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We show that the diophantine equation $n^\ell+(n+1)^\ell + ...+ (n+k)^\ell=(n+k+1)^\ell+ ...+ (n+2k)^\ell$ has no solutions in positive integers $k,n \ge 1$ for all $\ell \ge 3$.

Number Theory · Mathematics 2016-02-22 Simon Felten , Stefan Müller-Stach

In this paper, elliptic curves theory is used for solving the Diophantine equations X^3+Y^3+Z^3+aU^k=a_0U_0^{t_0}+...+a_nU_n^{t_n}, k=3,4 where n, ti are natural numbers and a, a_i are fixed arbitrary rational numbers. We try to transform…

Number Theory · Mathematics 2017-03-01 Farzali Izadi , Mehdi Baghalaghdam

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

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

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

Let $(L_n^{(k)})_{n\geq 2-k}$ be the sequence of $k$--generalized Lucas numbers for some fixed integer $k\ge 2$ whose first $k$ terms are $0,\ldots,0,2,1$ and each term afterwards is the sum of the preceding $k$ terms. In this paper, we…

Number Theory · Mathematics 2024-12-18 Herbert Batte , Florian Luca

In this paper we consider the Diophantine equation \begin{align*}b^k +\left(a+b\right)^k &+ \cdots + \left(a\left(x-1\right) + b\right)^k=\\ &=d^l + \left(c+d\right)^l + \cdots + \left(c\left(y-1\right) + d\right)^l, \end{align*} where…

Number Theory · Mathematics 2013-12-13 A. Bazsó , D. Kreso , F. Luca , Á. Pintér

The search of solutions of the Diophantine equation $x^3 + y^3 + z^3 = k$ for $k<1000$ has been extended with bounds of $|x|$, $|y|$ and $|z|$ up to $10^{15}$. The first solution for $k=74$ is reported. This only leaves $k=33$ and $k=42$…

Number Theory · Mathematics 2016-04-27 Sander G. Huisman

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

We solve the Diophantine equations $x^5 + y^5 = dz^p$ with $d=2, 3$ for a set of prime numbers of density 1/4, 1/2, respectively. The method consists in relating a possible solution to another Diophantine equation and solving the later by…

Number Theory · Mathematics 2011-03-29 Luis Dieulefait , Nuno Freitas

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

This work determine the entire family of positive integer solutions of the diophantine equation. The solution is described in terms of $\frac{(m-1)(m+n-2)}{2} $ or $\frac{(m-1)(m+n-1)}{2}$ positive parameters depending on $n$ even or odd.…

Number Theory · Mathematics 2014-02-24 Zahid Raza , Hafsa Masood Malik

Let $p$ be a prime number with $p>3$, $p\equiv 3\pmod{4}$ and let $n$ be a positive integer. In this paper, we prove that the Diophantine equation $(5pn^{2}-1)^{x}+(p(p-5)n^{2}+1)^{y}=(pn)^{z}$ has only the positive integer solution…

Number Theory · Mathematics 2020-02-27 Elif Kızıldere , Gökhan Soydan

We show the insolvability of the Diophantine equation $ax^d-y^2-z^2+xyz-b=0$ in $\mathbb{Z}$ for fixed $a$ and $b$ such that $a\equiv 1 \pmod {12}$ and $b=2^da-3$, where $d$ is an odd integer and is a multiple of $3$. Further, we…

Number Theory · Mathematics 2023-02-01 Om Prakash , Kalyan Chakraborty

We solve Diophantine equations of the type $ a \, (x^3 \!+ \! y^3 \!+ \! z^3 ) = (x \! + \! y \! + \! z)^3$, where $x,y,z$ are integer variables, and the coefficient $a\neq 0$ is rational. We show that there are infinite families of such…

Number Theory · Mathematics 2025-03-14 Bogdan A. Dobrescu , Patrick J. Fox

Using elementary number theory we study Diophantine equations over the rational integers of the following form, $y^2=(x+a)(x+a+k)(x+b)(x+b+k)$, $y^2=c^2x^4+ax^2+b$ and $y^2=(x^2-1)(x^2-\alpha^2)(x^2-(\alpha+1)^2).$ We express their integer…

Number Theory · Mathematics 2022-11-17 Konstantinos A. Draziotis

In this work, we give upper bounds for $n$ on the title equation. Our results depend on assertions describing the precise exponents of $2$ and $3$ appearing in the prime factorization of $T_{k}(x)=(x+1)^{k}+(x+2)^{k}+...+(2x)^{k}$. Further,…

Number Theory · Mathematics 2017-09-04 Attila Bérczes , István Pink , Gamze SavaŞ , Gökhan Soydan

Catalan's conjecture claims that the Diophantine equation $x^p-y^q=1$ admits the unique solution $3^2-2^3=1$ in integers $x,y,p,q \ge 2$. The conjecture has been finally proved by P. Mih\u{a}ilescu (2002) using the theory of cyclotomic…

Number Theory · Mathematics 2017-02-14 Paolo Leonetti

In this paper, we deal with two classes of Diophantine equations, $x^2+y^2+z^2+k_1yz+k_2zx+k_3xy=(3+k_1+k_2+k_3)xyz$ and $x^2+y^4+z^4+ky^2z^2+2xz^2+2xy^2=(7+k)xy^2z^2$, where $k_1,k_2,k_3,k$ are nonnegative integers. The former is known as…

Number Theory · Mathematics 2024-07-12 Yasuaki Gyoda , Kodai Matsushita

An open conjecture of Erdos and Moser is that the only solution of the Diophantine equation in the title is the trivial solution 1+2=3. Reducing the equation modulo k and k^2, we give necessary and sufficient conditions on solutions to the…

Number Theory · Mathematics 2011-07-13 Jonathan Sondow , Kieren MacMillan
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