论文标题
$ p^{3} _ {1} $ - 设置与立方三角形数字之间的意外会议
An unexpected meeting between the $P^{3}_{1}$-set and the cubic-triangular numbers
论文作者
论文摘要
一组$ M $阳性整数$ \ {x_ {1}，\ ldots，x_ {m} \} $称为$ p^{3} _ {1} $ - 尺寸$ m $的集合，如果该集合中任何三个元素的产品增加了一个cube integer。 a $ p^{3} _ {1} $ - set $ s $如果存在一个整数$ y \ not \ in s $中，则可以扩展，以至于$ s \ s \ cup \ {y \ {y \} $仍然是$ p^{3} _ {1} $ - set。现在，让我们考虑二磷剂方程$ u（u+1）/2 = v^{3} $，其整数解决方案会产生我们所谓的立方三角形数字。 The purpose of this paper is to prove simultaneously that the $P^{3}_{1}$-set $\{1,2,13\}$ is non-extendible and $n=1$ is the unique cubic-triangular number by showing that the two problems meet on the Diophantine equation $2x^{3}-y^{3}=1$ that we solve using $ p $ - 亚种分析。
A set of $m$ positive integers $\{x_{1},\ldots,x_{m}\}$ is called a $P^{3}_{1}$-set of size $m$ if the product of any three elements in the set increased by one is a cube integer. A $P^{3}_{1}$-set $S$ is said to be extendible if there exists an integer $y\not\in S$ such that $S\cup\{y\}$ still a $P^{3}_{1}$-set. Now, let consider the Diophantine equation $u(u+1)/2=v^{3}$ whose integer solutions produce what we called cubic-triangular numbers. The purpose of this paper is to prove simultaneously that the $P^{3}_{1}$-set $\{1,2,13\}$ is non-extendible and $n=1$ is the unique cubic-triangular number by showing that the two problems meet on the Diophantine equation $2x^{3}-y^{3}=1$ that we solve using $p$-adic analysis.
