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Old 2003-06-30, 13:30   #1
xilman
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Default 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?


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Old 2003-06-30, 15:21   #2
koal
 
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The easy one is really easy:

7! + 1 = 71^2

working on the hard one ... ;)
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Old 2003-06-30, 18:58   #3
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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! ?
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Old 2003-06-30, 19:41   #4
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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.
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Old 2003-07-01, 16:56   #5
Orgasmic Troll
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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
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Old 2003-07-01, 23:08   #6
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http://mathworld.wolfram.com/BrownNumbers.html

Bah! I shoulda known. ;)
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Old 2003-07-01, 23:18   #7
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Default 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
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Old 2003-07-02, 08:53   #8
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Default 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.

Perhaps I should return to the search.


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Old 2003-07-02, 08:55   #9
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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.

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Old 2003-07-20, 19:47   #10
Annunaki
 
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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..
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Old 2003-07-20, 19:50   #11
Annunaki
 
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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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