11

According to Wikipedia there are only five solvable quintic equations of the form $x^5+ax^2+b=0,~~a,b \in \mathbb{Q}$ (up to a scaling constant $s$).

$$x^5-2s^3x^2-\frac{s^5}{5}=0 $$ $$ x^5-100s^3x^2-1000s^5=0 $$ $$x^5-5s^3x^2-3s^5=0 $$ $$x^5-5s^3x^2+15s^5=0 $$ $$ x^5-25s^3x^2-300s^5=0 $$

But the source of this claim is a web-page (even if it's Harvard), not an article.

Thus, my questions are:

Why is this true, and what is the original source of this knowledge?

What are the solutions to these equations (in radical form, or possibly trigonometric/hyperbolic form)?

Edit

I found the proper citation on the linked Web-Page (for which I sincerely thank the author, if they ever visit this post).

The paper is On Solvable Quintics $X^5+aX+b$ and $X^5+aX^2+b$ by Blair K. Spearman and Kenneth S. Williams and the full text is available with open-access. I will see if my questions are answered by this paper and update the post.

Community
  • 1
Yuriy S
  • 31,474
  • 2
    This is an occasionally used visualization of solvable quintics. I don't know how to transform those equations into the form used in your source. – Jyrki Lahtonen Jun 27 '16 at 07:53
  • Some of these quintics can be expressed by trigonometric functions with arguments involving the golden ratio. See second answer below. – Tito Piezas III Jul 10 '16 at 18:07

2 Answers2

2

Like all solvable quintics, these can be expressed in terms of trigonometric and inverse trigonometric functions, and two of them have nice representations using the golden ratio,

$$\phi = \tfrac{1+\sqrt{5}}{2}$$

The unique real roots are given by,

I. $\quad x^5-5x^2-3=0 $

$$x=\frac{-2}{5^{1/4}}\left(\cos\Big(\tfrac{1}{5}\arccos\big(\frac{5^{3/4}}{2\,\phi}\big)\Big)-i\,\sin\Big(\tfrac{1}{5}\arcsin\big(\frac{5^{3/4}\phi\;i}{2}\big)\Big)\right)=-1.8087\dots$$

III. $\quad x^5-25x^2-300=0 $

$$x=2\cdot 5^{1/4}\left(\cos\Big(\tfrac{1}{5}\arccos\big(5^{1/4}\phi\big)\Big)-i\,\sin\Big(\tfrac{1}{5}\arcsin\big(\frac{5^{1/4}\;i}{\phi}\big)\Big)\right)=3.6287\dots $$

Tito Piezas III
  • 54,565
  • Not all solvable quintic can be expressed with trigonometric functions and their inverses. Only if the roots are all real is such a representation obtained. If instead there is only one real riot it would be expressed with hyperbolic functions and their inverses instead, or directly with radicals. – Oscar Lanzi Oct 12 '20 at 02:28
  • @OscarLanzi I just noticed this comment now. Actually even if a solvable quintic has one real root, you can bypass hypergeometric functions and express the quintic as a sum of TWO trigonometric functions (and their inverses) as shown by the examples above (which are one-real root quintics.) I'll try to give a more general formula later. – Tito Piezas III Jun 28 '23 at 20:19
  • Those examples are not conveniently rendered in terms of trig functions. To get answers in real variables you have to use hyperbolic functions because the arguments of the inverse hyperbolic cosines are greater than $1$ and the arguments of the inverse sines are imaginary. Using the hyperbolic/exponential relations the hyperbolic function combinations can then be converted to algebraic radicals. – Oscar Lanzi Jun 28 '23 at 23:01
  • @OscarLanzi Since we have the identity, $$\sinh\big(\tfrac15\operatorname{arcsinh}(x)\big)=-i \sin\big(\tfrac15\arcsin(x \sqrt{-1})\big)$$ thus one side is equally valid as the other. You prefer the LHS, but I used the RHS instead. To discuss which side of an equality is "convenient" or "better" is a matter of taste, and frankly a waste of time. – Tito Piezas III Jun 29 '23 at 09:09
1

