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 2003-06-30, 13:30 #1 xilman Bamboozled!     "πΊππ·π·π­" May 2003 Down not across 2·5,323 Posts Factorial problem Two sub-problems, one very easy, one very hard. 4! + 1 = 5^2 5! + 1 = 11^2 Easy question: what is the third in the series n!+1 = m^2 ? Very hard: what is the fourth? Paul
 2003-06-30, 15:21 #2 koal   Nov 2002 Vienna, Austria 2916 Posts The easy one is really easy: 7! + 1 = 71^2 working on the hard one ... ;)
 2003-06-30, 18:58 #3 Jwb52z     Sep 2002 11000101002 Posts Maybe it is because I haven't done math in years, but there seems as if there should be some kind of explanation with this problem. The reason I say this is that you don't give a rationale for why a certain thing should come next. For example, why is the next number 7! and not 6! ?
2003-06-30, 19:41   #4

"Richard B. Woods"
Aug 2002
Wisconsin USA

22×3×641 Posts

Quote:
 Originally Posted by Jwb52z why is the next number 7! and not 6! ?
6! + 1 = 721, which is not a square of an integer, so can not satisfy the right side of the equation n!+1 = m^2.

 2003-07-01, 16:56 #5 Orgasmic Troll Cranksta Rap Ayatollah     Jul 2003 641 Posts My first thought is that factorials have trailing zeros, so m^2 = 1 mod 10^k, so for say, 12! which has 2 trailing zeros, m has to end in either ..01, ..49 or ..99 hmm.. factorials have trailing zeros in any base, so m^2 = 1 mod b^k .. maybe somebody can use this to help in their search.. anyway, just some thoughts
 2003-07-01, 23:08 #6 Orgasmic Troll Cranksta Rap Ayatollah     Jul 2003 28116 Posts http://mathworld.wolfram.com/BrownNumbers.html Bah! I shoulda known. ;)
2003-07-01, 23:18   #7
Wacky

Jun 2003
The Texas Hill Country

44116 Posts
Re: Factorial problem

Quote:
 Originally Posted by xilman Two sub-problems, one very easy, one very hard. Very hard: what is the fourth? Paul
Quote:
 Originally Posted by TravisT Bah! I shoulda known. ;)
The moral to this story (problem) is :
Quote:
 Don't trust Paul

2003-07-02, 08:53   #8
xilman
Bamboozled!

"πΊππ·π·π­"
May 2003
Down not across

246268 Posts
Re: Factorial problem

Quote:
Originally Posted by Wackerbarth
[The moral to this story (problem) is :
Quote:
 Don't trust Paul
;)

I don't think I was being untrustworthy though, as I did warn you it was very hard. A proof that there is no fourth in the series would count as a solution to my problem. It would also result in a very publishable paper.

About 12-15 years ago I performed a search for a fourth solution without success. My search limit was around a million or so and although I didn't find a solution, the search did turn up a previously unknown bug in the /usr/games/primes program on SunOS. No idea whether Sun ever fixed it.

Paul

2003-07-02, 08:55   #9
xilman
Bamboozled!

"πΊππ·π·π­"
May 2003
Down not across

101001100101102 Posts

Quote:
 Originally Posted by TravisT http://mathworld.wolfram.com/BrownNumbers.html Bah! I shoulda known. ;)
Indeed. A rule of thumb is that when a mathematical problem is described as "very hard" a good way of approaching the problem is to start with a literature search. Which, of course, is exactly what you did.

Paul

 2003-07-20, 19:47 #10 Annunaki   Jul 2003 31 Posts may be that will give something may be not.. but as forst three examples have anly such n^2 that you can`t divide by 2, lets asume that so will be the 4th.. if so if x!+1=n^2 then x! must divide by 8..
 2003-07-20, 19:50 #11 Annunaki   Jul 2003 378 Posts hmm just figured out that all factorials past 4 should be divisible by 8.. but still may be this is a good point to start out.. may be it is even very wrong..:)

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