(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 $z_3$).
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
- 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?
- 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 $p=31$?
Best Answer
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
(found numerically by computing to high precision and applying GP's aldgep).