quadratic reciprocity law, Gauss needed the following lemma: If $p$ is a prime number ...
MikeTeX's user avatar
  • 677
7 votes
1 answer
183 views

2-adic valuation of $L(0,\chi)$ for a Dirichlet character

Let $\chi : (\mathbb Z/f\mathbb Z)^\times \to K = \mathbb Q(\mu_{\phi(f)})$ be a primitive Dirichlet character. Assume moreover that it is not quadratic, that is, $\chi^2$ is not the trivial character....
Asvin's user avatar
  • 7,646
7 votes
2 answers
237 views

Congruences of binomial sums

Let $a_n$ is a binomial sum, for example $$ a_n := \sum_{k} \binom{n-k}{k} \quad \text{or} \quad \sum_{k=0}^n\binom{n+k}{n-k}\binom{2k}{k} \quad \text{or} \quad \sum_{k=0}^n\sum_{\ell=0}^k\binom{n}{k}\...
Igor Pak's user avatar
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7 votes
1 answer
277 views

On $\prod^{(p-1)/2}_{i,j=1\atop p\nmid 2i+j}(2i+j)$ and $\prod^{(p-1)/2}_{i,j=1\atop p\nmid 2i-j}(2i-j)$ modulo a prime $p>3$

QUESTION: Is my following conjecture true? Conjecture. Let $p>3$ be a prime and let $h(-p)$ be the class number of the imaginary quadratic field $\mathbb Q(\sqrt{-p})$. Then $$\frac{p-1}2!!\prod^{...
Zhi-Wei Sun's user avatar
  • 14.5k
7 votes
1 answer
852 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
7 votes
1 answer
412 views

On $\varphi(m)\varphi(n)\equiv0\pmod{m+n}$

Euler's totient function $\varphi$ is multiplicative, and it plays important roles in number theory. QUESTION: Is it true that for each integer $m>6$ we have $\varphi(m)\varphi(n)\equiv0\pmod{m+n}$...
Zhi-Wei Sun's user avatar
  • 14.5k
7 votes
2 answers
779 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
7 votes
0 answers
180 views

Some conjectural congruences involving Domb numbers

The Domb numbers are given by $$D_n=\sum_{k=0}^n\binom{n}{k}^2\binom{2k}k\binom{2(n-k)}{n-k}\ \ \ (n=0,1,2,\ldots).$$ Such numbers have combinatorial interpretation, see, e.g., http://oeis.org/A002895....
Zhi-Wei Sun's user avatar
  • 14.5k
7 votes
0 answers
206 views

Does Morley's congruence characterize primes greater than $3$?

In 1895 Morley showed that $$\binom{p-1}{(p-1)/2}\equiv(-1)^{\frac{p-1}2}4^{p-1}\pmod{p^3}$$ for any prime $p>3$. In 2009, I formulated the following conjecture concerning the converse of Morley's ...
Zhi-Wei Sun's user avatar
  • 14.5k
7 votes
0 answers
210 views

How to prove the identity $\sum_{k=1}^\infty\frac{3H_{k-1}^2+4H_{k-1}/k}{k^2\binom{2k}k}=\frac{\pi^4}{360}$?

For each $n=0,1,2,\ldots$, the harmonic number $H_n$ is given by $$H_n:=\sum_{0<k\le n}\frac1k.$$ In 2016 I conjectured that $$\sum_{k=1}^\infty\frac{3H_{k-1}^2+4H_{k-1}/k}{k^2\binom{2k}k}=\frac{...
Zhi-Wei Sun's user avatar
  • 14.5k
6 votes
2 answers
240 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
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^...
6 votes
1 answer
562 views

Are there finitely many primes $x$ such that for a fixed odd prime $p$, $n=x^{p-1}+x^{p-2}+\dotsb + x+1$ is composite and $x \mid \phi(n)$?

