From F. Klein's books, It seems that one can find the roots of a quintic equation


(where $a,b,c,d,e\in\Bbb C$) by elliptic functions. How to get that?

Update: How to transform a general higher degree five or higher equation to normal form?

ziang chen
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    First you have to reduce it to Bring-Jerrard form. This is the one solvable in elliptic functions. Give me time to get my notes. (The one in [Mathworld](http://mathworld.wolfram.com/QuinticEquation.html) is missing some details.) – Tito Piezas III Oct 27 '13 at 03:08
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    If you don't know how to transform it to Bring-Jerrard form, ask it in the forum and I'll answer. – Tito Piezas III Oct 27 '13 at 15:49
  • @TitoPiezasIII Thanks! Can you add another answer here, or edit your solution? I don't think it is necessary to ask one more question – ziang chen Oct 27 '13 at 20:58
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    The combined answer will be too long. Besides, how to transform the general quintic to Bring-Jerrad form will be a good question, and will turn up in a google search for those looking. – Tito Piezas III Oct 27 '13 at 21:15
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    @TitoPiezasIII Thanks! see http://math.stackexchange.com/questions/542108/how-to-transform-a-general-higher-degree-five-or-higher-equation-to-normal-form – ziang chen Oct 27 '13 at 22:39
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    You may be interested in [this post](http://math.stackexchange.com/questions/1425349/) on how to solve the _Brioschi quintic_. – Tito Piezas III Sep 07 '15 at 15:08
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    The method is explained [there](http://www.diva-portal.org/smash/get/diva2:926660/FULLTEXT01.pdf) p.36, it is based on inverting the doubly periodic function $f_C(z) = \wp_j'(z)\wp_j(z)+i \sqrt{c}\wp_j(z)+2i\sqrt{c}$ (with $j=e^{2i\pi /3}$) to find its zeros $z_k$ and solve $x^5+cx+x=0$ as $x=\wp_j(z_k)$ – reuns Oct 29 '19 at 08:59

1 Answers1


To solve the general quintic using elliptic functions, one way is to reduce it to Bring-Jerrard form,

$$x^5-x+ d = 0\tag{1}$$

a transformation which can be done in radicals. (See this post.) To solve $(1)$, define,

$$k = \tan\left(\tfrac{1}{4}\arcsin\Big(\frac{16}{25\sqrt{5}\,d^2}\Big)\right)\tag{2}$$

$$p = i\frac{K(k')}{K(k)} = i\,\frac{\text{EllipticK[1-m]}}{\text{EllipticK[m]}}\tag3$$

with the complete elliptic integral of the first kind $K(k)$ and elliptic parameter $m=k^2$ (with $p$ also given in Mathematica syntax above).

Method 1: For $j=0,1,2,3,4$, let,

$$S_j={u}^j \frac{\sqrt{2}\,\eta(\tau_j)\,\eta^2(4\,\tau_j)}{\eta^3(2\,\tau_j)}$$

$$\tau_j = \tfrac{1}{10}(p+2j)$$

$$u=\zeta_8 = \exp(2 \pi i/8)$$


where $\eta(\tau)$ is the Dedekind eta function, then,

$$x = \frac{\pm 1}{2\cdot 5^{3/4}}\frac{(k^2)^{1/8}}{\sqrt{k(1-k^2)}}(S_0+S_5)(S_1+i\,S_4)(i\,S_2+S_3) \tag{4}$$

with the sign chosen appropriately.

$\color{blue}{\text{Remark}}$: The five roots $x_n$can be found by using $p_n = i\frac{K(k')}{K(k)}+16n$ for $n = 0,1,2,3,4$.

