# Steps to the Quintic

It is easy to solve a quadratic equation in Mathematica:

In[1]:=

Out[1]=

### Cubic equation

The cubic equation is also straightforward, but the answer is more complicated. It is shown in a very small font size in order to give an idea of how large it is without filling up your screen.

In[2]:=

Out[2]=

### Quartic equation

The quartic can also be solved, but now the answer is quite long, so in order to avoid choking your link, it is not displayed here.

In[3]:=

### Quintic equation

What about the quintic? Here is what happens if we try to solve it directly.

In[4]:=

Out[4]=

`Root` objects are an implicit way to represent the solution. They can be differentiated and expanded out in series, and with approximate numerical values for the coefficients, they immediately yield a numerical solution. Of course, we can solve a quintic with numerical coefficients immediately by using the built-in Mathematica function `NSolve`.

Ruffini (1799) and Abel (1826) proved that it is not possible to give an explicit solution for the general quintic equation with symbolic coefficients in terms of square roots, cube roots, and so on. Is there an explicit solution to the quintic with symbolic coefficients? Yes! In the late 1800s, several mathematicians constructed such solutions. However, it was necessary to go beyond the extraction of roots and to use elliptic and hypergeometric functions. Mathematica can handle these higher mathematical functions in the same way as ordinary trigonometric or exponential functions. Combined with Mathematica's algebraic capabilities, this makes it possible to implement various symbolic solutions to the quintic.

### The program

Here is a complete implementation of Hermite's solution of the quintic in Mathematica in terms of elliptic functions. In the general case it takes almost a trillion bytes to store the explicit formulas in terms of the symbolic coefficients. The program below provides a much shorter, yet complete, representation of the solution.

Hermite's solution proceeds in three steps. In the first step one eliminates the quartic and cubic terms; in the next step one eliminates the square term; in the final step one solves a reduced equation of the form:

`PrincipalTransform[p == 0, x, y]` eliminates the quartic and cubic terms of the generic quintic `p` of the form:

The `x` is the variable in `p`, and the `y` will be the variable of the new quintic. The output is a pair whose first element is the transformation in the form of a pure function and whose second element is the new quintic itself, called the principal quintic.

In[5]:=

`Psi[q, x, n]` calculates the sum of the `n`th powers of all roots of the quintic.

In[6]:=

`BringJerrardTransform[p == 0, y, z]` eliminates the quadratic term of the principle quintic `p` of the form:

```    5      2
b0 y + b3 y + b4 y + b5 = 0
```
The `y` is the variable of `p`, and the `z` will be the variable of the new quintic. Again, the output is a pair whose first element is the transformation in the form of a pure function and whose second element is the new quintic itself, this time called a Bring-Jerrard quintic.

In[7]:=

`CanonicalTransform[p == 0. z, t]` normalizes the linear term of the Bring-Jerrard quintic poly to be `-1.` The `z` is the variable of `p`, and the `t` will be the variable of the new quintic. Again, the output is a pair whose first element is the transformation in form of a pure function and whose second element is the new quintic, called the canonical quintic.

In[8]:=

`HermiteQuinticSolve` solves the canonical quintic.

In[9]:=

### An example

Start with the following quintic. To save space we use numerical coefficients:

In[10]:=

The first Tschirnhaus transformation eliminates the quartic and cubic terms:

In[11]:=

Out[11]=

After the second Tschirnhaus transformation the quadratic term is also gone:

In[12]:=

Out[12]=

The next step makes the coefficient of the linear term equal to `-1`:

In[13]:=

Out[13]=

Now we calculate the solution of this quintic:

In[14]:=

Out[14]=

To get the solutions of the original equation, we must reverse the three transformations that brought the original equation to the reduced one. The first step is a simple linear transformation which gives the roots of the Bring-Jerrard quintic:

In[15]:=

Out[15]=

Inverting the (nonlinear) Tschirnhaus transformations produces extraneous reults that must be dropped:

In[16]:=

These are the solutions of the principal quintic:

In[17]:=

Out[17]=

Finally we get the five roots of the original quintic:

In[18]:=

Out[18]=