Timeline for A hard diophantine equation: $m!+27=n^3$
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
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| Mar 3, 2023 at 10:42 | history | edited | Martin Sleziak | CC BY-SA 4.0 |
more descriptive title
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| May 13, 2012 at 17:19 | vote | accept | Roberto Bosch Cabrera | ||
| May 10, 2012 at 3:53 | comment | added | Mike Bennett | There's an excellent paper of Berend and Harmse [Trans. AMS 358 (2005)] which treats equations of the shape $m!=p(x)$ for various polynomials $p(x)$ (generalizing an old problem of Brocard). The (very nice) argument of GH is evident in section 4, applied to, for example, to $p(x)=x (x^2+1)$. | |
| May 10, 2012 at 3:43 | comment | added | Roberto Bosch Cabrera | @GH: Thank you, your proof is amazing. | |
| May 9, 2012 at 20:12 | comment | added | GH from MO | The equation has no solution: see the updated "EDIT" section in my response. | |
| May 9, 2012 at 2:14 | comment | added | user6976 | I think the question is interesting especially in view of GH's answer, and should stay open. | |
| May 8, 2012 at 21:19 | comment | added | GH from MO | In "EDIT" to my original response I demonstrate that $m<10^{12}$. | |
| May 8, 2012 at 19:06 | history | reopened |
user6976 GH from MO Neil Strickland Joël Michael Renardy |
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| May 8, 2012 at 8:28 | comment | added | Felix Goldberg | But the more interesting so... | |
| May 8, 2012 at 7:43 | comment | added | Noah Schweber | Strictly speaking, I don't think that's a Diophantine equation. . . | |
| May 8, 2012 at 5:28 | comment | added | Yemon Choi | Whe does this question arise? | |
| May 8, 2012 at 5:23 | history | closed |
Andreas Thom Will Sawin Will Jagy Andrés E. Caicedo Qiaochu Yuan |
too localized | |
| May 8, 2012 at 5:18 | answer | added | GH from MO | timeline score: 34 | |
| May 8, 2012 at 4:26 | history | asked | Roberto Bosch Cabrera | CC BY-SA 3.0 |