Adamchik transformation

The mathematical Adamchik Transformation is a type of mapping on Principal Quintic polynomials with the aim of transforming these polynomials into Bring Jerrard Quintic polynomials. This transformation is a special case of the Tschirnhaus Transformation and it uses a quartic transformation key. The Adamchik Transformation is developed by the mathematicians Victor Adamchik and David Jeffrey from the University of Southern California in the first years of the third millennium. And exactly this polynomial transformation is described in the essay Polynomial Transformations of Tschirnhaus, Bring and Jerrard by Adamchik and Jeffrey, and it was published in 2003.

Definition

The Adamchik Transformation transforms a Principal Quintic polynomial equation into a Bring Jerrard equation by using a quartic key, a key of fourth degree. The Principal Quintic is a special equation of fifth degree that contains a quintic as well as a quadratic, a linear and an absolute term but does neither contain a quartic nor a cubic term. Thus it fulfills the following pattern:

$x^5-u{x}^2+vx-w=0$

This Principal equation of fifth degree shall be transformed into a Bring Jerrard equation after this pattern:

$(x^4+a{x}^3+b{x}^2+cx+d{)}^5+m(x^4+a{x}^3+b{x}^2+cx+d)-n=0$

Along with the Polynomial Division in partial pattern, this accurately means that the left hand side of the under equation in this section must be able to be divided by the left hand side of the upper equation without any polynomial rest. By using this transformation the quadratic term of the Quintic system shall be eliminated. In this way the Bring Jerrard Quintic form appears.

The Bring Jerrard equation is a special fifth degree equation that does only contain a quintic, a linear and an absolute term. Along with the Abel Ruffini Theorem, the regular case of Bring Jerrard equations can not be solved in an elementary way. But that equation can be easily solved with the modular Elliptic Jacobi Theta Function and by using the elliptic modulus k in Legendre form and computing its Elliptic Nome function value. The determination of the elliptic modulus k for the corresponding Bring Jerrard equation is researched by Charles Hermite and Francesco Brioschi and published in the essay Sur la résolution de l'Équation du cinquiéme degré Comptes rendus from 1858.

Determination of the coefficients of the key

Basic pattern

Given is the mentioned pattern of the Adamchik Transformation in its parametrized form:

$x^5-u{x}^2+vx+w=0$

$y=x^4+a{x}^3+b{x}^2+cx+d$

$y^5+my-n=0$

To fulfill the mentioned polynomial division without any polynomial rest, and in the same way to create a Bring Jerrard that solves y and afterwards leads to the x solution, it is necessary to find out the relations between the coefficients of the quartic key and the coefficients of the principal quintic basic equation. The mathematicians Victor Adamchik and David Jeffrey worked out the exact equations that lead to the coefficients of the quartic key and of the Bring Jerrard final form.

Using a quartic key for that Tschirnhaus Transformation in the shape of the Adamchik Transformation is relevant. Only in a transformation key of fourth degree the coefficients of the key can stand in an elementary relation to the coefficients of the Principal Quintic so that a Bring Jerrard end equation can appear. In the case of using a cubic key the resulting coefficients of that key require the solving of an equation of sixth degree. And regular equations of sixth degree have no elementary solution. In this case the coefficient of the quartic transformation key must stand in a non elementary sixth degree radical relation to the coefficients of the Principal quintic equation so that the corresponding transformed quintic equation really can take the Bring Jerrard form as its form. But by using the quartic key this problem can be deactivated and there are coefficient combinations in elementary relation to the coefficients of the Principal equation that enable the appearance of a Bring Jerrard quintic as the corresponding Tschirnhaus transformed end quintic.

