#### mathbalarka

Math Team
Hello members of MMF,

One of our senior members of MMF, agentredlum, requested me to create such a topic but withdrawn his request soon. This subject has stayed very foggy to majority of peoples like us and no one ever really shown the effort to describe this subject elegantly enough. I will give it a try to do this.

Quintics are a mysterious subject. Unlike the other degrees < 5, quintics cannot be solved using elementary arithmetic operations and radicals. There are two proofs of this, one is by Abel and another is by Galois. Abel's theory is easier but it cannot specify which quintics are solvable and which are not, but Galois theory can be used to do the work. I will first describe the transformation of quintics to reduce it into several forms, then we will start the several methods to solve a quintic.

NOTE : Any user is allowed to reply in this topic and ask for clarification. For all other comments, I ask that you post them in the auxiliary topic created just for this purpose. Thank you!

• 2 people

#### mathbalarka

Math Team
Here, I will describe the transformation of a general quintic into a depressed quintic :

$$\displaystyle x^5 + a_1 x^4 + a_2 x^3 + a_3 x^2 + a_4 x + a_5 = 0 \;\;\; (1)$$

In our work throughout this topic, I will use the quintic above as your general quintic. We want to transform the quintic above into the depressed form :

$$\displaystyle y^5 + b_2 y^3 + b_3 y^2 + b_4 y + b_5 = 0 \;\;\; (2)$$

The first thing is to do is to write a linear relation between them :

$$\displaystyle y = x + k \;\;\; (3)$$

Finding k by substituting (3) into (1) is rather tedious, so we'll avoid the arithmetic over polynomial ring n = 5 by noting that in (2),

$$\displaystyle \sum y_i = 0$$

where yi for i = 1,2,3,4,5 are the five roots of (2). Now, since transformation (3) is showing a relation ship between the variables of quintic (2) and (1), it must also hold when we replace y in (3) by one of the roots of (2) and replace x by one of the roots of (1) corresponding to that of (2). Hence, we may write

$$\displaystyle y_i = x_i + k$$

Now, sum up all of these relation for i = 1,2,3,4,5 :

$$\displaystyle \sum y_i = \sum x_i + 5 \cdot k$$

Since the LHS is zero, and $$\displaystyle  \sum x_i = - a_1$$ by NI, $$\displaystyle k = -\frac{a_1}{5}$$.

So, we have the transformation and this way we can calculate the coefficients of the new depressed quintic which is much easier to handle than the original general form.

The principal quintic form will be described in the second post.

Balarka
.

• 2 people

#### mathbalarka

Math Team
OK, now for the principal quintic :

In order to transform (1) into the principle form :

$$\displaystyle z^5 + 5a z^2 + 5bz + c = 0 \;\; (4)$$

We need a quadratic Tschrinhausen transform :

$$\displaystyle z\, =\, x^2\, -\, ux\, +\, v \;\; (5)$$

Now, similar to the method used before, we will sum the identity up for all the roots :

$$\displaystyle \sum z_k = \sum x_k ^2 - u \sum x_k + 5v$$

Note that the LHS is zero. The sum $$\displaystyle \sum x_k^2$$ is $$\displaystyle a_1^2 - 2a_2$$ from elementary theory of Newton's identities.

Hence, we reckon that $$\displaystyle 5v = -a_1 u - a_1 ^2 + 2a_2$$. Now all we have to do is to derive the u in terms of the elementary symmetric polynomials.

Let us square 5 and sum it up again :

$$\displaystyle \sum z_k^2 = \sum (x_k ^2 - u x_k + v)^2$$

LHS is again zero since equation (4) doesn't have the 'x^3' term. Expanding the RHS gives :

$$\displaystyle \sum x_k ^4 + u^2 \sum x_k ^2 + 5v^2 - 2u \sum x_k ^3 + 2v \sum x_k^2 - 2uv \sum x_k = 0$$

Note that all the sums here are nothing but the Newton's symmetric formulas. After substituting the expressions obtained by NI, we get a quadratic

$$\displaystyle (2 a_1^2 - 5 a_2)u^2 - (4 a_1^3 - 13 a_1 a_2 + 15 a_3) u + (2 a_1^4 - 8 a_1^2 a_2 + 10 a_1 a_3 + 2 a_2^2 - 10 a_4) = 0$$

After solving this quadratic, we will get u and substituting this into v will yield the full transform (5).

In the next post, I will introduce how to avoid substituting (5) directly into (1) in order to calculate the coefficients of (4).

