简化Erdos-Moser方程1^n + 2^n +。. . + k^n =（k+1）^n模k和k^2

关于蕴涵Erdös-Moser猜想的强猜想

On stronger conjectures that imply the Erdös-Moser conjecture

Erdös-Moser猜想指出丢番图方程$S_k(m) = m^k$，其中$S_k(m)=1^k+2^k+...+(m-1)^k$，对于正整数$k$和$m$都没有解，其中$k \geq 2$。我们表明，更强的连续值的函数$S_k$，这似乎是更自然的，意味着埃尔德什-莫泽猜想。

幂和的$p$ -adic阶、Erdös-Moser方程和Bernoulli数

The $p$-adic Order of Power Sums, the Erdös-Moser Equation, and Bernoulli Numbers

Erdös-Moser方程是60多年前提出的一个丢番图方程，至今仍未得到解决。在本文中，我们认为这个问题的整除的权力和某些埃及分数方程。因此，我们表明，解决方案必须满足强整除性和限制性埃及分数方程。我们的研究使我们的结果伯努利数，并使我们能够激励莫泽的原始方法的问题。

幂和的整除性与广义Erdos-Moser方程

Divisibility of Power Sums and the Generalized Erdos-Moser Equation

使用初等方法，我们确定2的最高幂除以幂和1^n + 2^n +。. . + m^n，推广了Lengyel的公式，其中m本身是2的幂。Moree的结果的一个应用是：如果（a，m，n）是广义Erdos-Moser丢番图方程1 ^n +2 ^n+的任意解，则（a，m，n）是该结果的一个简单证明. . . + （m-1）^n = am^n，则m是奇数。

算术级数中的Erdős-Moser方程

Erdős-Moser Equation in Arithmetic Progressions

本文研究算术级数中的Erdős-Moser方程，证明k=2时该级数若有解必含两项或四项，且两种情况均存在无穷多可完全刻画的解。

Primary Pseudoperfect Numbers, Arithmetic Progressions, and the Erdős-Moser Equation

A primary pseudoperfect number (PPN) is an integer $K > 1$ such that the reciprocals of $K$ and its prime factors sum to 1. PPNs arise in studying perfectly weighted graphs and singularities of algebraic surfaces, and are related to Sylvester's sequence, Giuga numbers, Znám's problem, the inheritance problem, and Curtiss's bound on solutions of a unit fraction equation. Here we show $K \equiv 6 \pmod{6^2}$ if $6\mid K$, and uncover a remarkable $7$-term arithmetic progression of residues modulo $6^2\cdot8$ in the sequence of known PPNs. On that basis, we pose a conjecture which leads to a cond…

On the unsolvability of certain equations of Erdős-Moser type

Let $S_k(m):=\sum_{j=1}^{m-1}j^k$ denote a power sum. In 2011, Kellner proposed the conjecture that for $m>3$ the ratio $S_k(m+1)/S_k(m)$ is never an integer, or, equivalently, that for any positive integer $a$, the equation $aS_k(m)=m^k$ has no solutions in positive integers $k$ and $m$ with $m>3$. In this paper, we show that for many integers $a$ the equation $aT_k(m)=(2m+1)^k$, where $T_k(m):=\sum_{j=1}^m(2j-1)^k$, has no solutions in positive integers $k$ and $m$. This leads us to the conjecture that for $m>1$ the ratio $T_k(m+1)/T_k(m)$ is never an integer.