Timeline for A similar relationship between the generic cubic and the Lehmer quintic?
Current License: CC BY-SA 4.0
24 events
| when toggle format | what | by | license | comment | |
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| S Jan 25, 2023 at 16:44 | history | bounty ended | Tito Piezas III | ||
| S Jan 25, 2023 at 16:44 | history | notice removed | Tito Piezas III | ||
| Jan 25, 2023 at 16:43 | vote | accept | Tito Piezas III | ||
| Jan 21, 2023 at 10:24 | history | edited | Tito Piezas III | CC BY-SA 4.0 |
Added 2nd update re Hashimoto septic.
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| Jan 20, 2023 at 13:13 | answer | added | Peter Taylor | timeline score: 3 | |
| Jan 20, 2023 at 0:07 | comment | added | Tito Piezas III | @PeterTaylor Perhaps u can write a tentative answer? U can always edit it based on input from other people. I am curious as to what u found. | |
| Jan 19, 2023 at 11:57 | history | edited | Tito Piezas III | CC BY-SA 4.0 |
Changed wording
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| S Jan 19, 2023 at 11:55 | history | bounty started | Tito Piezas III | ||
| S Jan 19, 2023 at 11:55 | history | notice added | Tito Piezas III | Canonical answer required | |
| Jan 18, 2023 at 2:22 | history | edited | Tito Piezas III | CC BY-SA 4.0 |
Peter Taylor's alternative form
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| Jan 18, 2023 at 2:16 | comment | added | Tito Piezas III | @PeterTaylor Ah, clever! You also used the fact that $abcde = -1$ to get rid of the fifth roots. We can employ that further to make the relation include ALL roots and look more symmetric as, $$\frac1{a}-\frac1{ab}+\frac1{abc}-\frac1{abcd}+\frac1{abcde} = 0$$ I will update the post and credit that alternative form to you. | |
| Jan 17, 2023 at 22:20 | comment | added | Peter Taylor | I have a proof, but it's not very enlightening so I'll hold posting it as an answer for a few days in the hope that someone finds a more insightful one. Factor out $(a^4b^3c^2d)^{1/5}$ to get $1-\frac{1}{a}+\frac{1}{ab}-\frac{1}{abc}+\frac{1}{abcd}$ and compute in $\mathbb{Q}[n]/p(a)$. Sage code can be run online at sagecell.sagemath.org but a direct execution link is too long for a comment. | |
| Jan 17, 2023 at 20:13 | comment | added | KConrad | @TitoPiezasIII Oh! That helps a lot. Why not actually include this information in your post? Saying "for $n = -1$ we get $p = 7$" is quite opaque by comparison. Please put a direct relation between $n$ and $p$ in your post. | |
| Jan 17, 2023 at 17:31 | comment | added | Tito Piezas III | @CHUAKS: Exactly. However, for the cubic in the post, the discriminant is $(n^2-3n+9)^2$ though obviously it is just a minor negation of the variable. | |
| Jan 17, 2023 at 17:16 | comment | added | Tito Piezas III | @KConrad. For the cubic, I did mention in the post that, ".. the case $n=1$ yields $1^{1/7}"$, If you substitute $n=1$ into the cubic, you get the minimal polynomial for $2\cos(2\pi/7)$ and you have your prime $p=7$. For the quintic, if you substitute $n=-1$, you get the minimal polynomial for $2\cos(2\pi/11)$ and you have your prime $p=11$. | |
| Jan 17, 2023 at 16:29 | comment | added | KConrad | Thanks, but what is the connection between $n$ and $p$? You describe the quintic as "for $p \equiv 1 \bmod 10$" but there is no $p$ anywhere in the quintic. Imagine you saw the sentence "For a positive integer $k$, consider $a^2 + 1$." That is how your post reads when you bring up $n$ and $p$. I think there is some kind of information behind these polynomials that is being left unsaid. | |
| Jan 17, 2023 at 16:11 | history | edited | Tito Piezas III | CC BY-SA 4.0 |
Fixed mod format as was suggested. Added link.
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| Jan 17, 2023 at 16:01 | comment | added | Tito Piezas III | @KConrad My apologies. I will fix it. | |
| Jan 17, 2023 at 14:46 | comment | added | KConrad | Why do you write $p=6m+1$ or $p=10m+1$ when there is no visible role for that $m$ at all? At first I thought $m$ was a typo for $n$. Maybe you just mean $p\equiv 1\bmod 6$ and $p\equiv 1\bmod 10$, in which case I think it would be clearer to write “some prime $p\equiv 1\bmod 6$“, say, without mentioning an $m$ that never actually gets used. | |
| Jan 17, 2023 at 14:05 | history | edited | Tito Piezas III | CC BY-SA 4.0 |
Got rid of the 2nd question
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| Jan 17, 2023 at 14:03 | comment | added | Tito Piezas III | @PeterTaylor: Thanks for checking. I'll get rid of the 2nd question to simplify things, as it was just an afterthought anyway. So the conjectured property holds up so far? | |
| Jan 17, 2023 at 12:00 | history | edited | Tito Piezas III | CC BY-SA 4.0 |
Added second question
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| Jan 17, 2023 at 10:07 | comment | added | Peter Taylor | I haven't worked through the details, but Darmon's Note on a polynomial of Emma Lehmer is suggestive that the correct ordering might be given by the cyclic map between the roots $r \to \frac{n+2 + nr - r^2}{1 + (n+2)r}$ | |
| Jan 17, 2023 at 9:11 | history | asked | Tito Piezas III | CC BY-SA 4.0 |