• 2 people

#### mathbalarka

Math Team
Now, consider the quintic

$$\displaystyle x^5 + b_0 x^4 + b_1 x^3 + b_2 x^2 + b_3 x + b_4 = (x + u)^5 + a_0 (x + u)^4 + a_1 (x + u)^3 + a_2 (x + u)^2 + a_3 (x + u) + a_4 = 0$$

We will write the principal quintic (4) with z replaced by v :

$$\displaystyle v^5 + 5av^2 + 5bv + c = \prod (v - z_k) \;\; (6)$$

Now substitute (5) into the RHS of (6) to get :

$$\displaystyle v^5 + 5av^2 + 5bv + c = - \prod (x_k^2 - u x_k) = - \prod x_k \prod (x_k - u) = -a_4 \cdot b_4 \;\; (7)$$

It can be easily seen that $$\displaystyle b_4 = u^5 + a_0 u^4 + a_1 u^3 + a_2 u^2 + a_3 u + a_4$$ from the definition.

Substituting the above into (7) and solving for c will give the result :

$$\displaystyle c = - a_4 (u^5 + u^5 + a_0 u^4 + a_1 u^3 + a_2 u^2 + a_3 u + a_4) - 5a v^2 - 5bv$$

Similarly,

$$\displaystyle 5b = a_3 (u^4 + a_0 u^3 + a_1 u^2 + a_2 u + a_3) - a_4 (5 u^3 + 4 a_0 u^2 + 3 a_1 u + a_2) - 5 v^4 - 10 a v$$

$$\displaystyle 5a = - a_2 (u^3 + a_0 u^2 + a_1 u + a_2) + a_3 (4 u^2 + 3 a_0 u + 2 a_1) - a_4 (5u + 2a_0) - 10v^3$$

We can obtain these by differentiating and double differentiating (6), respectively.

The transformation of a principle quintic into a Brioschi Quintic will be explained in my next post. It would take a little knowledge in polyhedral symmetry, so be prepared!

• 2 people

#### mathbalarka

Math Team
We have been able to transform a general quintic into a principle quintic all the way to now. I will introduce how to transform the principle quintic (4) into the one-parameter Brioschi Quintic :

$$\displaystyle x^5 - 10 Z x^3 + 45 Z^2 x - Z^2 = 0$$

I said before that I would use Polyhedral Symmetry to geometrically describe the transformation but I and agentredlum decided together that we will save the Polyhedral Symmetry for Klein's method which will probably be described later and use the algebraic transformation instead.

Consider the functions

$$\displaystyle g(x) = x^3 + ax^2 - \frac{1}{5} \sum x_k \;\;\; h(z) = x^4 + bx^2 - \frac{1}{5} \sum x_k$$

We see that they have the property that $$\displaystyle \sum g = 0$$ and $$\displaystyle  \sum h = 0$$ where the sum is taken over the roots of (4) and g, h are short notations for g(x) and h(x). Interestingly, they have to property $$\displaystyle  \sum x_k g = 0$$ and $$\displaystyle  \sum x_k h = 0$$ if and only if $$\displaystyle  \sum x_k^4 + a \sum x_k ^3 = 0$$ and $$\displaystyle  \sum x_k ^ 5 + b \sum x_k^3 = 0$$ and both of these can be satisfied with proper values of a & b. Now consider the function :

$$\displaystyle \psi(x) = \alpha g(x) + \beta h(x) \;\; (8)$$

This have the property $$\displaystyle  \psi^2 = 0$$ if

$$\displaystyle \alpha \sum g^2 + 2 \alpha \beta \sum g h + \beta^2 \sum h^2 = 0$$

The homogeneous equation above can be satisfied with proper non-zero choices of $$\displaystyle \(\alpha, \beta$$\).

So, the properties we have got so far are :

$$\displaystyle \sum x_k = 0 \;\, \sum x_k^2 = 0 \;\, \sum \psi = 0 \;\, \sum \psi^2 = 0 \;\, \sum x_k \psi = 0 \;\; (9)$$

Now, from here, we can easily derive the Bring-Jerrard form but I'll save that for another day and proceed further to our current objective.

