20030630, 13:30  #1 
Bamboozled!
"πΊππ·π·π"
May 2003
Down not across
2·5,323 Posts 
Factorial problem
Two subproblems, 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 
20030630, 15:21  #2 
Nov 2002
Vienna, Austria
29_{16} Posts 
The easy one is really easy:
7! + 1 = 71^2 working on the hard one ... ;) 
20030630, 18:58  #3 
Sep 2002
1100010100_{2} 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! ?

20030630, 19:41  #4  
"Richard B. Woods"
Aug 2002
Wisconsin USA
2^{2}×3×641 Posts 
Quote:


20030701, 16:56  #5 
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 
20030701, 23:08  #6 
Cranksta Rap Ayatollah
Jul 2003
281_{16} Posts 
http://mathworld.wolfram.com/BrownNumbers.html
Bah! I shoulda known. ;) 
20030701, 23:18  #7  
Jun 2003
The Texas Hill Country
441_{16} Posts 
Re: Factorial problem
Quote:
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20030702, 08:53  #8  
Bamboozled!
"πΊππ·π·π"
May 2003
Down not across
24626_{8} Posts 
Re: Factorial problem
Quote:
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 1215 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. Perhaps I should return to the search. Paul 

20030702, 08:55  #9  
Bamboozled!
"πΊππ·π·π"
May 2003
Down not across
10100110010110_{2} Posts 
Quote:
Paul 

20030720, 19:47  #10 
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.. 
20030720, 19:50  #11 
Jul 2003
37_{8} 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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