Timeline for Solve in positive integers: $n!=m(m+1)$
Current License: CC BY-SA 3.0
22 events
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| Aug 16, 2023 at 13:59 | comment | added | user25406 | The equation $n!=m(m+1)$ can be rewritten $n!=2T_{m}$ in terms of triangular numbers. It's hard to believe there are other solutions than those mentioned in the comments, $2!=1*2$ and $3!=2*3$ for the simple reason that any factorial can be rewritten as a product of a bunch of twice a triangular number. Ex: $5!=1*2*3*4*5=1*(2*3)(4*5)=1*2T_{2}*2*T_{4}$. It's highly unlikely that a product of a bunch of twice a triangular numbers is itself twice triangular number. But maybe it can be proven by considering the triangular number formulation. | |
| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| S Aug 14, 2016 at 9:43 | history | suggested | The Thin Whistler | CC BY-SA 3.0 |
improved grammar
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| Aug 14, 2016 at 9:31 | review | Suggested edits | |||
| S Aug 14, 2016 at 9:43 | |||||
| Nov 1, 2013 at 5:52 | history | edited | Alexey Ustinov | CC BY-SA 3.0 |
Added refference to correct question
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| Nov 1, 2013 at 5:44 | history | edited | Alexey Ustinov | CC BY-SA 3.0 |
The tag "elementary-number-theory" was added
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| Feb 2, 2011 at 7:23 | answer | added | Tapio Rajala | timeline score: 19 | |
| Oct 9, 2010 at 9:19 | history | edited | Unknown |
edited tags
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| Oct 9, 2010 at 9:19 | history | edited | Unknown |
edited tags
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| Sep 20, 2010 at 6:38 | answer | added | Denis Serre | timeline score: 16 | |
| Sep 19, 2010 at 0:35 | vote | accept | Alexey Ustinov | ||
| Sep 18, 2010 at 23:46 | answer | added | Gjergji Zaimi | timeline score: 38 | |
| Sep 18, 2010 at 16:01 | comment | added | Felipe Voloch | I voted to close as too localized, which it is. But, on close inspection, maybe it's not easy. Following up on Dror's remark about ABC, maybe C. Stewart's weaker results towards ABC might be enough here. | |
| Sep 18, 2010 at 15:54 | comment | added | user6096 | This is very similar to Brocard's problem, which is unresolved... en.wikipedia.org/wiki/Brocard%27s_problem | |
| Sep 18, 2010 at 15:50 | comment | added | user6976 | If $n$ is a solution then $4n!+1$ is a square, but it looks like if $n\ne 2,3$, $4n!+1$ is not divisible by a square of a prime. | |
| Sep 18, 2010 at 14:56 | comment | added | Dror Speiser | Before someone posts an elementary solution, note that abc conjecture implies finitely many solutions since $rad(n!) = \prod_{p<n} p \sim e^n$ and $(e^n)^{1+\epsilon} < \sqrt{n!}$. | |
| Sep 18, 2010 at 14:20 | comment | added | muad | @Thierry Zell, I feel that way too - but I am not sure how can this idea be turned into a proof? | |
| Sep 18, 2010 at 14:05 | comment | added | Thierry Zell | I'm no specialist, but my guess would be that the two solutions above are the only ones. Rationale: both m and m+1 would need to have only small factors that somehow miraculously balance out to make n!. Seems too good to be true for a large value of n. For n up to 1000, there is no other solution. | |
| Sep 18, 2010 at 10:44 | comment | added | Alexey Ustinov | I can add one more $2!=1\cdot 2$ | |
| Sep 18, 2010 at 10:39 | comment | added | Joel David Hamkins | $3!=2\cdot3$ ? | |
| Sep 18, 2010 at 10:36 | comment | added | Robin Chapman | This one is a real problem :-) | |
| Sep 18, 2010 at 10:27 | history | asked | Alexey Ustinov | CC BY-SA 2.5 |