I. p = 11
The minimal polynomial of $x=2\cos\frac{2\pi}{11}$ is,
$$x^5 +x^4 −4x^3 −3x^2 +3x+1 = 0$$
As is known, Method 1 solves it in the form,
$$x = \frac15\left(A+y_1^{1/5}+y_2^{1/5}+y_3^{1/5}+y_4^{1/5}\right)$$
where the $y_k$ are the four roots of the quartic resolvent. There is a slight problem when all $y_k$ are complex. So alternatively, we can have a Method 2 using just one root $y_1$,
$$y_1= -\frac{ab}{11} =-\frac{11}{4} \left(89-25 \sqrt{5}-5 \sqrt{-(410+178 \sqrt{5})}\right) $$
which surprisingly has factors,
$$a=\frac{11}{4}\left(-1+5\sqrt{5}+\sqrt{-10(5+\sqrt{5})}\right)$$
$$b=\frac{11}{4}\left(-31+5\sqrt{5}-\sqrt{-10(85+31\sqrt{5})}\right)$$
and we can use this factorization to solve the minimal polynomial in a different way as,
$$x_k = -\frac{1}{5}\left(\frac{1}{\beta^{-1}}+\frac{1}{\beta^0}+\frac{11}{\beta^1}+\frac{a}{\beta^2}+\frac{b}{\beta^3} \right)$$
where $\color{blue}{\beta^5+y_1=0}$. Of course, this quintic in $\beta$ has five solutions which gives the five $x_k$. But since Method 2 takes the fifth root of only one complex number $y_1$, this avoids the problem of Method 1.
II. p = 29
The minimal polynomial of $x=2\big(\cos\frac{2\pi}{29} + \cos\frac{24\pi}{29}\big)$is,
$$x^7 + x^6 - 12x^5 - 7x^4 + 28x^3 + 14x^2 - 9x + 1 = 0$$
This can be solved by Method 2 as well. (I still have to locate my notes back from 2015.)
III. p = 43
The minimal polynomial of $x=2\big(\cos\frac{2\pi}{43} + \cos\frac{12\pi}{43} +\cos\frac{14\pi}{43}\big)$ is,
$$x^7 + x^6 - 18x^5 - 35x^4 + 38x^3 + 104x^2 + 7x - 49 = 0$$
Fortunately, I found my notes for this. Its sextic resolvent is,
$$43^{21} - 109288043513083876814190246493y + 343597088843572196970429y^2 - 781038748669395217y^3 + 1264067561247y^4 - 1479157y^5 + y^6 = 0$$
This can be greatly simplified and solved merely as a quadratic,
$$y^2 - 43(8265 - 7889r + 343r^2)y + 43^7 = 0$$
using the roots of $r^3+r^2-2r-1=0$, the minimal polynomial of $r = 2\cos\frac{2\pi}7$. (The sextic can be recovered by eliminating $r$ using resultants as in this WA calculation.) Like the quartic root, the sextic root $y$ also factors, but in two ways,
$$y_1 = \frac{43^4pq}{43} = \frac{43^4rs}{43}$$
where $(p,q,r,s)$ are the appropriate roots of,
$$43^3 - 1849p + 645p^2 + 41p^3 + 15p^4 - p^5 + p^6 = 0\\ 43^3 + 282983q + 511267q^2 + 615413q^3 + 511267q^4 + 282983q^5 + 43^3q^6 = 0$$
and,
$$43^2 - 1763r - 1553r^2 + 3277r^3 - 1553r^4 - 1763r^5 + 43^2r^6 = 0\\ 43^4 + 1577197s + 329337s^2 + 50161s^3 + 7659s^4 + 853s^5 + 43s^6= 0$$
For example, if we choose $y_1 \approx -22340.0362 + 520883.4167i$, then,
\begin{align} p &\approx 23.9788 - 280.9484i, \qquad q\approx -79435.9450 + 3360.6067i\\ r &\approx 887.1924 - 1622.2486i, \quad s\approx -10877.3102 + 5356.5441i \end{align}
The correct roots $(p,q,r,s)$ can be ascertained by the two relationships above. Note the sextics in $(q,r)$ are palindromic. (Now why is that?) Method 2 then yields,
$$x_k = \frac{1}{7}\left(\frac{1}{\beta^{-1}}-\frac{1}{\beta^0}+\frac{43}{\beta^1}+\frac{43\color{red}p}{\beta^2}+\frac{43^2r}{\beta^3}+\frac{43^2s}{\beta^4}+\frac{43^3\color{red}q}{\beta^5} \right)$$
where $\color{blue}{\beta^7-y_1=0}.$ Again, only one $7$th root extraction is needed.
IV. Questions
- Why can we factor the resolvent equation of degree $n$ into two equations with same degree $n$ and smaller coefficients? (Almost akin to how we factor composite integers into smaller integers.) Is it implicit in Galois theory that the resolvent is a "composite" algebraic number, or must certain conditions be satisfied?
- And why does one factor have palindromic coefficients? (For $p=11$, then $b/11^2$ is also a root of a palindrome.)
