Timeline for About the exact origin of a binomial congruence
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Feb 9 at 2:13 | comment | added | GH from MO | @ChristopheLeuridan There is a more direct proof: $(p-1)\dotsb (p-k)\equiv (-1)\dotsb(-k)=(-1)^k k!\pmod{p}$. | |
| Feb 8 at 23:21 | history | became hot network question | |||
| Feb 8 at 20:17 | vote | accept | Monk | ||
| Feb 8 at 15:54 | history | edited | Alexey Ustinov |
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| Feb 8 at 15:54 | answer | added | Alexey Ustinov | timeline score: 7 | |
| Feb 8 at 15:47 | comment | added | Monk | Thank you for your advice @MichaelLugo. We are confident that something helpful and accurate will be written here within 24h. | |
| Feb 8 at 15:35 | comment | added | Monk | Thank you @ChristopheLeuridan. It is so simple that its historical origin is very hard to find. | |
| Feb 8 at 15:25 | comment | added | Christophe Leuridan | A more known result is that $p$ divides $\binom{p}{k}$ whenever $1 \le k \le p-1$. The congruence you give is a simple consequence (by recursion on $k$), thanks to the recursion relation satisfied by binomial coefficients. So Pascal probably knew it. | |
| Feb 8 at 15:25 | comment | added | Michael Lugo | If you don't get an answer here you might try hsm.stackexchange.com | |
| Feb 8 at 15:21 | history | asked | Monk | CC BY-SA 4.0 |