It can be shown, that for a function $$\displaystyle \psi$$ defined as (8) and follows the properties (9), there always exists values p, q, r, a, b, t such that

$$\displaystyle p \psi^2 + 2 q x \psi + r x^2 - a \psi - b x + t = 0 \pmod{f(x)} \;\; (10)$$

Where $$\displaystyle f$$ is the principle quintic form (4). The proof is pretty elementary and I am not omitting it here; I'll surely post the proof if someone PMs for that reason. Now, if we sum up (10) for the roots of $$\displaystyle f(x)$$ and apply the properties (9), we see that t = 0 for obvious reasons. Now let $$\displaystyle \chi$$ and $$\displaystyle \phi$$ be linear functions(NOTE : Not affine functions of...) of $$\displaystyle x$$ and $$\displaystyle \psi$$, respectively. Hence, by (9),

$$\displaystyle \sum \chi = 0 \;\, \sum \chi^2 = 0 \;\, \sum \phi = 0 \;\, \sum \phi^2 = 0 \;\, \sum \chi \phi = 0 \;\; (11)$$

and from (10),

$$\displaystyle p \phi^2 + 2 q \chi \phi + r \chi^2 - a \phi - b \chi = 0 \pmod{f(x)} \;\; (12)$$

We may need another identity so I am introducing it before the real calculations :

$$\displaystyle m(p \chi^2 + 2 q \phi \chi + r \phi^2) = (d \chi + e \phi )^2 - c (a \chi + b \phi)^2 \;\; (13)$$

Where $$\displaystyle m = pb^2 - 2qab + ra^2, \; c = q^2 - pr, \; d = bp - aq, \; e = bq - ar$$

Now, let's apply a transformation to the principle quintic form :

$$\displaystyle y = \frac{d \chi + e \phi}{a \chi + b \phi}$$

And let the transformed quintic be

$$\displaystyle y^5 + a_1 y^4 + a_2 y^3 + a_3 y^2 + a_4 y + a_5 = 0 \;\; (14)$$

All we gave to do is to determine its coefficients. Here it goes :

$$\displaystyle y + \sqrt{c} = \frac{d \chi + e \phi + \sqrt{c} (a \chi + b \phi)}{a \chi + b \phi} (15)$$

Combining (12) and (13), we see that

$$\displaystyle m (a \chi + b \phi) = - (d \chi + e \phi )^2 + c (a \chi + b \phi)^2 \pmod{f(x)}$$

Hence, from (15),

$$\displaystyle \frac{m}{y + \sqrt{c}} = - d \chi - e \phi + \sqrt{c} (a \chi + b \phi) \pmod{f(x)}$$

Now from the properties (12), we see

$$\displaystyle \sum z = \sum z^2 = 0$$ where $$\displaystyle z = \frac{1}{y + \sqrt{c}} \;\; (16)$$

If we substitute $$\displaystyle y = \frac{1 - sqrt{c} z}{z}$$ in the quintic (14), we have

$$\displaystyle (1 - \sqrt{c} z)^5 + a_1 z (1 - \sqrt{c} z)^4 + a_2 z^2 (1 - \sqrt{c} z)^3 + a_3 z^3 (1 - \sqrt{c} z)^2 + a_4 z^4 (1 - \sqrt{c} z) + a_5 z^5 = 0$$

and since we have another value, namely $$\displaystyle - sqrt{c}$$ for the square root of c,

$$\displaystyle (1 + \sqrt{c} z)^5 + a_1 z (1 + \sqrt{c} z)^4 + a_2 z^2 (1 + \sqrt{c} z)^3 + a_3 z^3 (1 + \sqrt{c} z)^2 + a_4 z^4 (1 + \sqrt{c} z) + a_5 z^5 = 0$$

From (16), we see that the coefficients of z^4 and z^3 is 0 and from subtracting and summing those coefficients, we get the following 4 equations :

$$\displaystyle 5c^2 + 3 a_2 c + a_4 = 0 \;\, 10c + 3a_2 = 0 \;\, 4a_1 c + 2 a_3 = 0 \;\, 6 a_1 c + a_3 = 0$$

Hence, $$\displaystyle a_1 = a_3 = 0, \; a_2 = - 10 c/3$$ and $$\displaystyle a_4 = 5 c^2$$. The coefficient $$\displaystyle a_5$$ depends on the choice of $$\displaystyle \chi$$ and $$\displaystyle \phi$$.

