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The quintic can be transformed, in radicals, to the one-parameter Brioschi form,

$$w^5-10\alpha w^3+45\alpha^2w-\alpha^2 = 0\tag1$$

Letting $w = 1/(x^2+20)$ and it becomes the rather nice,

$$(x^2+20)^2(x-5)+\alpha^{-1}=0\tag2$$

I was wondering if someone else would like to explain the transformation, or know an alternative method?

This would complete the summary in this post.

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Since it has been several days and there is no answer, I guess it is ok to do it myself. To reduce the general quintic,

$$x^5+b_4x^4+b_3x^3+b_2x^2+b_1x+b_0=0\tag{1}$$

to the one-parameter Brioschi form,

$$w^5-10qw^3+45q^2w-q^2 = 0\tag{2}$$

which has a near-square discriminant,

$$D = 5^5q^8(-1+1728q)^2$$

is also done in two steps (like for the Bring-Jerrard form) .

Step 1: Transform the general quintic $(1)$ to the principal quintic (which is missing the $x^4,x^3$ terms) using a quadratic Tschirnhausen transformation $y=x^2+mx+n$ and is described in this post. One ends up with the form,

$$y^5+5ay^2+5by+c=0\tag{3}$$

Step 2: Find a second quadratic Tschirnhausen which establishes a relation between the roots of $(2)$ and $(3)$. A modern treatment can be found in Duke's and Tóth's "The Splitting of Primes in Division Fields of Elliptic Curves" (p.10). The transformation is,

$$y = \frac{pw+\lambda}{q^{-1}w^2-3}$$

where $\lambda$ is a quadratic purely in the coefficients of $(3)$, namely,

$$(a^4 + a b c - b^3)\lambda^2 - (11a^3\color{red}b - ac^2 + 2b^2c )\lambda+ (64a^2 b^2 - 27a^3c - b c^2) = 0\tag4$$

correcting a typo in the paper which is missing the variable in red. To find $p,q$, using any root $\lambda$, then,

$$p = \frac{ -8a \lambda^3 - 72b \lambda^2 - 72 c \lambda +j\, a^2}{a \lambda^2 + b \lambda + c}\tag5$$

$$q = \frac{1}{1728-j}\tag6$$

$$j = \frac{(a \lambda^2 - 3b \lambda - 3c)^3}{a^2(a c \lambda - b^2 \lambda - b c)}\tag7$$

provided $j\neq 0, 1728$. So it can be done using only quadratics. (Note: For the special case $a=0$, then a slightly different transformation can bypass division by zero.)

Example. Given,

$$y^5+5y^2+10y+2=0$$

so $a,b,c = 1,2,2$. To transform this to the form,

$$w^5-10qw^3+45q^2w-q^2 = 0$$

we let,

$$y = \frac{pw+\lambda}{q^{-1}w^2-3}$$

where,

$$\lambda = \tfrac{1}{3}\big(-17 + \sqrt{871}\big)$$

$$p = \tfrac{16}{27}\big(5071 - 179\sqrt{871}\big)$$

$$q = \tfrac{1}{43312^2}\big(615193 + 20894 \sqrt{871}\big)$$

using formulas $(4)-(7)$.

This completes the series of posts for the two reduced forms of the quintic. Now if someone knows how to reduce the general sextic into the Joubert sextic form...

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  • $\begingroup$ As per the footnote: Is there no straightforward way to remove the $x^5$ and $x^3$ terms from the general sextic? According to the footnotes on your Bring-Jerrard form post, one can even reduce it to $x^6+ax^2+bx+c$. $\endgroup$ – Simply Beautiful Art Jul 7 '18 at 20:56
  • $\begingroup$ @SimplyBeautifulArt: Yes, there is a one-step way to remove the $x^5$ and $x^3$ from the general sextic $F(y)=0$. One uses a quadratic Tschirnhausen $x = y^2+ay+b$ where the two unknowns $a,b$ allow two terms of the sextic to be eliminated. The transformation's coefficients are determined by a cubic. If the $x^5$ and $x^2$ are to be removed, then the coefficients are now determined by a quartic. And so on. $\endgroup$ – Tito Piezas III Jul 8 '18 at 4:23

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