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 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Jan 2, 2016 at 5:39 | comment | added | Włodzimierz Holsztyński | MO is not displaying $\LaTeX$ "\binom" nor \choose". Is it just me? | |
| Jan 2, 2016 at 5:21 | history | edited | Włodzimierz Holsztyński | CC BY-SA 3.0 |
"\bmod" instead of "\pmod"
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| Aug 30, 2010 at 20:54 | vote | accept | DoubleAW | ||
| Aug 30, 2010 at 14:45 | comment | added | fherzig | Maybe it helps to write $\binom m k \frac 1 {k+1}$ as $\binom {m+1}{k+1} \frac 1 {m+1}$. It's then also clear that only one term remains, without distinguishing cases. (In fact, there's a factor of $m$ missing in your second term.) When you wrote "$p$-integral" you meant "prime to $p$". Anyway, very nice. | |
| Aug 30, 2010 at 11:48 | history | edited | Franz Lemmermeyer | CC BY-SA 2.5 |
fixed a typo
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| Aug 30, 2010 at 11:40 | comment | added | DoubleAW | VERY cool. Is this the simplest way to solve it? Or are there ways that don't require Bernoulli numbers? | |
| Aug 30, 2010 at 11:01 | comment | added | Gjergji Zaimi | That's neat! Your answer combined with my observation gives the bijection between the two solution sets without having to check anything by hand or even explicitly writing down the solutions. | |
| Aug 30, 2010 at 10:33 | history | answered | Chris Wuthrich | CC BY-SA 2.5 |