Transformation patterns of the equations

If the solution of the given Principal Quintic equation is entered into another polynomial, the corresponding quintic equation for this polynomial contains other coefficients that are a rational combination of the coefficients of the quartic key and the coefficients of the Principal form. In the following, an algorithm that leads to the new coefficients, shall be shown:

We take the five complex solutions of the given Pincipal quintic equations and create the quartic transformation keys:

$y_1=x_1^4+a{x}_1^3+b{x}_1^2+c{x}_1+d$

$y_2=x_2^4+a{x}_2^3+b{x}_2^2+c{x}_2+d$

$y_3=x_3^4+a{x}_3^3+b{x}_3^2+c{x}_3+d$

$y_4=x_4^4+a{x}_4^3+b{x}_4^2+c{x}_4+d$

$y_5=x_5^4+a{x}_5^3+b{x}_5^2+c{x}_5+d$

Along with the Theorem of Francois Vieta, following equations are valid for the coefficients of the Principal quintic:

Zero set coefficient of the quartic term:

$x_1+x_2+x_3+x_4+x_5=0$

Zero set coefficient of the cubic term:

$x_1{x}_2+x_1{x}_3+x_1{x}_4+x_1{x}_5+x_2{x}_3+x_2{x}_4+x_2{x}_5+$

$+x_3{x}_4+x_3{x}_5+x_4{x}_5=0$

Coefficient of the quadratic term:

$x_1{x}_2{x}_3+x_1{x}_2{x}_4+x_1{x}_2{x}_5+x_1{x}_3{x}_4+x_1{x}_3{x}_5+x_1{x}_4{x}_5+$

$+x_2{x}_3{x}_4+x_2{x}_3{x}_5+x_2{x}_4{x}_5+x_3{x}_4{x}_5=u$

Coefficient of the linear term:

$x_1{x}_2{x}_3{x}_4+x_1{x}_2{x}_3{x}_5+x_1{x}_2{x}_4{x}_5+x_1{x}_3{x}_4{x}_5+x_2{x}_3{x}_4{x}_5=v$

Coefficient of the absolute term:

$x_1{x}_2{x}_3{x}_4{x}_5=w$

And all these formulas shall be combined in order to find out the coefficients of the Tschirnhaus transformed quintic that shall take a Bring Jerrard form. In this way the searched relations between the coefficients of the quartic transformation key and the coefficients of the given Principal Quintic fundamental equation do appear exactly.

Sums of the x powers

In order to simplify the equation system, we decide to only set up the pure power sums of the x value and then determinate the power sums of the y values and in the final step determinate the coefficient of the Tschirnhaus transformed quintic end equation. The list of the power sums for the mentioned given Vieta pattern shall be created. In order to do that we use the knowledge about the Newton's identities that are known with the terminus Girard Newton formulae as well:

These four equations are needed for determination of the quartic term of the Tschirnhaus transformed quintic end equation:

$x_1+x_2+x_3+x_4+x_5=0$

$x_1^2+x_2^2+x_3^2+x_4^2+x_5^2=0$

$x_1^3+x_2^3+x_3^3+x_4^3+x_5^3=3u$

$x_1^4+x_2^4+x_3^4+x_4^4+x_5^4=-4v$

Further four equations are needed for determination of the cubic term of that quintic end equation:

$x_1^5+x_2^5+x_3^5+x_4^5+x_5^5=5w$

$x_1^6+x_2^6+x_3^6+x_4^6+x_5^6=3u^2$

$x_1^7+x_2^7+x_3^7+x_4^7+x_5^7=-7uv$

$x_1^8+x_2^8+x_3^8+x_4^8+x_5^8=8uw+4v^2$

Further four equations are needed for determination of the quadratic term of that quintic end equation:

$x_1^9+x_2^9+x_3^9+x_4^9+x_5^9=3u^3-9vw$

$x_1^{10}+x_2^{10}+x_3^{10}+x_4^{10}+x_5^{10}=-10u^{2}v+5w^2$

$x_1^{11}+x_2^{11}+x_3^{11}+x_4^{11}+x_5^{11}=11u^{2}w+11u{v}^2$

$x_1^{12}+x_2^{12}+x_3^{12}+x_4^{12}+x_5^{12}=3u^4-24uvw-4v^3$

Further four equations are needed for determination of the linear term m of that quintic end equation:

$x_1^{13}+x_2^{13}+x_3^{13}+x_4^{13}+x_5^{13}=-13u^{3}v+13u{w}^2+13v^{2}w$

$x_1^{14}+x_2^{14}+x_3^{14}+x_4^{14}+x_5^{14}=14u^{3}w+21u^2{v}^2-14v{w}^2$

$x_1^{15}+x_2^{15}+x_3^{15}+x_4^{15}+x_5^{15}=3u^5-45u^{2}vw-15u{v}^3+5w^3$

$x_1^{16}+x_2^{16}+x_3^{16}+x_4^{16}+x_5^{16}=-16u^{4}v+24u^2{w}^2+48u{v}^{2}w+4v^4$

