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


into Bring-Jerrard normal form.

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


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


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






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).

  • $\begingroup$ 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! $\endgroup$ – Simply Beautiful Art Jun 25 '18 at 0:19
  • $\begingroup$ @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"). $\endgroup$ – 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 !

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

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