从毕达哥拉斯连续平方和到佩尔方程生成的三次连续和
From Pythagorean Runs to Pell-Generated Cubic Runs
- The University of Texas at Dallas（达拉斯德州大学）
机构由 AI 辅助整理，请以论文原文为准。
AI总结：
本文研究三次连续和恒等式，通过佩尔型方程构造无穷多个正整数解，并给出半长度序列的二阶线性递推关系。
AI中文摘要：
经典的毕达哥拉斯连续和是一个恒等式，其中一段连续的平方数之和等于紧随其后的另一段连续平方数之和。Boardman 的构造对每个指定的长度都给出了这样的连续和。我们考虑一个三次模拟，其中第二段允许有公共差 2。我们证明存在无穷多个正整数三元组 \\((m,A,B)\\)，满足 \\(B>A+2m-1\\)，使得 $$ \sum_{j=0}^{2m-1}(A+j)^3 = \sum_{j=0}^{m-1}(B+2j)^3. $$ 一个显式的无穷族由佩尔型方程 $$ 48329z^2-156m^2=161 $$ 获得，该方程可化为归一化形式的广义佩尔方程。排序条件确保步长为 2 的数列在连续段结束后严格开始。所得的半长度序列满足一个显式的二阶线性递推关系。
英文摘要：
A classical Pythagorean run is an identity in which a block of consecutive squares is equal to the immediately following block of consecutive squares. Boardman's construction gives such a run for every prescribed length. We consider a cubic analogue in which the second block is allowed to have common difference 2. We prove that there are infinitely many positive integer triples \((m,A,B)\), with \(B>A+2m-1\), such that $$ \sum_{j=0}^{2m-1}(A+j)^3 = \sum_{j=0}^{m-1}(B+2j)^3. $$ An explicit infinite family is obtained from the Pell-type equation $$ 48329z^2-156m^2=161, $$ which reduces to a generalized Pell equation in normalized form. The ordering condition ensures that the step-2 progression begins strictly after the consecutive block ends. The resulting sequence of half-lengths satisfies an explicit second-order linear recurrence.