Method 2: For $j=0,1,2,3,4$, let,

$$T_j =\left(\frac{\vartheta_2(0,w^j q^{1/5})}{\vartheta_3(0,w^j q^{1/5})}\right)^{1/2}$$

$$q = \exp(i \pi p_0)$$

$$w = \zeta_5 = \exp(2 \pi i/5)$$

$$T_5 =\frac{q^{5/8}}{(q^5)^{1/8}}\left(\frac{\vartheta_2(0,q^{5})}{\vartheta_3(0, q^{5})}\right)^{1/2}$$

with $\vartheta_n(0,q)$ as the Jacobi theta functions, then,

$$x = \frac{\pm{\zeta_8}}{2\cdot 5^{3/4}}\frac{(k^2)^{1/8}}{\sqrt{k(1-k^2)}}(T_0+T_5)(T_1-i\,T_4)(T_2+T_3) \tag{5}$$

and the sign of $\zeta_8$ chosen appropriately. Note that $(S_j)^8 = (T_j)^8$.

$\color{blue}{\text{Remark}}$: One can also find the other roots $x_i$, but is not as simple as in Method 1. (As one can see, you need much more than radicals to solve the general quintic.)

Example. Let,

$$x^5-x+1 = 0\tag6$$

so $d = \pm1$. Plugging it into $(2)$ gives $k \approx 0.072696$, and $m=k^2$, so $p \approx 2.550572\,i$. Since both methods use a square root, using the formulas one eventually finds,

$$x = \mp\,1.1673039\dots$$

$\color{blue}{\text{Afterword}}$ (Added June 2015): Given the nome $q = \exp(i \pi \tau)$, the two methods involve the function known either as the modular lambda function or elliptic lambda function, $\lambda(\tau)$ which has a beautiful q-continued fraction,

$$\big(\lambda(\tau)\big)^{1/8} = \frac{\sqrt{2}\,\eta(\tfrac{\tau}{2})\,\eta^2(2\tau)}{\eta^3(\tau)} = \left(\frac{\vartheta_2(0,q)}{\vartheta_3(0,q)}\right)^{1/2} = \cfrac{\sqrt{2}\,q^{1/8}}{1+\cfrac{q}{1+q+\cfrac{q^2}{1+q^2+\cfrac{q^3}{1+q^3+\ddots}}}}$$

studied by Ramanujan (who also had his own method to solve solvable quintics).

Tito Piezas III
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  • This gives just one of the five complex roots, right? It's clear that at most three are real, in which case, is this the largest, smallest, or middle one? – Ryan Reich Dec 23 '13 at 07:16
  • @Ryan: As you implied, the Bring-Jerrard quintic has $x_1^2+x_2^2+\dots+x_5^2=0$, hence not all its roots are real. I do not know if Hermite's method always gives the largest root (in absolute value), but you can tweak it to produce ALL five. I revised my answer to include an alternative method using the _Dedekind eta function_ which easily yields all the roots. – Tito Piezas III Dec 23 '13 at 22:51
  • @Tito Wow this is beautiful. Besides the fact that this brings together (apparently) distant fields of math, is there a reason why such solution is interesting besides what one can obtain with, I don't know, Newton's method? – lcv Nov 26 '15 at 21:42
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    @lcv: Thanks. I believe the primary utility of such results is to show connections (and in my opinion, _beautiful_ connections) between different mathematical objects. They are certainly not useful for numerical evaluations better than Newton's method. :) – Tito Piezas III Nov 27 '15 at 07:42
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    @TitoPiezasIII thanks for confirming my impression :) PS I agree: *beautiful* connections – lcv Nov 27 '15 at 08:13
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    Some notes for anyone trying this method, based on my own mistakes. (1) Mathematica and Wolfram Alpha expect the argument of the elliptic integral function to be $k^2$, not $k$. (2) In the formula for $S_j$, the power of $u$ is *not* doubled but the term in $\tau$ *is*. (3) Not accounting for their signs, the set of real and imaginary parts of $S_j$, for a given $n$, are simply permuted when $n$ is changed so it is only necessary to calculate them accurately once. – Jam Dec 19 '19 at 18:10
  • Can you link ramanujans own methods? – Alexander Conrad Apr 29 '21 at 14:51
  • @TitoPiezasIII do you have any references or explanations as to how to derive the methods 1 and 2 for solving the fifth degree polynomial? I've been trying to read Felix Klein's Lectures on the ikosahedron (which I'm guessing these methods may come from), but I still don't have a clear picture of how the icosahedron shows up in these methods. – fsuna064 Oct 20 '21 at 02:15