Timeline for On quintic roots $x_1^{1/5}+x_2^{1/5}+\dots+x_5^{1/5}$
Current License: CC BY-SA 3.0
23 events
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| Jul 17, 2023 at 15:06 | comment | added | Tito Piezas III | @NoamD.Elkies I revisited this by connecting it to the Emma Lehmer quintic. Care to take a peek at this new MSE post? | |
| Jul 11, 2017 at 14:09 | comment | added | davidoff303 | For instance, taking $p=11$ we obtain $$\left \{ S_1(g),S_2(g),S_3(g),S_4(g),S_5(g) \right \} =\\ \left \{ 2 \cos\frac{2\pi}{11},2 \cos\frac{4\pi}{11}, 2 \cos\frac{6\pi}{11}, 2 \cos\frac{8\pi}{11},2 \cos\frac{10\pi}{11}\right \}$$ and $f(x)=x^{125} + 15125x^{124}+ \ldots \equiv x^{125}+1 \bmod 5$. | |
| Jul 11, 2017 at 14:08 | comment | added | davidoff303 | It can be shown that $$s:=\left(\sqrt[5]{S_1(g)} + \sqrt[5]{S_2(g)} + \sqrt[5]{S_3(g)} + \sqrt[5]{S_4(g)} + \sqrt[5]{S_5(g)} \right)^5$$ is an algebraic integer, and let $f(x)\in \mathbb Z[x]$ be the minimal polynomial of $s$. Then, as I have checked for $p=11,31,41,61,71,101,131,151,181,191$, we have $f(x)\equiv {x^5}^l +1 \bmod 5$, where $l$ equals $3$ or $4$. | |
| Jul 11, 2017 at 14:08 | comment | added | davidoff303 | @ Noam D. Elkies Thank you, Noam. I checked some (perhaps interesting) heuristic facts. Let me try to explain what I mean. Let $p \equiv 1 \bmod 5$ be a prime number, and $g$ a primitive root modulo $p$. Put $$ S_1(g):= \sum_{k=0}^{\frac{p-6}5}\cos \left(\frac{2\pi}{p}g^{5k}\right) , S_2(g):= \sum_{k=0}^{\frac{p-6}5}\cos \left(\frac{2\pi}{p}g^{5k+1}\right) , S_3(g):= \sum_{k=0}^{\frac{p-6}5}\cos \left(\frac{2\pi}{p}g^{5k+2}\right) ,\\ S_4(g):= \sum_{k=0}^{\frac{p-6}5}\cos \left(\frac{2\pi}{p}g^{5k+3}\right) , S_5(g):= \sum_{k=0}^{\frac{p-6}5}\cos \left(\frac{2\pi}{p}g^{5k+4}\right) .$$ | |
| Jul 9, 2017 at 18:42 | comment | added | Noam D. Elkies | Come to think of it there's no need because the code's short enough. (I doubled the roots to make them algebraic integers, so the polynomial is in ${\bf Z}[x]$.)$$ $$ default(realprecision,500); v=2 * vector(5,n,cos(Pi*n/5.5)); for(n=1,5,v[n]=sign(v[n]) * abs(v[n])^(1/5)); z=exp(Pi * I/2.5); P=1; for(a=0,4,for(b=0,4,for(c=0,4,P *= x-(v * [z^a,z^b,z^c,z^(-a-b-c),1]~)^5))); round(P) | |
| Jul 9, 2017 at 17:48 | comment | added | Noam D. Elkies | What's your e-mail address (and why do you "really need to see" it)? | |
| Jul 9, 2017 at 13:29 | comment | added | davidoff303 | @Noam D. Elkies I really need to see this polynomial. Can you send me the paragraph of gp code ? | |
| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Aug 3, 2016 at 3:44 | comment | added | Tito Piezas III | @NoamD.Elkies: I'm afraid I'm not that familiar with gp. However, if you wish, you can answer that MSE question using that code. But only if you want to. I'm just happy to know its exact degree after wondering about it for years. | |
| Aug 3, 2016 at 1:57 | comment | added | Noam D. Elkies | You're welcome. Meanwhile, the number you ask about in that post (fifth power of $\sum_{n=1}^5 \cos(2\pi n/11)^{1/5}$) is indeed of degree $125$; if you really need to see the polynomial I can send you the paragraph of gp code . . . | |
| Aug 2, 2016 at 22:28 | comment | added | Tito Piezas III | @NoamD.Elkies: Ok. And thanks again for that quintic. I've been looking for something like it for more than two years already, as in this MSE post. | |
| Aug 2, 2016 at 21:20 | comment | added | Noam D. Elkies | No problem, and no need to apologize... | |
| Aug 2, 2016 at 20:57 | comment | added | Tito Piezas III | @NoamD.Elkies: Oops, I just saw your email. I've deleted my answer, and I'll wait for the one you planned that explains the general case. My sincere apologies. | |
| Aug 2, 2016 at 20:03 | comment | added | Noam D. Elkies | I thought you wanted me to post an answer including an explanation of why these identities exist. And yes, as I wrote in e-mail there are similar things for 7th roots and beyond, in fields with either a cyclic Galois group (such as the one contained in ${\bf Q}(\cos 2\pi/29)$) or dihedral Galois (I gave the examples of $x^5 + x^4 - 3x^3 - 3x^2 + 4x - 1$ and $x^7 - 26x^6 + 350x^5 - 2192x^4 + 6287x^3 - 6718x^2 - 162x - 1$ in the same email). | |
| Aug 2, 2016 at 19:34 | comment | added | Tito Piezas III | @NoamD.Elkies: Yes, great result! I've elaborated on your comment and gave more details as an answer below. I hope you don't mind. | |
| Aug 1, 2016 at 17:41 | history | edited | Tito Piezas III | CC BY-SA 3.0 |
Better formulas and examples
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| Aug 1, 2016 at 4:43 | comment | added | Noam D. Elkies | So you're looking for something like the roots of $x^5 + 6x^4 - x^3 - 32x^2 + 16x - 1$ (which are in ${\bf Q}(\cos \pi/11)$ and have fifth roots that sum to $v^{1/5}$ with $v^5 - 1370v^4 + 518385v^3 + 319010v^2 + 143005v + 1 = 0$)? | |
| Jul 31, 2016 at 18:14 | history | edited | Tito Piezas III | CC BY-SA 3.0 |
Gave illustrative example.
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| Jul 12, 2016 at 21:05 | history | edited | Tito Piezas III | CC BY-SA 3.0 |
added 17 characters in body
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| Jul 12, 2016 at 20:42 | history | edited | Tito Piezas III | CC BY-SA 3.0 |
Details.
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| Jul 12, 2016 at 20:28 | history | edited | Tito Piezas III | CC BY-SA 3.0 |
Major revision to incorporate new results.
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| Jan 15, 2015 at 0:38 | comment | added | Kimball | Your use of "degree of an algebraic number" is inconsistent. I guess you mean that number generate an extension of that degree, so you should say $F_3$ is a priori of degree at most or dividing 9, and similarly for $F_5$. Also, maybe it would be helpful to say what you tried. Did you try a computer search? Roots of quintics seem complicated to me, and this is what I would suggest. | |
| Jan 10, 2015 at 18:38 | history | asked | Tito Piezas III | CC BY-SA 3.0 |