How to remove the second two leading terms in the general quintic with just algebra?

Question

Motivated by How to transform a general higher degree five or higher equation to normal form?

The goal of the linked question is to transform the general quintic

$$x^5+ax^4+bx^3+cx^2+dx+e=0$$

into Bring-Jerrard normal form.

Tito Piezas III begins his answer with the quadratic Tschirnhausen transformation,

$$y=x^2+mx+n$$

and by using resultants which may be calculated by WolframAlpha, one can write the result as

$$y^5+c_1y^4+c_2y^3+c_3y^2+c_4y+c_5=0$$

where we proceed to make $c_1=c_2=0$.

However, it is not immediately obvious to me how one performs this step, particularly the process of eliminating $x$ and replacing it with $y$.

How can I perform this step without referring to resultants and anything outside of simple algebra?

Or, if it makes any difference, how can I go from

$$x^5+ax^4+bx^3+cx^2+dx+e=0$$

to

$$y^5+c_3y^2+c_4y+c_5=0$$

?

Answer 1

The original equation gives $\,x^5 = -ax^4-bx^3-cx^2-dx-e\,$, so $\,x^n\,$ can be expressed as a polynomial of degree (at most) $\,4\,$ in $\,x\,$ for $\,n \ge 5\,$.

It follows that the first few powers $\,k=1,2,3,4,5\,$ of $\,y\,$ can be written as:

$$ \begin{align} y^k &\,=\, a_{k,0} + a_{k,1}\,x + a_{k,2}\,x^2 + a_{k,3}\,x^3 + a_{k,4}\,x^4 \\ \end{align} $$

Eliminating $\,x,x^2,x^3,x^4\,$ between the $\,5\,$ equations gives a quintic in $\,y\,$, which can be done with "simple algebra" (albeit the calculations are tedious).

Answer 2

First note that the substitution can be rewritten as \begin{eqnarray*} y-n+\frac{m^2}{4} =\left(x+\frac{m}{2} \right)^2=X^2. \end{eqnarray*} Substitute $x=X-m/2$ into the quintic, to get \begin{eqnarray*} X^5+a'X^4+b'X^3 +c'X^2+d'X+e'=0 \\ X^5+b'X^3 +d'X=a'X^4+ c'X^2+e'. \end{eqnarray*} Now rearrange as above & square this equation, then substitute $X^2=y-n+m^2/4$ to obtain a quintic in $y$. ... Good luck Sir !