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

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.

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

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

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

?

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).

• Oh, I see. When raising $y$ to the $k$, we remove $x^5$ and higher terms with the original equation. One then proceeds to add these equations together, and then attempts to eliminate some terms. Hm, I'll try that! – Simply Beautiful Art Jun 25 '18 at 0:19
• @SimplyBeautifulArt we remove x^5 and higher terms with the original equation Right, then you can consider $x,x^2,x^3,x^4$ as independent "variables" and eliminate them between the equations, or consider the whole thing as a linear homogeneous system in "variables" $\,1,x,x^2,x^3,x^4\,$ and equate the determinant to $\,0\,$ in order for non-trivial solutions to exist (which could still qualify as "simple algebra" for certain values of "simple"). – dxiv Jun 25 '18 at 0:26

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 !

• The LHS of the last line contains odd powers of $X$. How do I convert these to $y$? – Simply Beautiful Art Jun 25 '18 at 0:13
• Square the equation ... there will only be $X^2$ terms ... & you will be able to substitute. – Donald Splutterwit Jun 25 '18 at 0:16
• While this is much cleaner, it doesn't look like it generalizes nicely to higher order substitutions. – Simply Beautiful Art Jun 26 '18 at 1:44