Further four equations are needed for determination of the absolute term n of that quintic end equation:

$x_1^{17}+x_2^{17}+x_3^{17}+x_4^{17}+x_5^{17}=17u^{4}w+34u^3{v}^2-51uv{w}^2-17v^{3}w$

$x_1^{18}+x_2^{18}+x_3^{18}+x_4^{18}+x_5^{18}=3u^6-72u^{3}vw-36u^2{v}^3+18u{w}^3+27v^2{w}^2$

$x_1^{19}+x_2^{19}+x_3^{19}+x_4^{19}+x_5^{19}=-19u^{5}v+38u^3{w}^2+114u^2{v}^{2}w+19u{v}^4-19v{w}^3$

$x_1^{20}+x_2^{20}+x_3^{20}+x_4^{20}+x_5^{20}=20u^{5}w+50u^4{v}^2-120u^{2}v{w}^2-80u{v}^{3}w-4v^5+5w^4$

The fact shall be noticed that the terms on the right side directly appear by creating the Taylor series after t on following term:

$-z\frac{\mathrm{d}}{\mathrm{d}z}\ln (1-u{z}^3+v{z}^4-w{z}^5)$

Sums of the y powers

First clue of the Tschirnhaus key

Now the synthesis is made:

The quintic end form of the Tschirnhaus transformation shall be a Bring Jerrard form. So the coefficient of the quartic term of the end form must be equal to zero. Therefore this equation must be valid:

$y_1+y_2+y_3+y_4+y_5=0$

Entering the Vieta clues of the x terms leads to this:

$(x_1^4+x_2^4+x_3^4+x_4^4+x_5^4)+a(x_1^3+x_2^3+x_3^3+x_4^3+x_5^3)+$

$+b(x_1^2+x_2^2+x_3^2+x_4^2+x_5^2)+c(x_1+x_2+x_3+x_4+x_5)+5d=0$

Result for the coefficient of the quartic term of the end quintic equation and in this way first clue of the equation system for determination of the Tschirnhaus transformation:

$3ua-4v+5d=0$

Second clue of the Tschirnhaus key

The quintic end form of the Tschirnhaus transformation as mentioned shall be a Bring Jerrard form. So the coefficient of the cubic term of the end form also must be equal to zero. Therefore this equation must be valid:

$y_1{y}_2+y_1{y}_3+y_1{y}_4+y_1{y}_5+y_2{y}_3+y_2{y}_4+y_2{y}_5+$

$+y_3{y}_4+y_3{y}_5+y_4{y}_5=0$

And therefore following equation must be valid too:

$y_1^2+y_2^2+y_3^2+y_4^2+y_5^2=0$

Entering the Vieta clues of the x terms now leads to that:

$(x_1^8+x_2^8+x_3^8+x_4^8+x_5^8)+2a(x_1^7+x_2^7+x_3^7+x_4^7+x_5^7)+$

$+(a^2+2b)(x_1^6+x_2^6+x_3^6+x_4^6+x_5^6)+(2ab+2c)(x_1^5+x_2^5+x_3^5+x_4^5+x_5^5)+$

$+(2ac+b^2+2d)(x_1^4+x_2^4+x_3^4+x_4^4+x_5^4)+(2ad+2bc)(x_1^3+x_2^3+x_3^3+x_4^3+x_5^3)+$

$+(2bd+c^2)(x_1^2+x_2^2+x_3^2+x_4^2+x_5^2)+2cd(x_1+x_2+x_3+x_4+x_5)+5d^2=0$

Result of that system and in this way second clue of the equation system for determination of the Tschirnhaus transformation:

$3u(2ad+2bc)-4v(2ac+b^2+2d)+5w(2ab+2c)+3u^2(a^2+2b)-14uva+5d^2+8uw+4v^2=0$

Third clue of the Tschirnhaus key

The quintic end form of the Tschirnhaus transformation shall be a Bring Jerrard form. So even the coefficient of the quadratic term of the end form also must be equal to zero. Therefore this equation must be valid:

