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Related papers: Reducing the Erdos-Moser equation 1^n + 2^n + . . …

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The Erd\"os-Moser conjecture states that the Diophantine equation $S_k(m) = m^k$, where $S_k(m)=1^k+2^k+...+(m-1)^k$, has no solution for positive integers $k$ and $m$ with $k \geq 2$. We show that stronger conjectures about consecutive…

Number Theory · Mathematics 2012-09-06 Bernd C. Kellner

We study the solutions of a Diophantine equation of the form $a^x+b^y=c^z$, where $a\equiv 2 \pmod 4$, $b\equiv 3 \pmod 4$ and $\gcd (a,b,c)=1$. The main result is that if there exists a solution $(x,y,z)=(2,2,r)$ with $r>1$ odd then this…

Number Theory · Mathematics 2015-05-13 Mihai Cipu , Maurice Mignotte

If $k$ is a sufficiently large positive integer, we show that the Diophantine equation $$n (n+d) \cdots (n+ (k-1)d) = y^{\ell}$$ has at most finitely many solutions in positive integers $n, d, y$ and $\ell$, with $\operatorname{gcd}(n,d)=1$…

Number Theory · Mathematics 2017-09-05 Michael A. Bennett , Samir Siksek

An important unsolved problem in Diophantine number theory is to establish a general method to effectively find all solutions to any given $S$-unit equation with at least four terms. Although there are many works contributing to this…

Number Theory · Mathematics 2025-03-04 Takafumi Miyazaki

Let $k,\ell\geq2$ be fixed integers and $C$ be an effectively computable constant depending only on $k$ and $\ell$. In this paper, we prove that all solutions of the equation $(x+1)^{k}+(x+2)^{k}+...+(\ell x)^{k}=y^{n}$ in integers $x,y,n$…

Number Theory · Mathematics 2019-09-16 Daniele Bartoli , Gökhan Soydan

In [1] it is shown that the Diophantine equation $(k!)^n+k^n=(n!)^k+n^k$ only has the trivial solution $n=k$, and $(k!)^n-k^n=(n!)^k-n^k$ only has the solutions $n=k$, $(n, k)=(1, 2),$ and $(2, 1)$. In this article we find all solutions of…

Number Theory · Mathematics 2021-05-25 Addea Gupta

Motivated by the recent result of Farhi we show that for each $n\equiv \pm 1\pmod{6}$ the title Diophantine equation has at least two solutions in integers. As a consequence, we get that each (even) perfect number is a sum of three cubes of…

Number Theory · Mathematics 2017-05-03 Maciej Ulas

From the Rhind Papyrus and other extant sources, we know that the ancient Egyptians were very iterested in expressing a given fraction into a sum of unit fractions, that is fractions whose numerators are equal to 1. One of the problems that…

General Mathematics · Mathematics 2009-12-15 Konstantine Zelator

We give bounds on the number of solutions to the Diophantine equation (X+1/x)(Y+1/y) = n as n tends to infinity. These bounds are related to the number of solutions to congruences of the form ax+by = 1 modulo xy.

Number Theory · Mathematics 2009-09-29 J. Brzezinski , W. Holsztynski , P. Kurlberg

Using elementary methods, we determine the highest power of 2 dividing a power sum 1^n + 2^n + . . . + m^n, generalizing Lengyel's formula for the case where m is itself a power of 2. An application is a simple proof of Moree's result that,…

Number Theory · Mathematics 2012-11-27 Kieren MacMillan , Jonathan Sondow

In this paper we extend a result of Hirata-Kohno, Laishram, Shorey and Tijdeman on the Diophantine equation $n(n+d)...(n+(k-1)d)=by^2,$ where $n,d,k\geq 2$ and $y$ are positive integers such that $\gcd(n,d)=1.$

Number Theory · Mathematics 2015-05-13 Szabolcs Tengely

The Erd\"{o}s-Moser equation $ \sum_{i=1}^{m - 1} i^k = m^k $ is a longstanding challenge in number theory, with the only known integer solution being $ (k,m) = (1,3) $. Here, we investigate whether other solutions might exist by using the…

Number Theory · Mathematics 2025-04-14 Guillaume Lambard

If the equation of the title has an integer solution with k>=2, then m>10^{10^6}. Leo Moser showed this in 1953 by amazingly elementary methods. With the hindsight of more than 50 years his proof can be somewhat simplified. We give a…

Number Theory · Mathematics 2020-08-28 Pieter Moree

Suppose that $n$ is a positive integer. In this paper, we show that the exponential Diophantine equation $$(n-1)^{x}+(n+2)^{y}=n^{z},\ n\geq 2,\ xyz\neq 0$$ has only the positive integer solutions $(n,x,y,z)=(3,2,1,2), (3,1,2,3)$. The main…

Number Theory · Mathematics 2020-03-31 Hairong Bai , Elif Kızıldere , Gökhan Soydan , Pingzhi Yuan

We study purely exponential Diophantine equations with four terms of consecutive bases. Notably, we prove that all solutions to the equation \[ n^x=(n+1)^y+(n+2)^z+(n+3)^w \] in positive integers $n,x,y,z$ and $w$ are given by…

Number Theory · Mathematics 2025-08-26 Maohua Le , Takafumi Miyazaki

We develop a parametric approach to study the Diophantine equation $\frac{k}{n} = \frac{1}{x} + \frac{1}{y} + \frac{1}{z}$, underlying the Erd\H{o}s--Straus ($k=4$), Sierpi\'nski ($k=5$), and related generalizations. We introduce and…

Number Theory · Mathematics 2026-03-24 Philemon Urbain Mballa

In number theory, the Erdos-Straus conjecture states that for all n >=2, the rational number 4/n can be expressed as the sum of three unit fractions. Paul Erdos and Ernst G. Straus formulated the conjecture in 1948. The restriction that the…

History and Overview · Mathematics 2019-01-01 Dagnachew Jenber Negash

In this paper we determine possible decompositions of Euler polynomials $E_k(x)$, i.e. possible ways of writing Euler polynomials as a functional composition of polynomials of lower degree. Using this result together with the well-known…

Number Theory · Mathematics 2013-12-16 D. Kreso , Cs. Rakaczki

We study solvability of the Diophantine equation \begin{equation*} \frac{n}{2^{n}}=\sum_{i=1}^{k}\frac{a_{i}}{2^{a_{i}}}, \end{equation*} in integers $n, k, a_{1},\ldots, a_{k}$ satisfying the conditions $k\geq 2$ and $a_{i}<a_{i+1}$ for…

Number Theory · Mathematics 2021-02-11 Szabolcs Tengely , Maciej Ulas , Jakub Zygadło

Let, for r>=2, (m_r(n)),n>=0, be Moser sequence such that every nonnegative integer is the unique sum of the form s_k+rs_l. In this article we give an explicit decomposition formulas of such form and an unexpectedly simple recursion…

Number Theory · Mathematics 2008-12-02 Vladimir Shevelev
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