July 4, 2023
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 powers that are sums of cubes of a three, five and seven term arithmetic progression). Under the assumptions $0 < r \leq 10^6$, $p \geq 5 $ a prime and $\gcd(x, r) = 1$, we show that solutions must satisfy $xy=0$. Moreover, we study the equation for prime exponents $2$ and $3$ in greater detail. Under the assumptions $r>0$ a positive integer and $\gcd(x, r) = 1$ we show that there are infinitely many solutions for $p=2$ and $p=3$ via explicit constructions using integral points on elliptic curves. We use an amalgamation of methods in computational and algebraic number theory to overcome the increased computational challenge. Most notable is a significant computational efficiency obtained through appealing to Bilu, Hanrot and Voutier's Primitive Divisor Theorem and the method of Chabauty, as well as employing a Thue equation solver earlier on.
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November 3, 2019
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 $(x-r)^3 + x^3 + (x+r)^3 = y^p$ and $(x-2r)^3 + (x-r)^3 + x^3 + (x+r)^3 + (x+2r)^3= y^p$ . The methodology in this paper makes use of the Primitive Divisor Theorem due to Bilu, Hanrot and Voutier for a complete resolution of the Diophantine equati...
January 16, 2019
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.
November 17, 2017
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 cubes. Recently, the equation $(x-r)^k + x^k + (x+r)^k$ has been studied for $k=4$ by Zhongfeng Zhang and for $k=2$ by Koutsianas. In this paper, we complement the work of Cassels, Koutsianas and Zhang by considering the case when $k=3$ an...
September 24, 2018
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 to Bilu, Hanrot and Voutier.
January 4, 2021
In this paper, using a deep result on the existence of primitive divisors of Lehmer numbers due to Y. Bilu, G. Hanrot and P. M. Voutier, we first give an explicit formula for all positive integer solutions of the Diophantine equation $(x-d)^2+x^2+(x+d)^2=y^n$ (*) when $n$ is an odd prime and $d=p^r$, $p>3$ a prime. So this improves the results on the papers of A. Koutsianas and V. Patel \cite{KP} and A. Koutsianas \cite{Kou}. Secondly, under the assumption of our first result...
February 21, 2018
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.
April 4, 2024
In this paper we study equation $$(x-dr)^5+\cdots+x^5+\cdots+(x+dr)^5=y^p$$ under the condition $\gcd(x,r)=1$. We present a recipe for proving the non-existence of non-trivial integer solutions of the above equation, and as an application we obtain explicit results for the cases $d=2,3$ (the case $d=1$ was already solved). We also prove an asymptotic result for $d\equiv 1, 7\pmod9$. Our main tools include the modular method, employing Frey curves and their associated modular ...
August 18, 2020
In this paper we determine the perfect powers that are sums of three fifth powers in an arithmetic progression. More precisely, we completely solve the Diophantine equation $$ (x-d)^5 + x^5 + (x + d)^5 = z^n,~n\geq 2, $$ where $d,x,z \in \mathbb{Z}$ and $d = 2^a5^b$ with $a,b\geq 0$.
January 9, 2024
This paper proposes an elementary solution to a special case of finding all perfect squares that can be written as sum of consecutive integer cubes. It is shown that there are no non-trivial solutions if the perfect square is a prime power, i.e., it is divisible by two different primes if a non-trivial one exists. Solution mostly depends on $v_{p}(x)$ and general forms of Pythagorean triples.
February 25, 2017
In this paper, we sharpen earlier work of the first author, Luca and Mulholland, showing that the Diophantine equation $$ A^3+B^3 = q^\alpha C^p, \, \, ABC \neq 0, \, \, \gcd (A,B) =1, $$ has, for "most" primes $q$ and suitably large prime exponents $p$, no solutions. We handle a number of (presumably infinite) families where no such conclusion was hitherto known. Through further application of certain {\it symplectic criteria}, we are able to make some conditional statements...