So, the quintic in (14) is actually

$$\displaystyle y^5 - \frac{10 c}{3} y^3 + 5 c^2 y + a_5 = 0$$

By means of a short Tschirnhausen transform and an elementary parameter Z, we can reduce this into

$$\displaystyle x^5 - 10 Z x^3 + 45 Z^2 x - Z^2 = 0 \;\; (17)$$

• 2 people

#### mathbalarka

Math Team
In this post, I will describe how to transform a Brioschi quintic into a Jacobi sextic :

Consider a quintic with roots of the form $$\displaystyle r \zeta^k - \frac{1}{r \zeta^k}$$ where $$\displaystyle \zeta$$ is the fifth root of unity. Multiplying out the factors, we see :

$$\displaystyle \prod $\chi - \( r \zeta^k - \frac{1}{r \zeta^k}$$$ = \chi^5 + 5 \chi^3 + 5 \chi - $$r^5 - \frac{1}{r^5}$$\)

NOTE : The index is running through 0 to 4. If we replace $$\displaystyle \chi$$ by $$\displaystyle \chi - 1$$,

$$\displaystyle \prod $\chi - \(1 + r \zeta^k - \frac{1}{r \zeta^k}$$$ = \chi^5 - 5 \chi^4 + 15 \chi^3 - 25 \chi^2 + 25 \chi - 125 \sigma \;\; (18)\)

Where $$\displaystyle \sigma = \frac{1}{125} \(11 - \frac{1}{r^5} + r^5$$\)

Now, if we replace $$\displaystyle \chi$$ by $$\displaystyle - \chi$$ in (18), multiply it by (18) and change the sign, we get :

$$\displaystyle $\prod \[ \chi - \(1 + r \zeta^k - \frac{1}{r \zeta^k}$$$\]$\prod \[ \chi - $$1 + r \zeta^k - \frac{1}{r \zeta^k}$$$ \] = \prod $\chi^2 - $$1 + r \zeta^k - \frac{1}{r \zeta^k}$$^2$\)

$$\displaystyle = \chi^{10} + 5 \chi^8 + 25 \chi^6 + 125(1- 10 \sigma) \chi^4 + 625 (1 - 10 \sigma) \chi^2 - 125^2 \sigma^2 \;\; (19)$$

Replacing $$\displaystyle \chi$$ by $$\displaystyle 5s/\rho$$ in (19) and multiplying by $$\displaystyle (\rho/5)^5 (s - \rho)$$ gives

$$\displaystyle (s - \rho) \prod $s - \frac{\rho}{5} \(1 + r \zeta^k - \frac{1}{r \zeta^k}$$^2$\)

$$\displaystyle s^6 - 10 \sigma \rho^3 s^3 - \(1 - 10 \sigma + 5 \sigma^2$$\rho^5 s + 5 \sigma^2 \rho^6 \;\; (20)\)

We recognize the coefficients of s^3 and s as the Polyhedral polynomials $$\displaystyle f$$ and $$\displaystyle H$$ of that of the icosahedron, respectively.

Hence, setting (20) as zero will introduce the Jacobi sextic :

$$\displaystyle s^6 - 10 f s^3 - H s + 5 f^2 = 0 \;\; (21)$$

The external root will be written as $$\displaystyle s_5$$. The other roots are of the form :

$$\displaystyle s_k = \frac{s_5}{5} \(1 + r \zeta^k - \frac{1}{r \zeta^k}$$^2 \; 0 \le k \le 4\)

All we have to do now is to relate the roots of this Jacobi sextic with that of the Brioschi quintic. I will describe O. Perron's results now :

Define the variable $$\displaystyle v_k$$ as

$$\displaystyle v_k = (s_5 - s_k)(s_{k+2} - s_{k+3})(s_{k+4} - s_{k+1})$$

Now we will calculate the product $$\displaystyle  \prod $v - \(\frac{v_k}{\sqrt{5}} - 4f$$$ \$\). By using some tedious algebra or by a clever application of Newton's identities, one can derive

$$\displaystyle \prod $v - \(\frac{v_k}{\sqrt{5}} - 4f$$$ = v^5 + 30 f^2 v^3 + 100 f^3 v^2 + 105 f^4 v + 36 f^5 + H^3\)

Replacing $$\displaystyle v$$ by $$\displaystyle \frac{w}{\sqrt{5}} - 4f$$ and multiplying everything by $$\displaystyle 25 \sqrt{5}$$ gives

$$\displaystyle \prod [w - v_k] = w^5 - 20 \sqrt{5} f w^4 + 950 f^2 w^3 - 4500 \sqrt{5} f^3 w^2 + 50625 f^4 w + 25 \sqrt{5} \(H^3 - 1728 f^5$$ = 0\)

Now substituting $$\displaystyle w = \sqrt{5} t^2$$ and dividing by $$\displaystyle 25 \sqrt{5}$$, we conclude that

$$\displaystyle t^{10} - 20f t^8 + 190 f^2 t^6 - 900 f^3 t^4 + 2025 f^4 t^2 = - H^3 + 1728 f^5$$

