The congruences tag has no usage guidance.

**6**

votes

**1**answer

207 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 ...

**6**

votes

**1**answer

470 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?

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 ...

**12**

votes

**1**answer

727 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 ...

**0**

votes

**0**answers

76 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 ...

**2**

votes

**0**answers

80 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 $A=a_1<\cdots<a_r$ ...

**8**

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. ...

**18**

votes

**2**answers

802 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) ...

**2**

votes

**1**answer

261 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 ...

**2**

votes

**1**answer

149 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 ...

**7**

votes

**1**answer

658 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 $a \in \mathbb{Z}/l\mathbb{Z}-\{0,1\},$ ...

**18**

votes

**0**answers

561 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
$$ \binom{2n}{n} \equiv 2\pmod p ? $$
...

**5**

votes

**1**answer

345 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 $r-s = 1$ ...

**0**

votes

**2**answers

312 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 ...

**1**

vote

**0**answers

52 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" ...

**1**

vote

**1**answer

190 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 ...

**4**

votes

**1**answer

143 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 ...

**9**

votes

**0**answers

208 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 ...

**-1**

votes

**1**answer

352 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 ...

**0**

votes

**0**answers

338 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 ...

**1**

vote

**2**answers

144 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' ...

**2**

votes

**1**answer

208 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 ...

**18**

votes

**0**answers

765 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 ...

**2**

votes

**0**answers

469 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 ...

**6**

votes

**3**answers

1k 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
...

**1**

vote

**2**answers

407 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 ...

**2**

votes

**4**answers

2k 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 ...

**15**

votes

**2**answers

2k 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 ...