Squares from Products of Consecutive Integers
Squares from Products of Consecutive Integers
复制标题
连续整数乘积的平方
DOI：
10.1080/00029890.2002.11919873
复制
发表时间：
2002
期刊：
影响因子：
--
通讯作者：
G. Woeginger
中科院分区：
文献类型：
--
作者：
A. J. Poorten;G. Woeginger
1.引言。请注意，1·2·3·4+1=52，2·3·4·5+1=112，3·4·5·6+1=192，...事实上，众所周知，任何四个连续整数的乘积与完美平方相差1。然而，一些实验很容易让人猜测，除了4之外，没有整数n，因此任何n个连续整数的乘积与完美平方的乘积只有取决于n的某个整数c=c(N)。这里有两个问题。首先是解释四个国家明显的特殊地位。我们证明，这件事并不比任何二次多项式都可以通过增加一个常数成为多项式的平方来完成的事实更深层次。其次，我们证明了不可能有n大于4且具有所述性质。
1. INTRODUCTION. Notice that 1· 2· 3· 4+ 1= 52, 2· 3· 4· 5+ 1= 112, 3· 4· 5· 6+ 1= 192,.... Indeed, it is well known that the product of any four consecutive integers differs by 1 from a perfect square. However, a little experimentation readily leads one to guess that there is no integer n, other than four, so that the product of any n consecutive integers differs from a perfect square by some integer c= c (n) depending only on n.There are two issues here. The first is to explain the apparently special status of four. We show that this matter lies little deeper than the fact that any quadratic polynomial can be completed by the addition of a constant to become the square of a polynomial. Second, we give a proof that there can be no n larger than four with the stated property.