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Related papers: Triples which are $D(n)$-sets for several $n$'s

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Necessary and sufficient conditions for the existence of an integer solution of the diophantine equation $m/n=1/x(\lambda)+1/y(\lambda)+1/z(\lambda)$ with $n=b+a\lambda$ are explicitly given for a,b coprime and a not a multiple of m . The…

General Mathematics · Mathematics 2024-04-03 Bernd R. Schuh

A set $A\subseteq\mathbb N$ is called $complete$ if every sufficiently large integer can be written as the sum of distinct elements of $A$. In this paper we present a new method for proving the completeness of a set, improving results of…

Combinatorics · Mathematics 2016-09-27 Vitaly Bergelson , David Simmons

In this paper, we find all the solutions of the Diophantine equation $P_\ell + P_m +P_n=2^a$, in nonnegative integer variables $(n,m,\ell, a)$ where $P_k$ is the $k$-th term of the Pell sequence $\{P_n\}_{n\ge 0}$ given by $P_0=0$, $P_1=1$…

Number Theory · Mathematics 2016-08-23 Jhon J. Bravo , Bernadette Faye , Florian Luca

In this paper, we consider dominating sets $D$ and $D'$ such that $D$ and $D'$ are disjoint and there exists a perfect matching between them. Let $DD_{\textrm{m}}(G)$ denote the cardinality of smallest such sets $D, D'$ in $G$ (provided…

Combinatorics · Mathematics 2017-09-01 William F. Klostermeyer , Margaret-Ellen Messinger , Alejandro Angeli Ayello

A rational Diophantine triple is a set of three nonzero rational a,b,c with the property that ab+1, ac+1, bc+1 are perfect squares. We say that the elliptic curve y^2 = (ax+1)(bx+1)(cx+1) is induced by the triple {a,b,c}. In this paper, we…

Number Theory · Mathematics 2020-12-22 Andrej Dujella , Juan Carlos Peral

Crux Mathematicorum with Mathematical Mayhem, is a problem solving journal published by the Canadian Mathematical Society. In the March 2010 issue(see reference[1]) ,the following problem was proposed:Determine all positive integers a,b,…

General Mathematics · Mathematics 2011-12-19 Konstantine Zelator

The subject matter of this work is the diophantine equation x^n+y^m=c(x^k)(y^l), where n,m,k,l,c are natural numbers.We investigate this equation from the point of view of positive integer solutions.A preliminary examination of sources such…

Number Theory · Mathematics 2010-06-10 Konstantine Zelator

We discuss properties of diophantine solutions of the Pythagoras equation, $a^2+b^2=c^2$, where the three numbers have no common factor. Some of the highlights are: (1) All triplets for which $c$ (called the `peak') is non-prime can be…

General Mathematics · Mathematics 2023-06-23 Palash B. Pal

Let $(P_n)_{n\ge 0}$ and $(Q_n )_{n\ge 0}$ be the Pell and Pell-Lucas sequences. Let $b$ be a positive integer such that $b\ge 2.$ In this paper, we prove that the following two Diophantine equations $P_{n}=b^{d}P_{m}+Q_{k}$ and…

Number Theory · Mathematics 2024-06-25 Kouèssi Norbert Adédji , Marija Bliznac Trebješanin

In this paper we solve the Diophantine equation $\binom{m}{l}-\binom{n}{k}=d$ (where m,n are positive integers unknowns) when (k,l)=(3,6) for various values of d and when (k,l)=(8,2) and d=1. As a byproduct of our results we will obtain…

Number Theory · Mathematics 2019-02-26 Nikos Katsipis

This manuscript introduces Diophantine labeling, a new way of labeling of the vertices for finite simple undirected graphs with some divisibility condition on the edges. Maximal graphs admitting Diophantine labeling are investigated and…

Combinatorics · Mathematics 2026-01-01 A. Nasr , A. Elsonbaty , M. A. Seoud , M. Anwar

Let $d$ be a square-free integer such that $d \equiv 15 \pmod{60}$ and the Pell's equation $x^2 - dy^2 = -6$ is solvable in rational integers $x$ and $y$. In this paper, we prove that there exist infinitely many Diophantine quadruples in…

Number Theory · Mathematics 2025-04-10 Shubham Gupta

Let R be a recursive subring of a number field. We show that recursively enumerable sets are diophantine for the polynomial ring R[Z].

Number Theory · Mathematics 2008-09-11 Jeroen Demeyer

Let $N$ be an odd perfect number. Then, Euler proved that there exist some integers $n, \alpha$ and a prime $q$ such that $N = n^{2}q^{\alpha}$, $q \nmid n$, and $q \equiv \alpha \equiv 1 \bmod 4$. In this note, we prove that the ratio…

Number Theory · Mathematics 2023-12-01 Yoshinosuke Hirakawa

In this paper we consider two new conjectures concerning $D(4)$-quadruples and prove some special cases which support their validity. The main result is a proof that $\{a,b,c\}$ and $\{a+1,b,c\}$ cannot both be $D(4)$-triples.

Number Theory · Mathematics 2024-06-25 Marija Bliznac Trebješanin

We study the Diophantine problem (decidability of finite systems of equations) in different classes of finitely generated solvable groups (nilpotent, polycyclic, metabelian, free solvable, etc), which satisfy some natural…

Group Theory · Mathematics 2020-03-25 Albert Garreta , Alexei Miasnikov , Denis Ovchinnikov

A square-free monomial ideal $I$ is called an {\it $f$-ideal}, if both $\delta_{\mathcal{F}}(I)$ and $\delta_{\mathcal{N}}(I)$ have the same $f$-vector, where $\delta_{\mathcal{F}}(I)$ ($\delta_{\mathcal{N}}(I)$, respectively) is the facet…

Commutative Algebra · Mathematics 2018-04-24 Jin Guo , Tongsuo Wu

A triple (a, b, c) of positive integers is called a Markoff triple iff it satisfies the Diophantine equation a2+b2+c2=abc . Recasting the Markoff tree, whose vertices are Markoff triples, in the framework of integral upper triangular 3x3…

Number Theory · Mathematics 2022-08-26 Norbert Riedel

We show that if {1, b, c, d} is a D(-1) diophantine quadruple with b<c<d and c=1+s^2, then the cases s=p^k, s=2p^k, c=p and c=2p^k do not occur, where p is an odd prime and k is a positive integer. For the integer d=1+x^2, we show that it…

Number Theory · Mathematics 2013-09-18 Anitha Srinivasan

A perfect number is a positive integer n such that n equals the sum of all positive integer divisors of n that are less than n. That is, although n is a divisor of n, n is excluded from this sum. Thus 6 = 1 + 2 + 3 is perfect, but 12 < 1 +…

Logic in Computer Science · Computer Science 2015-09-22 John Cowles , Ruben Gamboa
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