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.

Answer 1

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,

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

We can adopt a modified version of the Pythagorean condition when $c=0$, if $d$ and $e$ are both nonzero. If the Pythagorean condition has a double root at $c=0$, then $W=0$ is simultaneously satisfied. This can be checked by taking the derivative of the Pythagorean condition at $c=0$, which leads to

$$(e+16d^2)(d^2-e^2)^2+(2f)(2de)^2=(e-16d^2)(d^2+e^2)^2$$

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

---

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.)

Answer 2

Let $r_1$ through $r_5$ be any five numbers expressible with radicals (possibly including duplicates), and let $a \ne 0$. Define:

$$f(x) = A(x - r_1)(x - r_2)(x - r_3)(x - r_4)(x - r_5)$$

Equivalently:

$$f(x) = Ax^5 + Bx^4 + Cx^3 + Dx^2 + Ex + F$$

where: $$B = -A(r_1 + r_2 + r_3 + r_4 + r_5)$$ $$C = A(r_1r_2 + r_1r_3 + r_1r_4 + r_1r_5 + r_2r_3 + r_2r_4 + r_2r_5 + r_3r_4 + r_3r_5 + r_4r_5)$$ $$D = -A(r_1r_2r_3 + r_1r_2r_4 + r_1r_2r_5 + r_1r_3r_4 + r_1r_3r_5 + r_1r_4r_5 + r_2r_3r_4 + r_2r_3r_5 + r_2r_4r_5 + r_3r_4r_5)$$ $$E = A(r_1r_2r_3r_4 + r_1r_2r_3r_5 + r_1r_2r_4r_5 + r_1r_3r_4r_5 + r_2r_3r_4r_5)$$ $$F = -Ar_1r_2r_3r_4r_5$$

All solvable quintics can be written in this form. Finding $r_i$ given $(A, B, C, D, E, F)$ may be difficult, though.

---

For something that's a bit less of a “mathematician's answer”, one broad special case of solvable quintics is those that are the product of a quadratic and a cubic, which both have closed-form radical solutions.

$$f(x) = (ax^2 + bx + c)(x^3 + dx^2 + ex + f)$$ $$f(x) = ax^5 + (ad + b)x^4 + (ae + bd + c)x^3 + (af + be + cd)x^2 + (bf + ce)x + cf$$

Answer 3

I don't know the answer, but I have a good idea where to look.

If $f$ is an irreducible quintic, then its Galois group is a subgroup of $S_5$, and has order a multiple of five. Among these subgroups, the largest solvable one is of order $20$, and the other ones, of orders $10$ and $5$, are each contained in a subgroup of order $20$. So I'd say the closest thing to a generic solvable irreducible quintic is one with Galois group of order $20$.

Now in principle one can find conditions on the coefficients which are necessary and sufficient for $f$ to have that Galois group of order $20$, but I don't know how to do it. So, I can't tell you how to finish off this answer, but I've told you where I think the answer is.

Answer 4

Just a trick concerning the pythagorean triples @TitoPiezasIII

$$(yx(x+z))^2+z^2-(2x+z)^2=x(x+z)(x^2y^2+xy^2z-4)$$

It's also available for LFT .