Timeline for How to prove $(\phi-1)(\phi-2)...(\phi-p) = \sqrt{5} + p\left(\frac{1}{2}+A\sqrt{5}\right) \bmod p^2$?
Current License: CC BY-SA 4.0
17 events
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| Jun 15, 2020 at 7:27 | history | edited | CommunityBot |
Commonmark migration
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| Oct 28, 2019 at 8:16 | history | edited | ueir | CC BY-SA 4.0 |
eraborate
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| Oct 27, 2019 at 14:28 | history | edited | ueir | CC BY-SA 4.0 |
fix error
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| Oct 27, 2019 at 14:22 | vote | accept | ueir | ||
| Oct 27, 2019 at 13:43 | history | edited | ueir | CC BY-SA 4.0 |
fix error
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| Oct 27, 2019 at 9:30 | comment | added | Alexey Ustinov | Partial products of the form $(x-1)(x-2)...(x-k)$ have coefficients from oeis.org/A265165. This page refers to the article hal.archives-ouvertes.fr/hal-01236582v4/document where Theorem 7 gives some Supercongruences. Probably it will answer the question. | |
| Oct 27, 2019 at 9:18 | answer | added | Fedor Petrov | timeline score: 6 | |
| Oct 27, 2019 at 8:27 | history | edited | ueir | CC BY-SA 4.0 |
add explanation
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| Oct 27, 2019 at 8:21 | history | edited | ueir | CC BY-SA 4.0 |
add explanation
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| Oct 27, 2019 at 7:53 | history | edited | ueir | CC BY-SA 4.0 |
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| Oct 27, 2019 at 7:45 | history | edited | ueir | CC BY-SA 4.0 |
add explanation
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| Oct 27, 2019 at 3:26 | comment | added | ueir | I'm sorry for unclear question. I'll fix this. Thank you. | |
| Oct 27, 2019 at 3:09 | comment | added | WhatsUp | @AlexeyUstinov I have posted some clarification as an answer. Perhaps you'll be interested now. | |
| Oct 27, 2019 at 3:08 | answer | added | WhatsUp | timeline score: 4 | |
| Oct 27, 2019 at 2:09 | comment | added | Alexey Ustinov | The element $\phi\in\mathbb{F}_p(\sqrt5)$ is a linear combination of $1$ and $\sqrt 5$ with coefficients modulo $p$. So the product $(\phi-1)(\phi-2)...(\phi-p)$ is not well defined modulo $p^2$. You can replace $\phi$ by $\phi+tp$ with $t\in\mathbb{F}_p(\sqrt5)$ and it still be a root of $x^2=x+1\mod p.$ | |
| Oct 27, 2019 at 1:37 | history | edited | Alexey Ustinov | CC BY-SA 4.0 |
added 2 characters in body
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| Oct 27, 2019 at 0:24 | history | asked | ueir | CC BY-SA 4.0 |