Let \begin{equation} n =x^{p-1}+x^{p-2}+\dotsb + x+1 \end{equation} where $x$ and $p$ are odd primes. If $p$ is set to $5$, it appears $x=5$ is the only prime $x$ such that $n$ is composite and $x \...
ASP's user avatar
  • 319
6 votes
2 answers
637 views

Number of solutions of $am \equiv bn \pmod{q}$

Let $q$ be a (large) prime. Let $N$ be a positive integer of size${}\approx \sqrt{q}$. Let $\mathcal{M}$ be an arbitrary subset of $\{1, \dots, q\},$ such that $\mathcal{M}$ has cardinality $N$. ...
Kurisuto Asutora's user avatar
6 votes
1 answer
323 views

Corollaries of the halo conjecture that do not involve the eigencurve

In the theory of p-adic modular forms there is a certain construction called the Coleman-Mazur eigencurve. The spectral halo conjecture roughly states that if you remove a closed subdisc of the weight ...
user avatar
6 votes
0 answers
191 views

A conjecture involving $P_n=\prod_{k=1}^np_k$

For each positive integer $n$ let $P_n=\prod_{k=1}^n p_k$, where $p_k$ is the $k$th prime. Question. Is my following conjecture true? Conjecture. For any integer $n>1$, there are $k,m\in\{1,\...
Zhi-Wei Sun's user avatar
  • 14.5k
6 votes
0 answers
212 views

Two conjectural congruences for Franel numbers

Recall that the Franel numbers are given by $$f_n:=\sum_{k=0}^n \binom{n}{k}^3\ \ \ (n=0,1,\ldots).$$ Question. How to prove my following conjecture? Conjecture. For each odd prime $p$, we have $$\...
Zhi-Wei Sun's user avatar
  • 14.5k
5 votes
2 answers
376 views

"Efficient" way to build a table of multiplicative orders modulo $p$ of a fixed integer $a$

Given an integer $a$, I would like to build a table of entries $(p, \text{ord}_p(a))$, where $p$ runs over the prime numbers not dividing $a$ and not exceeding a fixed parameter $P$, and $\text{ord}_p(...
Fran's user avatar
  • 53
5 votes
2 answers
263 views

Modulo $x^2 + y^2 - 1$, is every homogeneous polynomial that is a square of a polynomial, necessarily of sum of squares of homogeneous polynomials?

I am hoping this question is alright for Math Overflow. I didn't get a definitive solution in Math Stack Exchange. Let $f(x, y) \in \mathbb{R}[x, y]$ be a homogeneous polynomial with real coefficients ...
Colin Tan's user avatar
  • 251
5 votes
1 answer
268 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
  • 24.2k
5 votes
1 answer
322 views

Reversing the CRT: Is $5$ tough?

Given odd primes $p\ne q$, by the CRT we can find an integer $x$ such that $x\equiv 2^{p-1}\pmod q$ and $x\equiv 2^{q-1}\pmod p$. Can this procedure be reversed? For which integers $x$ there exist ...
W-t-P's user avatar
  • 550
5 votes
4 answers
371 views

When does the following congruence identity hold?

Let $m$,$l$ be coprime integers where $m,l\geq 2$. For any integer $a$ and positive base $b \ (b\geq 2)$, let $ [a]_b $ denote the element of $\{0,\ldots, b-1\}$ that satisfies the equivalence $[a]...
Andrey Rukhin's user avatar
5 votes
1 answer
411 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
5 votes
2 answers
317 views

Permutations $\pi\in S_{p-1}$ with $\frac1{\pi(1)\pi(2)}+\frac1{\pi(2)\pi(3)}+\cdots+\frac1{\pi(p-2)\pi(p-1)}+\frac1{\pi(p-1)\pi(1)}\equiv0\pmod{p^2}$

A well known congruence of Wolstenholme states that $$\frac1{1^2}+\frac1{2^2}+\cdots+\frac1{(p-1)^2}\equiv0\pmod{p}$$ for any prime $p>3$. For each $n=3,4,\ldots$ we clearly have $$\frac1{1\times2}+...
Zhi-Wei Sun's user avatar
  • 14.5k