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Related papers: A New Generalization of Fermat's Last Theorem

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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 the problem about finding out perfect powers in an alternating sum of consecutive cubes. More precisely, we completely solve the Diophantine equation $(x+1)^3 - (x+2)^3 + \cdots - (x + 2d)^3 + (x + 2d + 1)^3 =…

Number Theory · Mathematics 2017-05-12 Pranabesh Das , Pallab Kanti Dey , B. Maji , S. S. Rout

Let $p$ be a prime number. In the early 2000s, it was proved that the Fermat equations with coefficients \[3x^p + 8y^p + 21z^p =0\quad \text{ and } \quad 3x^p + 4y^p + 5z^p=0 \] do not admit non-trivial solutions for a set of exponents $p$…

Number Theory · Mathematics 2016-06-15 Nuno Freitas , Alain Kraus

Assume that $n$ is a positive integer, $p_{j}$ ($j=1,2, \cdots, 6)$ are polynomials, $p$ is an irreducible polynomial, and $f$ is an entire function on $\mathbb{C}^{n}.$ Let $ L(f)=\sum_{j=1}^s q_{t_j}f_{z_{t_j}}$ and…

Complex Variables · Mathematics 2025-09-03 Tingbin Cao , Jun Wang , Zhuan Ye

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

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

Given positive integers a,b,c and d such that c and d are coprime we show that the primes p=c(mod d)dividing a^k+b^k for some k>=1 have a natural density and explicitly compute this density. We demonstrate our results by considering some…

Number Theory · Mathematics 2012-07-30 P. Moree , B. Sury

We refine a result of W.P. Li and Wang on the values of the form $ \lambda_1p_1 + \lambda_2p_2^{2} + \lambda_3p_3^{2} + \mu_1 2^{m_1} +...+ \mu_s 2^{m_s}, $ where $p_1,p_2,p_3$ are prime numbers, $m_1,..., m_s$ are positive integers,…

Number Theory · Mathematics 2012-12-27 Alessandro Languasco , Valentina Settimi

For every $n \geq 2$ we determine the asymptotic formula for the number of integer triples $(a,b,c)$ of bounded absolute value such that the generalised Fermat equation given by $ax^n+by^n+cz^n=0$ is everywhere locally soluble. We compute…

Number Theory · Mathematics 2025-01-30 Peter Koymans , Ross Paterson , Tim Santens , Alec Shute

Let $p$ and $q$ be distinct primes such that $q+1 | p-1$. In this paper we find all integer solutions $a$, $b$ to the equation $1/a + 1/b = (q+1)/pq$ using only elementary methods.

History and Overview · Mathematics 2019-05-09 Jeremiah W. Johnson

We prove that for each odd prime p, positive integer alpha, and non-negative integers beta and gamma, the Diophantine equation X^{2N} + 2^{2 alpha} 5^{2 beta} p^{2 gamma} = Z^5 has no solution with X, Z, N in Z^+, N > 1, and gcd(X,Z) = 1.

Number Theory · Mathematics 2014-09-09 Eva G. Goedhart , Helen G. Grundman

To give a parametrization of the Diophantine equation $A^{3}+B^{3}=C^{3}+D^{3}$ in terms of integral binary quadratic forms in a constructive way.

History and Overview · Mathematics 2021-02-22 Dom Fosse

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

Let $(a,b,c)$ be a primitive Pythagorean triple. Set $a=m^2-n^2$,$b=2mn$, and $c=m^2+n^2$ with $m$ and $n$ positive coprime integers, $m>n $ and $ m \not \equiv n \pmod 2$. A famous conjecture of Je\'{s}manowicz asserts that the only…

Number Theory · Mathematics 2020-10-17 Amir Ghadermarzi

From some works of P. Furtw\"angler and H.S. Vandiver, we put the basis of a new cyclotomic approach to Fermat's last theorem for p>3 and to a stronger version called SFLT, by introducing governing fields of the form Q(exp(2 i pi/q-1)) for…

Number Theory · Mathematics 2011-04-14 Georges Gras , Roland Quême

Recently, Grynkiewicz et al. [{\it Israel J. Math.} {\bf 193} (2013), 359--398], using tools from additive combinatorics and group theory, proved necessary and sufficient conditions under which the linear congruence $a_1x_1+\cdots…

Discrete Mathematics · Computer Science 2020-10-13 Khodakhast Bibak , Bruce M. Kapron , Venkatesh Srinivasan

In this paper we are interested in solving the Fermat-type equations x^5+y^5=dz^p where d is a positive integer and p a prime number $\ge 7$. We describe a new method based on modularity theorems which allows us to improve all the results…

Number Theory · Mathematics 2008-06-11 Nicolas Billerey , Luis Dieulefait

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

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

Given two relatively prime numbers $a$ and $b$, it is known that exactly one of the two Diophantine equations has a nonnegative integral solution $(x,y)$: $$ ax + by \ =\ \frac{(a-1)(b-1)}{2}\quad \mbox{ and }\quad 1 + ax + by \ =\…

Number Theory · Mathematics 2025-09-11 Hung Viet Chu , Rishabh Gulecha , Sicheng Guo , Nathanael Johnson , Steven J. Miller , Yeju Shin
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