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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.

Yuriy S
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3 Answers3

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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.

Tito Piezas III
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  • I've seen the first case in some old mathforum discussion. I think you were there as well. So, no such simple solution for the general $a$? I've also seen your paper on sextic resolvents and I have a question - how do we guess the roots, such that their squares are rational? – Yuriy S Jul 01 '16 at 07:58
  • @YuriyS: Do you really want the formula for general $a$, or would you rather know a simple method to solve the solvable quintic? That way, you'll be able to deal with any new quintics yourself. If so, ask a question on "how to solve the solvable quintic using its quartic lagrange resolvents" and I'll gladly answer it. – Tito Piezas III Jul 07 '16 at 20:27
  • Tito, no thank you. 'Simple' is a relative word. I'm collecting parametric solvable quintics with relatively short solutions, which can be easily written in closed form. Think De Moivre quintic. I've found several, but the search continues. – Yuriy S Jul 07 '16 at 21:01
  • @YuriyS: Ok, I have a collection of my own. Another famous one is the Emma Lehmer quintic. Do you have it already? – Tito Piezas III Jul 07 '16 at 21:30
  • Ttio, no this is the first time I've heard of it. I've found several papers now, including Spearman Williams paper. It relates four roots to one root. But I don't see a formula for the one root. And another case here Thank you for mentioning it. – Yuriy S Jul 07 '16 at 21:49
  • @YuriyS: I've added another answer re the general family to which the Lehmer quintic belongs to. – Tito Piezas III Jul 07 '16 at 21:59
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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

Justin Benfield
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    I'm aware. I meant an easier formula for this special case, like for De Moivre quintic – Yuriy S Jun 28 '16 at 12:49
  • @YuriyS: I imagine it's possible to deduce a somewhat simpler formula for the Brumer quintic from the general formula for solvable quintics, but it's likely still very complicated, and the aforementioned derivation is presumably a very nontrivial application of Galois Theory. – Justin Benfield Jun 29 '16 at 01:13
  • @JustinBenfield: The Brumer quintic has two free parameters. If, for convenience, we suppress one parameter, then yes it is possible to find a rather simple formula. Kindly see other answer for this post. – Tito Piezas III Jul 01 '16 at 05:03
  • @TitoPiezasIII Since the formula is not homogeneous, suppressing one parameter is not just a "convenience", it is a loss of generality (unless I missed something) – Ewan Delanoy Jul 05 '16 at 08:10
  • @EwanDelanoy: Yes, it is a loss of generality. However, I should have added that assuming $a=1$ (or any value) is an illustrative example, as the formulas for $a=0,1,2,3,\dots$ are very similar in form. I could have used general $a$, but the poster did mention particular cases would suffice. – Tito Piezas III Jul 07 '16 at 20:20
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(Addendum to my answer.) Since the primary objective of the OP is to find solvable quintics with a "simple" solution, then we can add the "depressed" multi-parameter family,

$y^5+10cy^3+10dy^2+5ey+f = 0\tag{1}$

where the coefficients obey the quadratic in $f$,

$(c^3 + d^2 - c e) \big((5 c^2 - e)^2 + 16 c d^2\big) = (c^2 d + d e - c f)^2 \tag{2}$

Solve for $f$. Define this quintic's Lagrange resolvent as,

$$(z^2+u_1z-c^{5})(z^2+u_2z-c^{5}) = 0$$

where the $u_i$ are the two roots of the quadratic,

$$u^2-fu+(4c^5-5c^3e-4d^2e+ce^2+2cdf) = 0$$

then the solution to $(1)$ is,

$y = z_1^{1/5}+z_2^{1/5}+z_3^{1/5}+z_4^{1/5}\tag{3}$

Note that appropriate choices of the $3$ free parameters $c,d,e$ can yield rational $f$.

Example: A particular case is the Lehmer quintic,

$$x^5 + n^2x^4 - (2n^3 + 6n^2 + 10n + 10)x^3 + (n^4 + 5n^3 + 11n^2 + 15n + 5)x^2 + (n^3 + 4n^2 + 10n + 10)x + 1=0$$

Let $x = (y-n^2)/5$ to transform it to depressed form $(1)$. Its transformed coefficients then obey $(2)$.

Tito Piezas III
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