当前位置: X-MOL 学术J. Number Theory › 论文详情
Our official English website, www.x-mol.net, welcomes your feedback! (Note: you will need to create a separate account there.)
On a Diophantine equation of Erdős and Graham
Journal of Number Theory ( IF 0.7 ) Pub Date : 2020-12-01 , DOI: 10.1016/j.jnt.2020.05.006
Szabolcs Tengely , Maciej Ulas , Jakub Zygadło

Abstract We study solvability of the Diophantine equation n 2 n = ∑ i = 1 k a i 2 a i , in integers n , k , a 1 , … , a k satisfying the conditions k ≥ 2 and a i a i + 1 for i = 1 , … , k − 1 . The above Diophantine equation (of polynomial-exponential type) was mentioned in the monograph of Erdős and Graham, where several questions were stated. Some of these questions were already answered by Borwein and Loring. We extend their work and investigate other aspects of Erdős and Graham equation. First of all, we obtain the upper bound for the value a k given in terms of k only. This mean, that with fixed k our equation has only finitely many solutions in n , a 1 , … , a k . Moreover, we construct an infinite set K , such that for each k ∈ K , the considered equation has at least five solutions. As an application of our findings we enumerate all solutions of the equation for k ≤ 8 . Moreover, by applying greedy algorithm, we extend Borwein and Loring calculations and check that for each n ≤ 10 4 there is a value of k such that the considered equation has a solution in integers n + 1 = a 1 a 2 … a k . Based on our numerical calculations we formulate some further questions and conjectures.

中文翻译:

关于 Erdős 和 Graham 的丢番图方程

摘要 我们研究丢番图方程 n 2 n = ∑ i = 1 kai 2 ai 的可解性,整数 n , k , a 1 , … , ak 满足条件 k ≥ 2 和 aaii + 1 for i = 1 , … , k - 1 . Erdős 和 Graham 的专着中提到了上述丢番图方程(多项式指数型),其中提出了几个问题。Borwein 和 Loring 已经回答了其中一些问题。我们扩展了他们的工作并研究了 Erdős 和 Graham 方程的其他方面。首先,我们获得仅根据 k 给出的值 ak 的上限。这意味着,在固定 k 的情况下,我们的方程在 n , a 1 , … , ak 中只有有限多个解。此外,我们构造了一个无限集 K ,使得对于每个 k ∈ K ,所考虑的方程至少有五个解。作为我们发现的应用,我们列举了 k ≤ 8 方程的所有解。此外,通过应用贪心算法,我们扩展了 Borwein 和 Loring 计算,并检查对于每个 n ≤ 10 4 都有一个 k 值,使得所考虑的方程具有整数 n + 1 = a 1 a 2 … ak 的解。基于我们的数值计算,我们提出了一些进一步的问题和猜想。
更新日期:2020-12-01
down
wechat
bug