(This is a natural continuation of a [previous post](https://mathoverflow.net/questions/438594/).)


**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](https://mathoverflow.net/questions/438701/), 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 $z\_3$).




---


**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.




---


\*\*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 $p=31$?



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


