Can De Moivre's quintic be extended to other F20 quintics?

Question

For the DeMoivre's quintic $x^5 - 5ax^3 + 5a^2x - b = 0$ exist formulas for the relations between the roots. When two roots, say $x_1$ and $x_2$, are known then one can find the other roots with the formulas:

$\frac{x_1^3 – 3ax_1 – ax_2}{x_1x_2 + a}, \frac{x_2^3 – 3ax_2 – ax_1}{x_1x_2 + a},\frac{(x_1 + x_2)(3a – (x_1^2 + x_2^2))}{x_1x_2 + a}.$

I discovered these formulas on the internet in the article ‘DeMoivre's quintic and a theorem of Galois’ by B.K. Spearman and K.S. Williams.

An irreducible DeMoivre's quintic has Galois group F20. Does anyone know if there are analogous formulas for other quintics with Galois group F20? Thank you in advance.

Answer 1

Yes, this is possible. When you are familiar with the analytical computation of the roots of a quintic using the Lagrange resolvent, then this answer is likely to interest you. It presents an algorithm for the solution of the problem.

When $x_0…x_4$ are the roots of the quintic, $r_1…r_4$ the roots of the resolvent and $\omega= e^{2\pi i/5}$ then:
    $x_0 = r_1 + r_2 + r_3 + r_4$
    $x_1 = \omega^4r_1 + \omega^3r_2 + \omega^2r_3 + \omega r_4$
    $x_2 = \omega^3r_1 + \omega r_2 + \omega^4r^3 + \omega^2 r_4$
    $x_3 = \omega^2r_1 + \omega^4r_2 + \omega r_3 + \omega^3r_4$
    $x_4 = \omega r_1 + \omega^2r_2 + \omega^3r^3 + \omega^4 r_4$.
With $r_1=r_4=0$ respectively $r_2=r_3=0$ we get De Moivre’s equation (DM-quintic):
   $y^5 - 5ay^3 + 5a^2y – b = 0$
with $a=r_1 r_4$ or $a=r_2 r_3$ and with roots as in the table below.
   $y_0 = r_1 + r_4$    or  $y_0 = r_2 + r_3$
   $y_1 = \omega^4r_1 + \omega r_4$   or  $y_1 = \omega^3r_2 + \omega^2r_3$
   $y_2 = \omega^3r_1 + \omega^2 r_4$   or  $y_2 = \omega r_2 + \omega^4r_3$
   $y_3 = \omega^2r_1 + \omega^3r_4$   or  $y_3 = \omega^4r_2 + \omega r_3$
   $y_4 = \omega r_1 + \omega^4r_4$   or  $y_4 = \omega^2r_2 + \omega^3r_3$
Now we can convert the roots of the general quintic to de roots of the DM-quintic and back by solving the next to systems of equations.
   $a_1x_0^4 + a_2x_0^3 + a_3x_0^2 + a_4x_0 + a_5 = y_0$
   $a_1x_1^4 + a_2x_1^3 + a_3x_1^2 + a_4x_1 + a_5 = y_1$
   $a_1x_2^4 + a_2x_2^3 + a_3x_2^2 + a_4x_2 + a_5 = y_2$
   $a_1x_3^4 + a_2x_3^3 + a_3x_3^2 + a_4x_3 + a_5 = y_3$
   $a_1x_4^4 + a_2x_4^3 + a_3x_4^2 + a_4x_4 + a_5 = y_4$

   $b_1y_0^4 + b_2y_0^3 + b_3y_0^2 + b_4y_0 + b_5 = x_0$
   $b_1y_1^4 + b_2y_1^3 + b_3y_1^2 + b_4y_1 + b_5 = x_1$
   $b_1y_2^4 + b_2y_2^3 + b_3y_2^2 + b_4y_2 + b_5 = x_2$
   $b_1y_3^4 + b_2y_3^3 + b_3y_3^2 + b_4y_3 + b_5 = x_3$
   $b_1y_4^4 + b_2y_4^3 + b_3y_4^2 + b_4y_4 + b_5 = x_4$
The calculation of, for example, $x_3$ from $x_1$ and $x_2$ then looks like this:
   $y_1=g(x_1)=a_1x_1^4+a_2x_1^3+a_3x_1^2+a_4x_1+a_5$
   $y_2=g(x_2)=a_1x_2^4+a_2x_2^3+a_3x_2^2+a_4x_2+a_5$
   $y_3=\frac{y_1^3-a(3y_1+y_2)}{y_1y_2+a}=\frac{g(x_1)^3-a(3g(x_1)+g(x_2)}{g(x_1)(g(x_2)+a}$
   $x_3=h(y_3)=h(b_1y_3^4+b_2y_3^3+b_3y_3^2+b_4y_3+b_5)$.

The values we find for the a’s and b’s all have the form $c_1+c_2 \sqrt{p}$, where $c_1,c_2$ and p are rational numbers. The conversion via $r_1 r_4$ matches the positive root and the conversion via $r_2 r_3$ the negative root of p, or vice versa. Both conversions give the same result.

To eliminate $\sqrt{p}$ we split $a_1$ into $a_{11}$ and $a_{12} \sqrt{p}, a_2$ into $a_{21}$ and $a_{22}\sqrt{p}$ etc. So:
   $g_1(x)=a_{11} x_1^4+a_{21} x_1^3+a_{31} x_1^2+a_{41} x_1+a_{51}$,
   $g_2(x)=a_{12} x_1^4+a_{22} x_1^3+a_{32} x_1^2+a_{42} x_1+a_{52}$,

   $h_1(y)=b_{11} y_1^4+b_{21} y_1^3+b_{31} y_1^2+b_{41} y_1+b_{51}$,
   $h_2(y)=b_{12} y_1^4+b_{22} y_1^3+b_{32} y_1^2+b_{42} y_1+b_{52}$.

