How to solve a quintic polynomial equation?

Question

I know that not all quintics are solvable. But how do I identify the class of solvable ones?

Answer 1

If you haven't reached that part of Galois theory yet, there is still a "simple" elementary way to test if a quintic is solvable. First, reduce it to depressed form (without the $x^4$ term).

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I. Method 1:

Theorem (by Watson, 1930s): "Given an irreducible quintic with rational coefficients,

$$x^5 + 10c x^3 + 10d x^2 + 5 e x + f = 0\tag1$$

If the sextic,

$$3125p^6 - 625(3c^2 + e)p^4 + 25(15c^4 + 8c d^2 - 2c^2e + 3e^2 - 2d f)p^2 + p\sqrt{D} + (-25c^6 - 40c^3d^2 - 16d^4 + 35c^4e + 28c d^2e - 11c^2e^2 + e^3 - 2c^2d f - 2d e f + c f^2)=0$$

with discriminant $D$,

$$D=-3200c^3d^2e^2 - 2160d^4e^2 + 6400c^4e^3 + 5760c d^2e^3 - 2560c^2e^4 + 256e^5 + 5120c^3d^3f + 3456d^5f - 11520c^4d e f - 10080c d^3e f + 4480c^2d e^2f - 640d e ^3f + 3456c^5f^2 + 2640c^2d^2f^2 - 1440c^3e f^2 + 360d^2e f^2 + 160c e^2f^2 - 120c d f^3 + f^4$$

has a root $p$ such that $p^2$ is rational, then $(1)$ is a solvable quintic."

See "Commentary on an unpublished lecture by G. N. Watson on solving the quintic" by Bruce Berndt. This is easily implemented in Mathematica and is useful when dealing with parametric quintics.

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(Edit: July 17, 2023.) As asked by user dxiv, Watson's method in Berndt's paper has a stumbling block since it has division by $p$ (the sextic root). So what happens when $p=0$? I found the solution was to bring Watson's method to its logical conclusion and construct the quartic resolvent, then check if there were any cancellation of divisors. Fortunately there was.

Thus, given the quintic,

$$x^5 + 10c x^3 + 10d x^2 + 5 e x + f = 0\tag1$$

if the coefficients obey,

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

then the constant term of Watson's sextic resolvent is zero, hence its root $p=0$. (Note: Since the condition is just a quadratic in $f$, it is quite easy to find solvable quintics by finding $c,d,e$ such that $f$ is rational.)

More aesthetically, if $c\neq 0$, the quintic's coefficients form a "square",

$$\big(c^3 + d^2 - c e\big) \big(25 c^4 - 10 c^2 e + e^2 + 16 c d^2\big) = \big(c^2 d + d e - c f\big)^2\tag2$$

The solution of $(1)$ is then,

$$x = z_1^{1/5}+z_2^{1/5}+z_3^{1/5}+z_4^{1/5}$$

where the $z_i$ are the four roots of the two quadratics in $z$,

$$z^2+\frac12\left(f\pm\sqrt{-16 c^5 + 20 c^3 e + 16 d^2 e - 4 c e^2 - 8 c d f + f^2}\right)z-c^5=0$$

which are purely in the coefficients of the quintic.

Example. Let,

$$x^5 + 10x^3 + 110x^2 - 15x + 478 = 0$$

which obeys $(2)$ so Watson's sextic has root $p=0$. Then the two quadratics are,

$$z^2+\left(239+95\sqrt5\right)z-1=0$$ $$z^2+\left(239-95\sqrt5\right)z-1=0$$

The real root of the quintic is then,

$$\qquad x = z_1^{1/5}+z_2^{1/5}+z_3^{1/5}+z_4^{1/5} = -4.509908339\dots$$

P.S. The case $p=0$ in fact is special. More on that in this post.

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II. Method 2:

If you are in a rush and just want to determine if a particular equation is solvable, you can find the order of its Galois group using this online Magma calculator. For example, to test the solvable but irreducible $x^5-5x+12=0$, copy and paste the command,

Z := Integers(); P < x > := PolynomialRing(Z); f := x^5-5*x+12; G, R := GaloisGroup(f); G;

One then finds the order is $10$, hence that quintic is solvable. All groups with order $<60$ are, though there are solvable groups with order $>60$.

Note: Don't forget the asterisk between the numerical coefficient and the variable, like this: 5*x.

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III. Addendum on Method 1 (by Oscar Lanzi):

We can reduce Method 1 to a simple rational root test by "root squaring".

Render the sextic equation above as

$\alpha p^6 + \beta p^4 + \gamma p^2 + p\sqrt{D} + \delta = 0$

Separate the odd degree term from the even degree ones:

$\alpha p^6 + \beta p^4 + \gamma p^2 + \delta = - p\sqrt{D}$

And square. Only even powers of $p$ now appear so we have a polynomial equation in $p^2$::

$(\alpha p^6 + \beta p^4 + \gamma p^2 + \delta)^2=p^2D$

$\alpha^2(p^2)^6 + 2\alpha \beta(p^2)^5 + (2\alpha \gamma + \beta^2)(p^2)^4 + 2(\alpha \delta + \beta \gamma)(p^2)^3 + (2\beta \delta + \gamma^2)(p^2)^2 + (2 \gamma \delta - D)(p^2) + \delta^2 = 0$

You may then apply the standard rational root test to this equation. If it passes, the conditions for solvability of the quintic are satisfied.