Related papers: Sum of consecutive powers as a perfect power

We prove that the equation $(x-2r)^3 + (x-r)^3 + x^3 + (x+r)^3 + (x+2r)^3= y^p$ only has solutions which satisfy $xy=0$ for $1\leq r\leq 10^6$ and $p\geq 5$ prime.

Using only elementary arguments, Cassels and Uchiyama (independently) determined all squares that are sums of three consecutive cubes. Zhongfeng Zhang extended this result and determined all perfect powers that are sums of three consecutive…

We study the equation $(x-4r)^3 + (x-3r)^3 + (x-2r)^3+(x-r)^3 + x^3 + (x+r)^3+(x+2r)^3 + (x+3r)^3 + (x+4r)^3 = y^p$, which is a natural continuation of previous works carried out by A. Arg\'{a}ez-Garc\'{i}a and the fourth author (perfect…

We prove that the equation $(x-3r)^3+(x-2r)^3 + (x-r)^3 + x^3 + (x+r)^3 + (x+2r)^3+(x+3r)^3= y^p$ only has solutions which satisfy $xy=0$ for $1\leq r\leq 10^6$ and $p\geq 5$ prime. This article complements the work on the equations…

Let $k \ge 2$ be even, and let $r$ be a non-zero integer. We show that for almost all $d \ge 2$ (in the sense of natural density), the equation $$ x^k+(x+r)^k+\cdots+(x+(d-1)r)^k=y^n, \qquad x,~y,~n \in \mathbb{Z}, \qquad n \ge 2, $$ has no…

Let $A$ be a finite set of integers. We show that if $k$ is a prime power or a product of two distinct primes then $$|A+k\cdot A|\geq(k+1)|A|-\lceil k(k+2)/4\rceil$$ provided $|A|\geq (k-1)^{2}k!$, where $A+k\cdot A=\{a+kb:\ a,b\in A\}$. We…

In 1997 we proved that if $n$ is of the form $$ 4k, \quad 8k-1\quad {\rm or} \quad 2^{2m+1}(2k-1)+3, $$ where $k,m\in \mathbb N,$ then there are no positive rational numbers $x,y,z$ satisfying $$ xyz = 1, \quad x+y+z = n. $$ Recently, N. X.…

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 =…

We present and analyze an algorithm to enumerate all integers $n\le x$ that can be written as the sum of consecutive $k$th powers of primes, for $k>1$. We show that the number of such integers $n$ is asymptotically bounded by a constant…

The main aim of this article is to find all solutions of the Diophantine equation $x^2 + p^k=y^n$ where $p \equiv 1 \pmod 4$, $\frac{p-1}{3}$ is a perfect square and the class number of $\mathbb{Z}[\sqrt{-p}]$ is $2$. In this article, I…

We study the equation $F_n + F_m = y^p$, where $F_n$ and $F_m$ are respectively the $n$-th and $m$-th Fibonacci numbers and $p \ge 2$. We find all solutions under the assumption $n \equiv m \pmod{2}$.

In this paper, we determine all primitive solutions to the equation $(x+r)^2 +(x+2r)^2 +\cdots +(x+dr)^2 = y^n$ for $2\leq d\leq 10$ and for $1\leq r\leq 10^4$. We make use of a factorization argument and the Primitive Divisors Theorem due…

We investigate the average number of representations of a positive integer as the sum of $k + 1$ perfect $k$-th powers of primes. We extend recent results of Languasco and the last Author, which dealt with the case $k = 2$ [6] and $k = 3$…

We will prove several congruences modulo a power of a prime such as $$ \sum_{0<k_1<...<k_{n}<p}\leg{p-k_{n}}{3} {(-1)^{k_{n}}\over k_1... k_{n}}\equiv {lll} -{2^{n+1}+2\over 6^{n+1}} p B_{p-n-1}({1\over 3}) &\pmod{p^2} &{if $n$ is odd}…

We determine primitive solutions to the equation $(x-r)^2 + x^2 + (x+r)^2 = y^n$ for $1 \le r \le 5,000$, making use of a factorization argument and the Primitive Divisors Theorem due to Bilu, Hanrot and Voutier.

We use Gale duality for polynomial complete intersections and adapt the proof of the fewnomial bound for positive solutions to obtain the bound (e^4+3) 2^(k choose 2) n^k/4 for the number of non-zero real solutions to a system of n…

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…

We show that if p is an odd prime then $$\sum_{k=0}^{p-1}E_kE_{p-1-k}=1 (mod p)$$ and $$\sum_{k=0}^{p-3}E_kE_{p-3-k}=(-1)^{(p-1)/2}2E_{p-3} (mod p),$$ where E_0,E_1,E_2,... are Euler numbers. Moreover, we prove that for any positive integer…

We prove that the equation ${ (x - y)^4 + x^4 + (x + y)^4 = z^n }$ has no integer solutions ${ x, y, z}$ with ${ \gcd(x, y) = 1 }$ for all integers ${ n > 1 }$. We mainly use a modular approach with two Frey ${ \mathbb{Q} }$-curves defined…

Let $k$ be a positive integer. In this paper, using the modular approach, we prove that if $k\equiv 0 \pmod{4}$, $30< k<724$ and $2k-1$ is an odd prime power, then under the GRH, the equation $x^2+(2k-1)^y=k^z$ has only one positive integer…