$y_1{y}_2{y}_3+y_1{y}_2{y}_4+y_1{y}_2{y}_5+y_1{y}_3{y}_4+y_1{y}_3{y}_5+y_1{y}_4{y}_5+$

$+y_2{y}_3{y}_4+y_2{y}_3{y}_5+y_2{y}_4{y}_5+y_3{y}_4{y}_5=0$

And therefore following equation must also be valid:

$y_1^3+y_2^3+y_3^3+y_4^3+y_5^3=0$

Entering the Vieta clues of the x terms now leads to that:

$(x_1^{12}+x_2^{12}+x_3^{12}+x_4^{12}+x_5^{12})+3a(x_1^{11}+x_2^{11}+x_3^{11}+x_4^{11}+x_5^{11})+$

$+(3a^2+3b)(x_1^{10}+x_2^{10}+x_3^{10}+x_4^{10}+x_5^{10})+(a^3+6ab+3c)(x_1^9+x_2^9+x_3^9+x_4^9+x_5^9)+$

$+(3a^{2}b+6ac+3b^2+3d)(x_1^8+x_2^8+x_3^8+x_4^8+x_5^8)+(3a{b}^2+3a^{2}c+6ad+6bc)(x_1^7+x_2^7+x_3^7+x_4^7+x_5^7)+$

$+(6abc+3a^{2}d+b^3+6bd+3c^2)(x_1^6+x_2^6+x_3^6+x_4^6+x_5^6)+(3a{c}^2+3b^{2}c+6abd+6cd)(x_1^5+x_2^5+x_3^5+x_4^5+x_5^5)+$

$+(6acd+3b{c}^2+3b^{2}d+3d^2)(x_1^4+x_2^4+x_3^4+x_4^4+x_5^4)+(3a{d}^2+6bcd+c^3)(x_1^3+x_2^3+x_3^3+x_4^3+x_5^3)+$

$+(3b{d}^2+3c^{2}d)(x_1^2+x_2^2+x_3^2+x_4^2+x_5^2)+3c{d}^2(x_1+x_2+x_3+x_4+x_5)+5d^3=0$

Result of that system and in this way third clue of the equation system for determination of the Tschirnhaus transformation:

$(-10u^{2}v+5w^2)(3a^2+3b)+(3u^3-9vw)(a^3+6ab+3c)+(8uw+4v^2)(3a^{2}b+6ac+3b^2+3d)-$ $-7uv(3a{b}^2+3a^{2}c+6ad+6bc)+3u^2(6abc+3a^{2}d+b^3+6bd+3c^2)+$ $+5w(3a{c}^2+3b^{2}c+6abd+6cd)-4v(6acd+3b{c}^2+3b^{2}d+3d^2)+$ $+3u(3a{d}^2+6bcd+c^3)+33u{v}^{2}a+33u^{2}wa+5d^3+3u^4-24uvw-4v^3=0$

Setting up the quartic Tschirnhaus key

This fundamental pattern is given:

$x^5-u{x}^2+vx+w=0$

$y=x^4+a{x}^3+b{x}^2+cx+d$

According to the found out clues, every Tschirnhaus transformation from the Principal Quintic form to the Bring Jerrard quintic form must fulfill following equation system for the coefficients of the quartic Tschirnhaus key in the general pattern:

$3ua-4v+5d=0$

$3u(2ad+2bc)-4v(2ac+b^2+2d)+5w(2ab+2c)+3u^2(a^2+2b)-14uva+5d^2+8uw+4v^2=0$

$(-10u^{2}v+5w^2)(3a^2+3b)+(3u^3-9vw)(a^3+6ab+3c)+(8uw+4v^2)(3a^{2}b+6ac+3b^2+3d)-$ $-7uv(3a{b}^2+3a^{2}c+6ad+6bc)+3u^2(6abc+3a^{2}d+b^3+6bd+3c^2)+$ $+5w(3a{c}^2+3b^{2}c+6abd+6cd)-4v(6acd+3b{c}^2+3b^{2}d+3d^2)+$ $+3u(3a{d}^2+6bcd+c^3)+33u{v}^{2}a+33u^{2}wa+5d^3+3u^4-24uvw-4v^3=0$

The mathematicians Professor Victor Adamchik and Professor David Jeffrey tried to solve this equation system so that the coefficients of the quartic Tschirnhaus key stand in elementary relation to the coefficients of the given Principal Quintic equation. In order to do this, they decided to use a certain special trick that contains the setting up of a further condition. In the mentioned second clue of that system only one of the four coefficients from the quartic Tschirnhaus key appears just in linear power, and that is the coefficient c accurately. So it is possible to set up a combination of the unknown coefficients a, b and d so that the second clue is always valid independent of which c value you enter. Adamchik and Jeffrey decided to insert this condition into the equation system.

