找回密码
 欢迎注册
查看: 12523|回复: 2

[提问] 阶乘加1为平方数

[复制链接]
发表于 2015-12-14 20:53:55 | 显示全部楼层 |阅读模式

马上注册,结交更多好友,享用更多功能,让你轻松玩转社区。

您需要 登录 才可以下载或查看,没有账号?欢迎注册

×
比如说:
\(4!+1=5^{2}\)
\(5!+1=11^{2}\)
\(7!+1=71^{2}\)
只有这三个解吗?
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
 楼主| 发表于 2015-12-15 13:23:15 | 显示全部楼层
原来1876年就有人提过了
http://math.fau.edu/richman/Interesting/WebSite/Number71.pdf

Brocard’s problem (1876) is to find numbers m such that m! + 1 is a square. Only
three such numbers m are known, 4, 5, and 7. For the largest of these, the square is
712 = 7! + 1. So 71 is the largest number known whose square is a factorial plus one. A
search by Berndt and Galway in 2000 showed that there are no further numbers m with
fewer than nine digits. Thus any number greater than 71 whose square is a factorial plus
one must have over 90;000 digits.
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
 楼主| 发表于 2015-12-15 13:24:37 | 显示全部楼层
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
您需要登录后才可以回帖 登录 | 欢迎注册

本版积分规则

小黑屋|手机版|数学研发网 ( 苏ICP备07505100号 )

GMT+8, 2024-4-25 12:22 , Processed in 0.044288 second(s), 16 queries .

Powered by Discuz! X3.5

© 2001-2024 Discuz! Team.

快速回复 返回顶部 返回列表