Timeline for Binomial coefficient congruence modulo $p^n$
Current License: CC BY-SA 4.0
23 events
| when toggle format | what | by | license | comment | |
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| Mar 12, 2023 at 18:21 | comment | added | Ofir Gorodetsky | I see, you guessed the filename from the EUDML broken link, cool. It could be that these two scans are all there is on-line (I am not sure I've seen the remaining 1965 paper with my own eyes, could be I became aware of it only through zbMath/MathSciNet, I can't recall exactly, these sites have a short review of it). | |
| Mar 12, 2023 at 13:31 | comment | added | darij grinberg | Ah, thank you for that one! Then hdml.di.ionio.gr/pdfs/journals/994.pdf is the second one. The scans are horrible and I can only download them with "wget --no-check-certificate", but it's a start... | |
| Mar 12, 2023 at 13:27 | comment | added | Ofir Gorodetsky | @darijgrinberg No, sorry. Google does find the last one: "On congruences in number-theory", see hdml.di.ionio.gr/pdfs/journals/1014.pdf (note the link gives a security warning). | |
| Mar 12, 2023 at 3:07 | comment | added | darij grinberg | Do you have the Kazandzidis papers? The EUDML links are going nowhere right now... | |
| Oct 27, 2022 at 14:20 | vote | accept | Vlad Matei | ||
| Oct 26, 2022 at 18:18 | comment | added | Vlad Matei | Great! At least this shows that what I was trying to attempt for another problem clearly does not work. Thank you for looking into it | |
| Oct 26, 2022 at 18:16 | history | edited | Ofir Gorodetsky | CC BY-SA 4.0 |
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| Oct 26, 2022 at 18:11 | history | edited | Ofir Gorodetsky | CC BY-SA 4.0 |
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| Oct 26, 2022 at 18:03 | history | edited | Ofir Gorodetsky | CC BY-SA 4.0 |
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| Oct 26, 2022 at 18:00 | comment | added | Ofir Gorodetsky | @VladMatei I managed to prove above, using existing results, that your congruence fails for $n=4$, $a=2$ and $b=1$ and most primes up to $10^9$ (only 2 exceptions). For $n \le 3$ it holds by existing congruences you are familiar with. | |
| Oct 26, 2022 at 17:54 | history | edited | Ofir Gorodetsky | CC BY-SA 4.0 |
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| Oct 26, 2022 at 17:47 | history | edited | Ofir Gorodetsky | CC BY-SA 4.0 |
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| Oct 26, 2022 at 17:40 | history | edited | Ofir Gorodetsky | CC BY-SA 4.0 |
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| Oct 26, 2022 at 17:35 | history | edited | Ofir Gorodetsky | CC BY-SA 4.0 |
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| Oct 26, 2022 at 17:27 | history | edited | Ofir Gorodetsky | CC BY-SA 4.0 |
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| Oct 26, 2022 at 16:33 | comment | added | Ofir Gorodetsky | @DavidESpeyer I've updated the answer, thanks. | |
| Oct 26, 2022 at 16:32 | history | edited | Ofir Gorodetsky | CC BY-SA 4.0 |
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| Oct 26, 2022 at 16:27 | comment | added | Vlad Matei | Thank you Ofir! I was aware of all those results but they do not help for this | |
| Oct 26, 2022 at 16:24 | comment | added | Vlad Matei | Thank you for the remark! This does not exclude the $p^n$. I would be happy with an example showing that is also false; in the example you had $n=2$ so mod 5^2 it works out | |
| Oct 26, 2022 at 16:19 | history | edited | Ofir Gorodetsky | CC BY-SA 4.0 |
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| Oct 26, 2022 at 16:01 | comment | added | David E Speyer | Note that the congruence as stated in the OP is false: We have $\binom{ap^n}{bp^n} \equiv \binom{a p^{n-1}}{b p^{n-1}} \bmod p^{3n}$, but we only have $\binom{ap^n}{bp^n} \equiv \binom{a}{b} \bmod p^{3}$. Indeed, I checked $p=5$, and we have $\binom{50}{25} \equiv \binom{10}{5} \bmod 5^6$, but $\binom{50}{25} \equiv \binom{2}{1}$ only modulo $5^3$. | |
| Oct 26, 2022 at 15:37 | history | edited | Ofir Gorodetsky | CC BY-SA 4.0 |
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| Oct 26, 2022 at 15:23 | history | answered | Ofir Gorodetsky | CC BY-SA 4.0 |