$3ua-4v+5d=0$

$(-8va+6ub+10w)c+6uad-4v{b}^2-8vd+10wab+3u^2{a}^2+6u^{2}b-14uva+5d^2+8uw+4v^2=0$

$(-10u^{2}v+5w^2)(3a^2+3b)+(3u^3-9vw)(a^3+6ab+3c)+(8uw+4v^2)(3a^{2}b+6ac+3b^2+3d)-$ $-7uv(3a{b}^2+3a^{2}c+6ad+6bc)+3u^2(6abc+3a^{2}d+b^3+6bd+3c^2)+$ $+5w(3a{c}^2+3b^{2}c+6abd+6cd)-4v(6acd+3b{c}^2+3b^{2}d+3d^2)+$ $+3u(3a{d}^2+6bcd+c^3)+33u{v}^{2}a+33u^{2}wa+5d^3+3u^4-24uvw-4v^3=0$

$-8va+6ub+10w=0$

Now there is an equation system of four clues and four unknowns that even can be solved in an elementary way. Because now in the third clue a cubic equation system in relation to c cam be solved without concerning the impact on the second clue. Accurately this is the Adamchik transformation, and the corresponding quartic key is the Adamchik key if the now inserted fourth clue shall be valid. And exactly this equation system can be simplified in the following way:

$4va-3ub=5w$

$3u^2{a}^2-3uva+25wab-10v{b}^2=5uw-2v^2$

$d=\frac{4}{5}v-\frac{3}{5}ua$

$u{c}^3+(5wa-4vb+3u^2)c^2+(uv{a}^2+6uwa+5w{b}^2-8uvb+3u^3+vw)c+$ $+u^3{a}^3+vw{a}^3-2u^2{b}^3+2uva{b}^2-2u^2{a}^{2}d+10d^3-4u^{2}v{a}^2+vwab+$ $+3uvad+2u^{2}wa+2u^{2}vb-v^{2}d+u^4-4v^3+10uvw=0$

With this recipe the coefficients of the quartic Adamchik Tschirnhaus key can be set up.

Two simple calculation examples

And in the next step the coefficients of the Bring Jerrard Quintic endform must be determined. Before doing this we show two very simple structured calculation examples on which the now mentioned recipe for the quartic Adamchik transformation key really works:

Calculation example one:

$x^5-5x^2+15x-7=0$

$y=x^4+x^3+\frac{5}{3}{x}^2-\frac{2}{3}x+9$

$y^5-\frac{338000}{81}y-\frac{8788000}{243}=0$

Given coefficients of the Principal Quintic equation:
$u=5,v=15,w=7$
Quadratic fundamental system:
$60a-15b=35$ $75a^2-225a+175ab-150b^2=-275$ $d=12-3a$
$a=1$	$b=\frac{5}{3}$	$d=9$
Cubic resolvent:
$5c^3+10c^2-\frac{1240}{9}c-\frac{2560}{27}=0$
$c=-\frac{2}{3}$

Calculation example two:

$x^5+10x^2+15x-4=0$

$y=x^4-\frac{1}{2}{x}^3+\frac{5}{3}{x}^2+\frac{19}{3}x+9$

$y^5-\frac{10440125}{1296}y-\frac{177482125}{3888}=0$

Given coefficients of the Principal Quintic equation:
$u=-10,v=15,w=4$
Quadratic fundamental system:
$60a+30b=20$ $300a^2+450a+100ab-150b^2=-650$ $d=12+6a$
$a=-\frac{1}{2}$	$b=\frac{5}{3}$	$d=9$
Cubic resolvent:
$-10c^3+190c^2-\frac{11435}{18}c-\frac{95}{54}=0$
$c=\frac{19}{3}$

