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We investigate a variation of Nicomachus's identity in which one term in the cubic sum is replaced by a different cube. Specifically, we study the Diophantine identity \[ \sum_{j=1}^{n} j^3 + x^3 - k^3 = \left( \sum_{j=1}^{n} j + x - k…

General Mathematics · Mathematics 2026-02-20 Hajrudin Fejzić

It is important in drawing techniques to find combinations of two straight lines and their angle bisectors whose slopes are all rational numbers. This problem is reduced to solving the Diophantine equation $(a-c)^2(b^2+1) = (b-c)^2(a^2+1).$…

Number Theory · Mathematics 2025-01-03 Takashi Hirotsu

We prove that for given integers b and c, the diophantine equation x^2+bx+c=y^2, has finitely many integer solutions(i.e. pairs in ZxZ),in fact an even number of such solutions(including the zero or no solutions case).We also offer an…

General Mathematics · Mathematics 2008-03-28 Konstantine "Hermes" Zelator

We exactly determine the integral solutions to a previously untreated infinite family of cubic Thue equations of the form $F(x,y)=1$ with at least $5$ such solutions. Our approach combines elementary arguments, with lower bounds for linear…

Number Theory · Mathematics 2014-02-11 Michael A. Bennett , Amir Ghadermarzi

The Diophantine equation $A^4+hB^4=C^4+hD^4$, where $h$ is a fixed arbitrary positive integer, has been investigated by some authors. Currently, by computer search, the integer solutions of this equation are known for all positive integer…

Number Theory · Mathematics 2017-02-23 Farzali Izadi , Mehdi Baghalagdam

Let a1,..., a9 be non-zero integers and n any integer. Suppose that a1 + ... + a9 = n (mod 2) and (ai, aj) = 1 for 1 <= i < j <= 9. We will prove that (i) if not all of the aj's are of the same sign, then the cubic diagonal equation a1p1^3…

Number Theory · Mathematics 2007-05-23 Desmond Leung

In this paper we investigate Diophantine equations of the form $T^2=G(\overline{X}),\; \overline{X}=(X_{1},\ldots,X_{m})$, where $m=3$ or $m=4$ and $G$ is specific homogenous quintic form. First, we prove that if…

Number Theory · Mathematics 2015-02-26 Maciej Gawron , Maciej Ulas

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

In this paper, we solve Diophantine equation in the tittle in nonnegative integers m,n, and a. In order to prove our result, we use lower bounds for linear forms in logarithms and and a version of the Baker-Davenport reduction method in…

Number Theory · Mathematics 2018-01-01 Zafer Şiar , Refik Keskin

In this paper, we use a variety of classical and new research methods for ternary exponential Diophantine equations and extensive use of computer calculations to study the conjecture of R. Scott and R. Styer which asserts that for any fixed…

Number Theory · Mathematics 2026-04-22 Takafumi Miyazaki , Reese Scott , Robert Styer

This paper is a continuation of [1], in which I studied Harvey Friedman's problem of whether the function f(x,y) = x^2 + y^3 satisfies any identities; however, no knowledge of [1] is necessary to understand this paper. We will break the…

General Mathematics · Mathematics 2009-10-13 Roger Tian

In this paper, we use the method of Thue and Siegel, based on explicit Pade approximations to algebraic functions, to completely solve a family of quartic Thue equations. From this result, we can also solve the diophantine equation in the…

Number Theory · Mathematics 2018-07-12 Chen Jian Hua , Paul Voutier

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

Given odd, coprime integers $a$, $b$ ($a>0$), we consider the Diophantine equation $ax^2+b^{2l}=4y^n$, $x, y\in\Bbb Z$, $l \in \Bbb N$, $n$ odd prime, $\gcd(x,y)=1$. We completely solve the above Diophantine equation for…

Number Theory · Mathematics 2020-01-15 Andrzej Dąbrowski , Nursena Günhan , Gökhan Soydan

We give sufficient conditions to determine the existence of nontrivial solutions to the Fermat equation $x^3+y^3=kz^3$ over $\mathbb{Q}(\sqrt{d})$ by constructing a relationship with the points on the elliptic curve $y^2=x^3-432d^3k^2$ over…

Number Theory · Mathematics 2025-05-21 Alejandro Argaez-Garcia , Javier Diaz-Vargas , Luis Eli Pech-Moreno

In this paper we find a parametric solution to the hitherto unsolved problem of finding three positive integers such that their sum, the sum of their squares and the sum of their cubes are simultaneously perfect squares.

Number Theory · Mathematics 2019-08-27 Ajai Choudhry

In this paper we show that Diophantine problem for quadratic equations in Baumslag-Solitar groups $BS(1,k)$ and in wreath products $A \wr \mathbb{Z}$, where $A$ is a finitely generated abelian group and $\mathbb{Z}$ is an infinite cyclic…

Group Theory · Mathematics 2023-05-02 Olga Kharlampovich , Laura Lopez , Alexei Miasnikov

The classical Pell equation can be extended to the cubic case considering the elements of norm one in $Z[\sqrt[3]{r}]$, which satisfy $x^3 + r y^3 + r^2 z^3 - 3 r x y z = 1$. The solution of the cubic Pell equation is harder than the…

Number Theory · Mathematics 2022-03-11 Simone Dutto , Nadir Murru

In this paper we consider the Diophantine equation $x^2+q^{2m}=2y^p$ where $m,p,q,x,y$ are integer unknowns with $m>0,$ $p$ and $q$ are odd primes and $\gcd(x,y)=1.$ We prove that there are only finitely many solutions $(m,p,q,x,y)$ for…

Number Theory · Mathematics 2015-06-26 Szabolcs Tengely

For two relatively prime positive integers $a, b\in \mathbb{N}$, it is known that exactly one of the two Diophantine equations $$ax + by \ =\ \frac{(a-1)(b-1)}{2}\ \mbox{ and }\ 1 + ax + by \ =\ \frac{(a-1)(b-1)}{2}$$ has a nonnegative…

Number Theory · Mathematics 2025-12-16 Hung Viet Chu , Steven J. Miller , Garrett Tresch