Using the Lehmer quintic to solve $11$-degree equations and higher?

Question

(This is a natural continuation of a previous post.)

I. Quintic method

Given the Lehmer quintic,

$$x^5 + n^2x^4 - (2n^3 + 6n^2 + 10n + 10)x^3 + (n^4 + 5n^3 + 11n^2 + 15n + 5)x^2 + (n^3 + 4n^2 + 10n + 10)x +1 = 0$$

From this post, we saw its roots $x_i$ can be ordered such that,

$$(x_1^4\, x_2^3\, x_3^2\, x_4)^{1/5} + (x_2^4\, x_3^3\, x_4^2\, x_5)^{1/5} + \dots + (x_5^4\, x_1^3\, x_2^2\, x_3)^{1/5} = 0$$

Or equivalently,

$$\frac1{x_1}-\frac1{x_1 x_2}+\frac1{x_1 x_2 x_3}-\frac1{x_1 x_2 x_3 x_4}+\frac1{x_1x_2x_3x_4x_5} = 0$$

This ordering is useful, since using the same order of roots, it turns out that,

$$(x_1\,x_2^4\,x_3^2\,x_4^7)^{1/11} + (x_2\,x_3^4\,x_4^2\,x_5^7)^{1/11} + \dots + (x_5\,x_1^4\,x_2^2\,x_3^7)^{1/11} = z_1$$ $$(x_1^2\,x_2\,x_3^7\,x_4^4)^{1/11} + (x_2^2\,x_3\,x_4^7\,x_5^4)^{1/11} + \dots + (x_5^2\,x_1\,x_2^7\,x_3^4)^{1/11} = z_2$$ $$(x_1^4\,x_2^7\,x_3\,x_4^2)^{1/11} + (x_2^4\,x_3^7\,x_4\,x_5^2)^{1/11} + \dots +(x_5^4\,x_1^7\,x_2\,x_3^2)^{1/11} = z_3$$ $$(x_1^7\,x_2^2\,x_3^4\,x_4)^{1/11} + (x_2^7\,x_3^2\,x_4^4\,x_5)^{1/11} + \dots +(x_5^7\,x_1^2\,x_2^4\,x_3)^{1/11} = z_4$$

where the $z_i$ are now roots of four different $11$-deg equations.

Notice that $z_1$ and $z_4$ are "complementary", with the $x_1, x_4$ of their starting terms just swapping exponents, likewise with the $x_2, x_3$. (The same can be said for $z_2$ and $z_3$).

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II. Example

Let $n=-1$ and we have

$$x^5 + x^4 - 4x^3 - 3x^2 + 3x + 1 = 0$$

with roots $x_i = 2\cos\frac{2\pi k}{11}$ in the same order $k= 1,4,5,2,3.$ Using the four expressions above, these yield the four quintics,

\begin{align} -1 + 529 y + 5361 y^2 + 756 y^3 - 377 y^4 &+ y^5\\ -1 - 439 y - 45701 y^2 - 5536 y^3 - 135 y^4 &+ y^5\\ -1 - 4806 y - 5771 y^2 - 1543 y^3 - 47 y^4 &+ y^5\\ -1 + 408 y + 2215 y^2 + 1724 y^3 - 740 y^4 &+ y^5 \end{align}

such that,

$$y_1^{1/11}+y_2^{1/11}+y_3^{1/11}+y_4^{1/11}+y_5^{1/11} = z_i$$

are roots of the four $11$-degree equations,

\begin{align} -16954 - 2387 z + 3762 z^2 + 2200 z^3 + 704 z^4 & + 187 z^5 - 110 z^6 - 33 z^7 + 22 z^8 + z^{11}\\ 9908 + 8987 z + 7029 z^2 + 4499 z^3 + 1188 z^4 & - 418 z^5 - 352 z^6 - 33 z^7 + 22 z^8 + z^{11}\\ 7213 - 572 z - 1804 z^2 + 2563 z^3 + 1430 z^4 & - 418 z^5 - 352 z^6 - 33 z^7 + 22 z^8 + z^{11}\\ -18043 + 7777 z + 253 z^2 + 3168 z^3 - 385 z^4 & - 418 z^5 - 110 z^6 - 33 z^7 + 22 z^8 + z^{11} \end{align}

Incidentally, the unexplained phenomenon of shared coefficients (also found in the cubic method) appears again.

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**III. Questions

1. For the quintic with one root "fixed", there are now $(n-1)! = 24$ permutations, creating 24 different quintics. Why is it there are only four useful ones (I checked) that can solve an $11$-deg equation?
2. And what would be one ($a,b,c,d)$ such that $(x_1^a\,x_2^b\,x_3^c\,x_4^d)^{1/p}$ and its Galois conjugates can be used to solve $p=31$?

Answer 1

Question 1: An irreducible Lehmer quintic $L(n)$ has Galois group ${\bf Z} / 5 {\bf Z}$. Let $\sigma$ be a generator, and order the roots $x_i$ ($i \bmod 5$) so that $x_{i+1} = \sigma(x_i)$ for each $i$.

Note that $\prod_i x_i = -1$ because $L(n)$ has constant coefficient $1$. Thus $\prod_i x_i^{a_i} = (-1)^c \prod_i x_i^{a_i+c}$ for any integer $c$. You chose $c$ that makes $a_0 = 0$, but other choices may be more useful. For sums of (1/11)th powers, taking $c=6$ converts the exponents $(4,2,7,0,1)$ to $(10,8,2,6,7)$, proportional to the 5th roots of unity $\bmod 11$ (namely $c_i \equiv -3^i \bmod 11$ for $i=0,1,2,3,4$). The four choices of exponents correspond to the four isomorphisms from the Galois group to the group of 5th roots of unity $\bmod 11$.

Question 2: For (1/31)st powers we likewise need 5th roots of unity $\bmod 31$, which are $(1,2,4,8,16)$ (or ($0,1,3,7,15$) if you insist on having a zero exponent). Again there are four choices, again with "shared coefficients" including some zero coefficients, such as

x^31 + 744*x^26 - 620*x^25 - 4960*x^24 - 16058*x^23 + 27528*x^22 + 186372*x^21 + 13392*x^20 + 57908*x^19 - 2151400*x^18 + 6700526*x^17 + 6967374*x^16 + 33644796*x^15 - 86990712*x^14 - 125974204*x^13 - 244592976*x^12 + 559706116*x^11 + 1329731980*x^10 + 1937968348*x^9 + 760060790*x^8 - 615794292*x^7 - 539362428*x^6 + 376540942*x^5 + 474549116*x^4 + 105617372*x^3 - 15563860*x^2 - 3550678*x + 645413

(found numerically by computing to high precision and applying GP's aldgep).