Brumer quintic polynomials - is there a general formula for the roots?

Question

There exist a family of quintic polynomials, called Brumer's polynomials (or Kondo-Brumer), which have the form:

$$x^5+(a-3)x^4+(-a+b+3)x^3+(a^2-a-1-2b)x^2+bx+a,~~~a,b \in \mathbb{Q}$$

According to Wikipedia these polynomials are solvable in radicals.

Is there a general formula for roots of these polynomials? Or at least the closed form for some special cases?

I searched the web, but only found papers discussing the group properties (for example here) or other properties of Brumer's polynomials. Nothing about the roots.

Edit

I'm starting a bounty, and I would like either of these things:

• General solution (at least one root), depending on $a,b$ - only if such a solution exists, and is short enough to write here in closed form.

• Some methods for obtaining this solution - again, if it will lead to a form of the solution more compact and simpler than the general way to solve an arbitrary solvable quintic.

• Solutions to some special cases (for some values of $a,b$ with $b \neq 0$)

• A proof that no such simple solution is possible for this family of quintics.

Answer 1

Given the solvable Kondo-Brumer quintic,

$$x^5 + (a - 3)x^4 + (-a + b + 3)x^3 + (a^2 - a - 1 - 2b)x^2 + b x + a = 0\tag1$$

One in fact can make explicit formulas for these. For simplicity, assume $a=1$, so

$$x^5 - 2x^4 + (b + 2)x^3 + (-1 - 2b)x^2 + b x + 1=0\tag2$$

Define the four roots $z_i$ of its quartic Lagrange resolvent,

$$z^2 + \tfrac{1}{2}\big((597 + 225 b + 200 b^2) \color{blue}+ 5\sqrt{5} (43 + 33 b + 20 b^2)\big)z +{c_1}^5=0\tag3$$

$$z^2 + \tfrac{1}{2}\big((597 + 225 b + 200 b^2) \color{blue}- 5\sqrt{5} (43 + 33 b + 20 b^2)\big)z +{c_2}^5=0\tag4$$

which was factored into two quadratics for convenience, and,

$$c_1=-\tfrac{1}{2}\big((2+5b)\color{blue}-\sqrt{5}(4-b)\big)\tag5$$ $$c_2=-\tfrac{1}{2}\big((2+5b)\color{blue}+\sqrt{5}(4-b)\big)\tag6$$

Note the constant term of the quartic is a nice fifth power,

$$(c_1 c_2)^5=(5b^2+15b-19)^5$$

We can then give the relatively "simple" solution,

$$x = \frac{1}{5}\Big(2+z_1^{1/5}+z_2^{1/5}+z_3^{1/5}+z_4^{1/5}\Big)\tag7$$

Example: Let $b=-1$, then,

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

a quintic which was also solved by Ramanujan. Its resolvent using $(3),(4)$ is,

$$z^4 + 572z^3 + 70444z^2 + 1600203z - 29^5=0$$

then,

$$x= \frac{1}{5}\Big(2+z_1^{1/5}+z_2^{1/5}+z_3^{1/5}+z_4^{1/5}\Big) = -0.90879\dots$$

If all the $z_i$ are real, as in the example, then it is a simple matter of taking fifth roots of real numbers which will then yield a real root of the quintic.

Answer 2

The roots of any (irreducible) solvable quintic can be found, using methods due to George Paxton Young in 1888. An explicit 3-page formula based on those methods was given by Daniel Lazard in 2004. Source: https://en.wikipedia.org/wiki/Quintic_function