According to the linked paper On Solvable Quintics $X^5+aX+b$ and $X^5+aX^2+b$ by Blair K. Spearman and Kenneth S. Williams, the solutions are as follows (we take $s=5$ in the first equation and $s=1$ for the rest):

$$x=\omega^{j}u_1+\omega^{2j}u_2+\omega^{3j}u_3+\omega^{4j}u_4$$

$\omega$ - the fifth root of unity.

$$x^5-5x^2-3=0 $$

$$u_1=\left(-\frac{1}{4}+\frac{\sqrt{5}}{20}-\frac{1}{100} \sqrt{150+30\sqrt{5}}+\frac{1}{50} \sqrt{150-30\sqrt{5}} \right)^{1/5}$$

$$u_2=\left(-\frac{1}{4}-\frac{\sqrt{5}}{20}-\frac{1}{100} \sqrt{150+30\sqrt{5}}-\frac{1}{50} \sqrt{150-30\sqrt{5}}\right)^{1/5}$$

$$u_3=\left(-\frac{1}{4}-\frac{\sqrt{5}}{20}+\frac{1}{100} \sqrt{150+30\sqrt{5}}+\frac{1}{50} \sqrt{150-30\sqrt{5}}\right)^{1/5}$$

$$u_4=\left(-\frac{1}{4}+\frac{\sqrt{5}}{20}+\frac{1}{100} \sqrt{150+30\sqrt{5}}-\frac{1}{50} \sqrt{150-30\sqrt{5}}\right)^{1/5}$$

$$x^5-5x^2+15=0 $$

$$u_1=\left(\frac{5}{4} + \frac{13\sqrt{5}}{20} - \frac{7}{100} \sqrt{750+330\sqrt{5}} \right)^{1/5}$$

$$u_2=\left(\frac{5}{4} - \frac{13\sqrt{5}}{20} - \frac{7}{100} \sqrt{750-330\sqrt{5}} \right)^{1/5}$$

$$u_3=\left(\frac{5}{4} - \frac{13\sqrt{5}}{20} + \frac{7}{100} \sqrt{750-330\sqrt{5}} \right)^{1/5}$$

$$u_4=\left(\frac{5}{4} + \frac{13\sqrt{5}}{20} + \frac{7}{100} \sqrt{750+330\sqrt{5}} \right)^{1/5}$$

$$ x^5-25x^2-300=0 $$

$$u_1=\left(-\frac{25}{2} - \frac{5\sqrt{5}}{2} - \frac{5}{2} \sqrt{30+6\sqrt{5}} \right)^{1/5}$$

$$u_2=\left(-\frac{25}{2} + \frac{5\sqrt{5}}{2} - \frac{5}{2} \sqrt{30-6\sqrt{5}} \right)^{1/5}$$

$$u_3=\left(-\frac{25}{2} + \frac{5\sqrt{5}}{2} + \frac{5}{2} \sqrt{30-6\sqrt{5}} \right)^{1/5}$$

$$u_4=\left(-\frac{25}{2} - \frac{5\sqrt{5}}{2} + \frac{5}{2} \sqrt{30+6\sqrt{5}} \right)^{1/5}$$

$$ x^5-100x^2-1000=0 $$

$$u_1=-2^{6/5},~~~~u_2=-2^{7/5},~~~~u_3=2^{3/5},~~~~u_4=-2^{4/5}$$

$$x^5-250x^2-625=0 $$

$$u_1=(-125+50 \sqrt{5})^{1/5}$$

$$u_2=\frac{(-375-175 \sqrt{5})^{1/5}}{2^{1/5}}$$

$$u_3=\frac{(-375+175 \sqrt{5})^{1/5}}{2^{1/5}}$$

$$u_4=(-125-50 \sqrt{5})^{1/5}$$

Yuriy S
  • 31,474
  • How should I proceed in order to get the roots of the polynomial? Substitute j as 0,1,2,3,4 and the respective $u's$ but what root of unity $w$ should be? – Mr. N Dec 19 '19 at 13:44
  • @Mr.N, I'm not sure. I think it would be better if you look up the linked paper, maybe it contains more information – Yuriy S Dec 19 '19 at 15:02