Determination of the coefficients of the Bring Jerrard end form

Given information about the coefficients of the Bring Jerrard form

The coefficients of the Bring Jerrard Quintic end form m and n shall be determined. In order to do that, we again take the Girard Newton identities for the x terms and the y terms and combine them. This is the end form with the unknown coefficients:

$y^5+my-n=0$

Along with the Vieta sentence sollowing expressions are valid:

$y_1{y}_2{y}_3{y}_4+y_1{y}_2{y}_3{y}_5+y_1{y}_2{y}_4{y}_5+y_1{y}_3{y}_4{y}_5+y_2{y}_3{y}_4{y}_5=m$

$y_1{y}_2{y}_3{y}_4{y}_5=n$

According to the Girard Newton identities following formulas are valid:

$y_1^4+y_2^4+y_3^4+y_4^4+y_5^4=-4m$

$y_1^5+y_2^5+y_3^5+y_4^5+y_5^5=5n$

These are the previously mentioned formulas for the key on the five complex x solutions:

$y_1=x_1^4+a{x}_1^3+b{x}_1^2+c{x}_1+d$

$y_2=x_2^4+a{x}_2^3+b{x}_2^2+c{x}_2+d$

$y_3=x_3^4+a{x}_3^3+b{x}_3^2+c{x}_3+d$

$y_4=x_4^4+a{x}_4^3+b{x}_4^2+c{x}_4+d$

$y_5=x_5^4+a{x}_5^3+b{x}_5^2+c{x}_5+d$

And those are the also previously mentioned formulas for the x power sums:

These four equations are needed for determination of the quartic term of the Tschirnhaus transformed quintic end equation:

$x_1+x_2+x_3+x_4+x_5=0$ $x_1^2+x_2^2+x_3^2+x_4^2+x_5^2=0$ $x_1^3+x_2^3+x_3^3+x_4^3+x_5^3=3u$ $x_1^4+x_2^4+x_3^4+x_4^4+x_5^4=-4v$

$x_1^5+x_2^5+x_3^5+x_4^5+x_5^5=5w$ $x_1^6+x_2^6+x_3^6+x_4^6+x_5^6=3u^2$ $x_1^7+x_2^7+x_3^7+x_4^7+x_5^7=-7uv$ $x_1^8+x_2^8+x_3^8+x_4^8+x_5^8=8uw+4v^2$

$x_1^9+x_2^9+x_3^9+x_4^9+x_5^9=3u^3-9vw$ $x_1^{10}+x_2^{10}+x_3^{10}+x_4^{10}+x_5^{10}=-10u^{2}v+5w^2$ $x_1^{11}+x_2^{11}+x_3^{11}+x_4^{11}+x_5^{11}=11u^{2}w+11u{v}^2$ $x_1^{12}+x_2^{12}+x_3^{12}+x_4^{12}+x_5^{12}=3u^4-24uvw-4v^3$

$x_1^{13}+x_2^{13}+x_3^{13}+x_4^{13}+x_5^{13}=-13u^{3}v+13u{w}^2+13v^{2}w$ $x_1^{14}+x_2^{14}+x_3^{14}+x_4^{14}+x_5^{14}=14u^{3}w+21u^2{v}^2-14v{w}^2$ $x_1^{15}+x_2^{15}+x_3^{15}+x_4^{15}+x_5^{15}=3u^5-45u^{2}vw-15u{v}^3+5w^3$ $x_1^{16}+x_2^{16}+x_3^{16}+x_4^{16}+x_5^{16}=-16u^{4}v+24u^2{w}^2+48u{v}^{2}w+4v^4$

$x_1^{17}+x_2^{17}+x_3^{17}+x_4^{17}+x_5^{17}=17u^{4}w+34u^3{v}^2-51uv{w}^2-17v^{3}w$ $x_1^{18}+x_2^{18}+x_3^{18}+x_4^{18}+x_5^{18}=3u^6-72u^{3}vw-36u^2{v}^3+18u{w}^3+27v^2{w}^2$ $x_1^{19}+x_2^{19}+x_3^{19}+x_4^{19}+x_5^{19}=-19u^{5}v+38u^3{w}^2+114u^2{v}^{2}w+19u{v}^4-19v{w}^3$ $x_1^{20}+x_2^{20}+x_3^{20}+x_4^{20}+x_5^{20}=20u^{5}w+50u^4{v}^2-120u^{2}v{w}^2-80u{v}^{3}w-4v^5+5w^4$