Looking at above carefully, we reckon that this is actually

$$\displaystyle \( t^5 - 10f t^3 + 45 f^2 t$$^2 = - H^3 + 1728 f^5\)

Applying the icosahedral identity to the RHS and taking square root of both sides gives

$$\displaystyle t^5 - 10 f t^3 + 45 f^2 t - T = 0 \;\; (22)$$

Which is just the Brioschi normal form in two variables. We can obtain (17) by substituting $$\displaystyle t = y T/f^2$$ and introducing $$\displaystyle Z = f^5/t^2$$.

Hence, the roots of Brioschi quintic and the Jacobi sextic are related as

$$\displaystyle t_k = \sqrt{\frac{v_k}{\sqrt{5}}}$$

The next thing would be to solve the Jacobi sextic in terms of Weierstrass elliptic function which is analogues to that of the roots of cubic in terms of Trigonometric functions. The difference is that we perform period division by 3 on the periodic trigonometric function for a cubic, and we perform period division by 5 on the doubly periodic elliptic functions for a quintic.

• 2 people

#### mathbalarka

Math Team
In this post, I will describe the solution of Jacobi sextic in terms of Weierstrass elliptic function. But before that, we need to transform the coefficients of the Jacobi sextic (21), namely the polyhedral invariants, into the elliptic invariants :

First, let us write the Brioschi normal form (22) like this

$$\displaystyle t^5 + \frac{10}{\Delta} + \frac{45}{\Delta^2} t - \frac{216 g_3}{\Delta^3} = 0 \;\; (23)$$

Where $$\displaystyle g_2, g_3$$ are elliptic invariants and $$\displaystyle \Delta = g_2^3 - 27 g_3^2$$ is the elliptic discriminant. However, (23) implies

$$\displaystyle f = - \frac{1}{\Delta} \;\, H = \frac{12g_2}{\Delta^2} \;\, T = \frac{216 g_3}{\Delta^3}$$

Hence, the Jacobi sextic in the elliptic form would be :

$$\displaystyle s^6 + \frac{10}{\Delta} s^3 - \frac{12 g_2}{\Delta^2} s - \frac{5}{\Delta^2} = 0 \;\; (24)$$

Now we can begin our work. First let me introduce an interesting property of the Weierstrass elliptic function :

$$\displaystyle \frac{\sigma(5 u)}{\sigma^{(25)}(u)} = \frac{1}{82944} \left | \begin{matrix} \wp' & \wp'' & \wp''' & \wp^{\text{IV}} \\ \wp'' & \wp''' & \wp^{\text{IV}} & \wp^{\text{V}} \\ \wp''' & \wp^{\text{IV}} & \wp^{\text{V}} & \wp^{\text{VI}} \\ \wp^{\text{IV}} & \wp^{\text{V}} & \wp^{\text{VI}} & \wp^{\text{VII}} \end{matrix} \right | \;\; (25)$$

(NOTE : The sigma function is the Weierstrass sigma) Note that the determinant vanishes at the zeros of $$\displaystyle \sigma$$, explicitly, at

$$\displaystyle u_{m, n} = \frac{2m \omega + 2n \omega'}{5} \;\; (26)$$

Where $$\displaystyle \omega, \omega'$$ are the two periods of $$\displaystyle \wp$$ and m,n are integers mod 5 not both zero.

Expanding the determinant (25) and setting it to zero, we get the 12-th degree polynomial :

$$\displaystyle \( (\wp'')^2 - 12 \wp (\wp')^2$$^3 - 16 (\wp')^4 \wp'' $$(\wp'')^2 - 12 \wp (\wp')^2$$ - 64 (\wp')^8 = 0 \;\; (27)\)

The 24 roots of (26) relates to the 12 roots of (24) by the relation $$\displaystyle \wp_{m, n} = \wp_{-m,-n}$$ due to the mod 5 arithmetic of m,n.