In the formulas for $y_3$ and $x_3$ above, we replace:
   $g(x)$ by $g_1(x)+g_2(x)\sqrt{p}$,
   $h(y)$ by $h_1(y)+h_2(y)\sqrt{p}$,
   $a$ by $u+v\sqrt{p}$.

Then we combine the conversions with the positive and the negative value of $\sqrt{p}$ into a function $f(x)$ whose output is the average of two equal results. The root of p disappears because the positive and negative values cancel each other out. In this way we find a rational relationship between the roots of the genaral F20-quintic.

For the equation $x^5-s_1x^4+s_2x^3-s_3x^2+s_4x-s_5=0$ the value of p can be calculated with the formula
   $p=5(s_2^2+12s_4+4\theta)$,
where $\theta$ is the rational root of the resolvent sextic.

That all coefficients are rational can be easily proved by showing that they are fixed by the permutations of the roots.

For the quintic $x^5 - 10x^3-20x^2 - 1505x - 7412 = 0$ the values of the a’s, b’s, u, v and p are as below.

$a_{11}=0$  $a_{12}=+\frac{34}{59020}$  $b_{11}=0$     $b_{12}=0$

$a_{21}=0$  $a_{22}=-\frac{251}{59020}$  $b_{21}= +\frac{8291}{9569}$   $b_{22}=+\frac{5872}{9569}$

$a_{31}=0$  $a_{32}=+\frac{1760}{59020}$    $b_{31}=+\frac{1154}{9569}$  $b_{32}=+\frac{523}{9569}$

$a_{41}=\frac{1}{2}$  $a_{42}=-\frac{21431}{29510}$   $b_{41}=+\frac{19928}{9569}$ $b_{42}=+\frac{7257}{9569}$

$a_{51}=0$   $a_{52}=-\frac{46324}{59020}$   $b_{51}= -\frac{216}{9569}$  $b_{52}=+\frac{1262}{9569}$

v = 1, u = -1, p = 2

These values, along with $x_1$ en $x_2$, are the input to the function $f(x)$ that calculates $x_3$ from $x_1$ and $x_2$. To find $x_4$ we need to swap $x_1$ and $x_2$. The value of $x_5$ then follows from the sum of the roots.

The function $f(x)$ performs the following calculations:
$x_3=h(y_3)=h(\frac{g(x_1)^3-a(3g(x_1)+g(x_2)}{g(x_1)g(x_2)+a})$

where $g(x)=g_1(x)+g_2(x)\sqrt{p}$,
   $h(y)=h_1(y)+h_2(y)\sqrt{p}$,
   $a=u+v\sqrt{p}$.
As mentioned above this calculations are performed twice: once with the positive root and once with de negative root of p. In the final result $\sqrt{p}$ is squared or has disappeared.

The calculations are best performed with mathematical software. I myself performed the calculations for over 40 equations where p is not a square in Excel (with 64-bit floating point numbers) and converted the results after rounding to rationals. For some equations the accuracy turned out to be insufficient.

G.J.Th. van Berkel

Answer 2

Not an answer to the question, just clarifying a bit de Moivre polynomials.

If $x$ satisfies the equation $t^5-1=0$, then $x=u t + \frac{v}{t}$ will satisfy the equation obtained by eliminating $t$ from the two equalities. One way is to consider the resultant in $t$ of the polynomials $t^5-1$ and $t ( u t + \frac{v}{t}-x)$ ( faster than Groebner bases). This will get us the equation

$$x^5 - 5 (u v) x^3 + 5 (u v)^2 x - (u^5 + v^5)=0$$

It should be easy to pass from form posted to this one. Also, given two roots $u + v$ and $u \xi + \frac{v}{\xi}$, one can get all the other roots, since $\xi$ is a primitive root of $1$ of order $5$.

It seems we could do a similar thing by considering for $p$ prime the equation obtained by eliminating $t$ from $t^p-1$, $x= u t + v/t$ .

$\bf{Added:}$ Assume that the roots of an equation of degree $n$ are labelled by $\mathbb{Z}/n$, and the Galois group $G$ is contained in the group of affine transformations of $\mathbb{Z}/n$, $t \mapsto \alpha t + \beta$. Then every transformation in $G$ that fixes two roots ( that is, two elements of $\mathbb{Z}/n$) is identity, that is, fixes all the roots. It follows that: every other root is contained in the field generated by the two roots.

This is a theorem of Galois. It has to do with Galois group of an irreducible polynomial of degree $p$. Since the group $G$ acts transitively on the $p$ roots, it has order divisible by $p$. No the theorem ( of Cauchy) says that $G$ has an element of order $p$, who most be a cycle of length $p$ ( again use $p$ prime). Now we have the affine structure on the indexing set for the roots. It we moreover assume that $G$ is solvable, then $G$ must be contained in the groups of affine transforms of $\mathbb{Z}/p$ ( this seems to be a theorem in group theory).

So it seems yes, whenever we have the Galois group $F_{20}$ ( or even containd in $F_{20}$ -- this is equivalent to "solvable") then every two roots generate all the others rationally. Of course, one would like to see more details, and examples.