Timeline for Why are the only numbers $m$ for which $n^{m+1}\equiv n \bmod m$ also the only numbers such that $\displaystyle\sum_{n=1}^{m}{n^m}\equiv 1 \bmod m$?
Current License: CC BY-SA 2.5
11 events
| when toggle format | what | by | license | comment | |
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| Sep 1, 2010 at 23:24 | comment | added | Chris Wuthrich | almost all of it. Can you expand ? | |
| Sep 1, 2010 at 18:24 | comment | added | M.S | which part of above answer is not clear? | |
| Sep 1, 2010 at 18:21 | history | edited | M.S | CC BY-SA 2.5 |
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| Sep 1, 2010 at 7:03 | history | edited | M.S | CC BY-SA 2.5 |
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| Sep 1, 2010 at 6:13 | history | edited | M.S | CC BY-SA 2.5 |
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| Sep 1, 2010 at 5:43 | history | edited | M.S | CC BY-SA 2.5 |
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| Sep 1, 2010 at 5:01 | history | edited | M.S | CC BY-SA 2.5 |
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| Sep 1, 2010 at 2:53 | comment | added | Gerry Myerson | I've attempted to clean up the TeX, but remain baffled by the math. | |
| Aug 31, 2010 at 23:57 | history | edited | Gerry Myerson | CC BY-SA 2.5 |
improved formatting, corrected spelling, fixed grammar
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| Aug 31, 2010 at 23:04 | history | edited | M.S | CC BY-SA 2.5 |
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| Aug 31, 2010 at 22:54 | history | answered | M.S | CC BY-SA 2.5 |