We will now set a variable defined as

$$\displaystyle y_{m, n} = \wp_{2m, 2n} - \wp_{m, n} = \wp\(\frac{4m \omega + 4n \omega}{5}$$ - \wp$$\frac{2m \omega + 2n \omega}{5}$$\)

Using the formula $$\displaystyle \wp(u) - \wp(2u) = \frac{\wp'(u) \wp'''(u) = \wp''(u)^2}{4 \wp'(u)^2}$$ and some simplification on the derivatives we reckon,

$$\displaystyle y = y_{m, n} = \frac{\wp''_{m,n}^2 - 12 \wp_{m,n} \wp'_{m,n}^2}{4 \wp'_{m,n}} \;\; (28)$$

Substituting this into (27) and dividing by $$\displaystyle 64 (\wp')^6$$ gives

$$\displaystyle y^3 - \wp'' y - (\wp')^2 = 0 \;\; (29)$$

Also, by solving for $$\displaystyle (\wp'')^2$$ in (28),

$$\displaystyle (\wp'')^2 = (4y + 12 \wp)(\wp')^2 = (4y + 12 \wp)(y^3 - \wp'' y) \;\; (30)$$

By the closed form expression of the derivatives of $$\displaystyle \wp$$, we have two equations now

$$\displaystyle y^3 - (6 \wp^2 - g_2/2) y - (4 \wp^3 - g_2 \wp - g_3) = 0 \;\; (31) \\ (6 \wp^2 - g_2/2 )^2 = (4y + 12 \wp)(y^3 - 6 \wp^2 y + g_2 y /2) \;\; (32)$$

Let us define an auxiliary variable z defiend as

$$\displaystyle z = 2 \wp + y$$

so that the equations (30) and (31) becomes (respectively) :

$$\displaystyle z^3 - (3y^2 + g_2) z + 2 g_3 = 0 \;\; (32) \\ 9z^4 - 6(5y^2 + g_2)z^2 + (5y^4 - 2 g_2 y^2 + g_2^2) = 0 \;\; (33)$$

Using quadratic formula for (33), we get

$$\displaystyle 3 z^2 = 5 y^2 + g_2 + 2y \sqrt{5 y^2 + 3 g_2} \;\; (34)$$

Now let us write (32) as

$$\displaystyle z(z^2 - 3y^2 - g_2) = 2g_2 \;\; (35)$$

Squaring (35) and substituting (34) gives

$$\displaystyle \(5y^2 + g_2 + 2y \sqrt{5y^2 + 3g_2}$$$$-4y^2 - 2g_2 + 2y\sqrt{5y^2+3g_2}$$^2 = 108 g_3^2\)

Multiplying out the factors and dividing both sides by 4 gives the sextic

$$\displaystyle 5 y^6 - 2 y^5 \sqrt{5y^2 + 3 g_2} + g_2^3 = 27 g_2^3$$

Using the definition of the elliptic discriminant, this is equivalent to

$$\displaystyle 5 y^6 + \Delta = 2 y^5 \sqrt{5y^2 + 3 g_2}$$

Squaring both sides, rearranging the equation by the powers of y gives

$$\displaystyle 5y^{12} - 12 g_2 y^{10} + 10 \Delta y^6 + \Delta^2 = 0$$

If we let $$\displaystyle y = \frac{1}{s^2}$$ and multiply everything by $$\displaystyle \frac{s^6}{\Delta^2}$$, we get the Jacobi sextic (24)

$$\displaystyle s^6 + \frac{10}{\Delta} y^3 - \frac{12 g_2}{\Delta^2} s - \frac{5}{\Delta^2} = 0$$

Hence, we reckon that the roots of (24) are actually of the form :

$$\displaystyle \sqrt{s_{m,n}} = \frac{1}{\wp\(\frac{4m\omega + 4n \omega'}{5}$$ - \wp$$\frac{2m\omega + 2n\omega'}{5}$$}\)

After going through all the 12 roots, we see that the correct ones are :

$$\displaystyle \sqrt{s_5} = \frac{1}{\wp\(\frac{2\omega}{5}$$ - \wp$$\frac{4\omega}{5}$$}\)

$$\displaystyle \sqrt{s_k} = \frac{1}{\wp\(\frac{2\omega' + 48k\omega}{5}$$ - \wp$$\frac{4\omega' + 96 k \omega}{5}$$} \;\;\; 0 \le k \le 4\)

Hence, we have the roots of the Jacobi sextic, undoing all the transformation will give the equivalent roots of general quintic equation, and this way, the Kiepert's algorithm ends.

•  R. B. King, Beyond the quartic equation
•  S. Bessels, One step beyond the solvable equation

• 2 people

#### WMHalsdorf

Would be nice to see the equations.

#### Maschke

Quintics are a mysterious subject. Unlike the other degrees < 5, quintics cannot be solved using elementary arithmetic operations and radicals.
Yet the problem was solved by Galois, a political radical. Go figure.

• 1 person