(This question was inspired by this post.)
I. Quintic
Given a solvable quintic
$$F(x)=0\tag1$$
the problem is to transform it to the de Moivre form
$$y^5+5ay^3+5a^2y+b=0\tag2$$
using only a third-degree Tschirnhausen transformation
$$y=x^3+c_1x^2+c_2x+c_3\tag3$$
Eliminating $x$ between $(1)$ and $(3)$ yields the quintic
$$y^5+d_1y^4+d_2y^3+d_3y^2+d_4y+d_5=0.\tag4$$ The three unknowns $c_i$ allow us to solve the system $$d_1 = d_3 =0$$ $$d_2^2=5d_4$$
Its final equation turns out to be of $12$-deg, but seems to be factorable and solvable when $(1)$ is solvable. And since the quintic roots $x,y$ are related by
$$y=x^3+c_1x^2+c_2x+c_3=2\sqrt{-a}\;\cos\left(-\tfrac{2\pi\,k}{5}+\tfrac{1}{5}\,\arccos\big(\tfrac{-b}{2\sqrt{-a^5}}\big)\right)\tag5$$
then such transformations may yield unusual trigonometric identities like,
$$\color{brown}{x^3-\phi\,x^2-\tfrac{7+\sqrt{5}}{2}x+\tfrac{5+4\sqrt{5}}{5} = 2\,\phi\sqrt{\tfrac{11}{5}}\cos\left(-\tfrac{6\pi}{5}+\tfrac{1}{5}\,\cos^{-1}\big(\tfrac{-89-25\sqrt{5}}{44\sqrt{11}}\big)\right)=-4.7985\dots}$$
where $x=2\cos\big(\tfrac{2\pi}{11}\big)$ and the golden ratio $\phi=\frac{1+\sqrt{5}}{2}$. Similar relations in radicals can be found using other quintics.
II. Septic
Similarly, given the solvable septic $$F(x) =0\tag6$$ we transform it to the de Moivre quintic's seventh-degree analogue $$y^7+7ay^5+14a^2y^3+7a^3y+b=0\tag7$$ using a fifth-degree Tschirnhausen $$y=x^5+c_1x^4+c_2x^3+c_3x^2+c_4x+c_5.\tag8$$ Eliminating $x$ between $(6)$ and $(8)$ yields the septic $$y^7+d_1y^6+d_2y^5+d_3y^4+d_4y^3+d_5y^2+d_6y+d_7=0.\tag9$$ The five unknowns $c_i$ allow us to solve the system $$d_1 = d_3 = d_5 =0$$ $$2d_2^2=7d_4$$ $$d_2^3=49d_6$$ The septic roots $x,y$ are then related by $$y=x^5+c_1x^4+c_2x^3+c_3x^2+c_4x+c_5=2\sqrt{-a}\;\cos\left(-\tfrac{2\pi\,k}{7}+\tfrac{1}{7}\,\arccos\big(\tfrac{-b}{2\sqrt{-a^7}}\big)\right)\tag{10}$$
III. Questions:
- If $F(x)=0$ is a solvable equation, will the coefficients $c_i$ of the Tschirnhausen transformation turn out to be radicals? (I'm of the opinion they are.)
- For $p=5$, it seems the degree of the $c_i$'s minimal polynomial are quadratics. For $p=7$, are the $c_i$ roots of cubics?
