The most generic radicals-solvable quintic

Question

It's well known that it is impossible to solve a generic quintic equation in terms of radicals involving its coefficients. However: what's the "most generic" quintic equation that is still possible to solve using radicals?

If we think of a quintic as $ax^5+bx^4+cx^3+dx^2+ex+f = 0$ then "the most generic" here would mean such an equation that describes the largest solvable subset of quintics. There certainly exist some subsets of quintics that are solvable, a good example is $ax^5+b=0$ , but I want to know what's the most generic look of a solvable quintic is.

A fascinating book: http://people.reed.edu/~jerry/Quintic/quintic.pdf — Jean Marie, Feb 12 '22 at 13:43
I'm now curious if the set of solvable quintics is connected (viewed as a subset of R^6), simply connected, etc. — Beren Gunsolus, Feb 12 '22 at 13:56
Have you seen this question and its answers https://math.stackexchange.com/q/1019449/305862 ? — Jean Marie, Feb 15 '22 at 08:42
Would something like this post help? Also this paper if a polynomial can be depressed into a trinomial. — Тyma Gaidash, Feb 19 '22 at 15:56
@BerenGunsolus This question doesn't make sense: One can only ask whether a quintic is solvable when it has rational coefficients. — David E Speyer, Feb 23 '22 at 14:53
Suppose I mean for it to be solvable with radicals, allowing use of the coefficients. In this sense x^5=e is "solvable" — Beren Gunsolus, Feb 23 '22 at 23:16
The answer is if one of the roots is made from radicals, then so must the other $4$ since factoring the general quintic using a radical root will give a general quartic which always has radical roots. Right? — Тyma Gaidash, Feb 24 '22 at 16:55
@JeanMarie There is broad family of solvable quintics connected to Pythagorean triples. Kindly see answer below. — Tito Piezas III, Jan 12 '25 at 18:35
@DavidESpeyer This family of solvable quintics and their relation to Pythagorean triples might be of interest. Kindly see answer. — Tito Piezas III, Jan 12 '25 at 18:36

Tito Piezas III · Answer 1 · 2025-01-13T05:26:31.357

Given the quintic,

$$y^5+Ay^4+By^3+Cy^2+Dy+E=0\tag1$$

there is an extremely broad solvable class with four free parameters $(A,B,C,D)$ and $E$ is only a quadratic with respect to them. As such, we can create a nice relationship between solvable quintics and Pythagorean triples.

I. Watson's resolvent (1930s)

For purposes of simplicity, by using a simple linear substitution, we can transform $(1)$ into the depressed form (missing the $x^4$ term),

$$x^5 + 10c x^3 + 10d x^2 + 5 e x + f = 0$$

If the coefficients obey,

$$W:=-25c^6 - 40c^3d^2 - 16d^4 + 35c^4e + 28c d^2e - 11c^2e^2 + e^3 - 2c^2d f - 2d e f + c f^2 = 0$$

then the quintic is solvable in radicals. The condition is simply the constant term $W$ of Watson's sextic resolvent where we set to $W=0$. Those interested in the details can refer to this 2016 post.

II. Pythagorean triples

Or for a more aesthetic form, if the coefficients of the quintic obey the Pythagorean triple,

$$(c^3 - 25 c^4 - 16 c d^2 - c e + 10 c^2 e + d^2 - e^2)^2 + (2c^2 d - 2c f + 2d e)^2 = (c^3 + 25 c^4 + 16 c d^2 - c e - 10 c^2 e + d^2 + e^2)^2$$

then the quintic is solvable in radicals. However the price to pay for this second form is the condition $c\neq0$ since, expanded out, the equation factors as $c\times W=0$. In fact, one can see that if $c=0$, then the above reduces to the well-known identity,

$$(d^2-e^2)^2+(2de)^2 =(d^2+e^2)^2$$

So the primary condition is $W=0$ (valid for any $c$), and the Pythagorean form is only when $c\neq0$.

III. Example

Given $(c,d,e)=(1,2,-3)$ then $W$ factors as $(f-28)(f+36)=0$, so another nice feature of these quintics is they come in pairs,

$$x^5 + 10c x^3 + 10d x^2 + 5 e x + f =0\\ \\ \,x^5 + 10 x^3 + 20 x^2 - 15x +28 = 0\\ \,x^5 + 10 x^3 + 20 x^2 - 15x - 36 = 0$$

Using the formulas above, their $(c,d,e,f)$ yield the triple,

$$120^2+64^2 = 136^2$$

IV. Magma command

For those who wish to test the quintics, there is an online Magma calculator and the command is,

R < x > := PolynomialRing (RationalField ());
f := -36 - 15 * x + 20 * x^2 + 10 * x^3 + x^5;
G := GaloisGroup (f);
G;
IsSolvable (G);

Just copy and paste, press the submit button, and it will evaluate the command. (Don't forget the asterisks between the coefficient and the variable.)

@Somos Thanks for the edit, Somos. Looks much nicer now. By the way, given A360187 (or a Generalized Somos-5 sequence), then its even entries $p = 3,113,57123,\dots,$ appear in a recent post of mine involving $p^4-2q^4=r^2$ though $q = 2,84,6214,\dots,$ doesn't have an OEIS entry yet. Would $q$ be also a bisection of some Somos-5 sequence? — Tito Piezas III, Jan 13 '25 at 01:27

The most generic radicals-solvable quintic

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