Coefficient of the linear term

Now all these formulas are put together:

$-4m=y_1^4+y_2^4+y_3^4+y_4^4+y_5^4$

Inserting the values of y leads to this:

$-4m=(x_1^{16}+x_2^{16}+x_3^{16}+x_4^{16}+x_5^{16})+4a(x_1^{15}+x_2^{15}+x_3^{15}+x_4^{15}+x_5^{15})+(6a^2+4b)(x_1^{14}+x_2^{14}+x_3^{14}+x_4^{14}+x_5^{14})+$

$(4a^3+12ab+4c)(x_1^{13}+x_2^{13}+x_3^{13}+x_4^{13}+x_5^{13})+(a^4+12a^{2}b+12ac+6b^2+4d)(x_1^{12}+x_2^{12}+x_3^{12}+x_4^{12}+x_5^{12})+$

$+(4a^{3}b+12a{b}^2+12a^{2}c+12ad+12bc)(x_1^{11}+x_2^{11}+x_3^{11}+x_4^{11}+x_5^{11})+$

$+(6a^2{b}^2+4a^{3}c+24abc+12a^{2}d+4b^3+12bd+6c^2)(x_1^{10}+x_2^{10}+x_3^{10}+x_4^{10}+x_5^{10})+$

$+(4a{b}^3+12a{c}^2+12a^{2}bc+4a^{3}d+24abd+12b^{2}c+12cd)(x_1^9+x_2^9+x_3^9+x_4^9+x_5^9)+$

$+(12a{b}^{2}c+12a^{2}bd+6a^2{c}^2+b^4+12b{c}^2+24acd+12b^{2}d+6d^2)(x_1^8+x_2^8+x_3^8+x_4^8+x_5^8)+$

$+(12ab{c}^2+12a{b}^{2}d+12a^{2}cd+12a{d}^2+4b^{3}c+24bcd+4c^3)(x_1^7+x_2^7+x_3^7+x_4^7+x_5^7)+$

$+(4a{c}^3+6a^2{d}^2+24abcd+6b^2{c}^2+4b^{3}d+12b{d}^2+12c^{2}d)(x_1^6+x_2^6+x_3^6+x_4^6+x_5^6)+$

$+(12ab{d}^2+12a{c}^{2}d+4b{c}^3+12b^{2}cd+12c{d}^2)(x_1^5+x_2^5+x_3^5+x_4^5+x_5^5)+$

$+(12ac{d}^2+6b^2{d}^2+12b{c}^{2}d+c^4+4d^3)(x_1^4+x_2^4+x_3^4+x_4^4+x_5^4)+(4a{d}^3+12bc{d}^2+4c^{3}d)(x_1^3+x_2^3+x_3^3+x_4^3+x_5^3)+$

$+(4b{d}^3+6c^2{d}^2)(x_1^2+x_2^2+x_3^2+x_4^2+x_5^2)+4c{d}^3(x_1+x_2+x_3+x_4+x_5)+5d^4$

Inserting the values of the sums of the x powers leads to that:

$-4m=(6a^2+4b)(14u^{3}w+21u^2{v}^2-14v{w}^2)+$

$+(4a^3+12ab+4c)(-13u^{3}v+13u{w}^2+13v^{2}w)+$

$+(a^4+12a^{2}b+12ac+6b^2+4d)(3u^4-24uvw-4v^3)+$

$+(4a^{3}b+12a{b}^2+12a^{2}c+12ad+12bc)(11u^{2}w+11u{v}^2)+$

$+(6a^2{b}^2+4a^{3}c+24abc+12a^{2}d+4b^3+12bd+6c^2)(-10u^{2}v+5w^2)+$

$+(4a{b}^3+12a{c}^2+12a^{2}bc+4a^{3}d+24abd+12b^{2}c+12cd)(3u^3-9vw)+$

$+(12a{b}^{2}c+12a^{2}bd+6a^2{c}^2+b^4+12b{c}^2+24acd+12b^{2}d+6d^2)(8uw+4v^2)+$

$-7uv(12ab{c}^2+12a{b}^{2}d+12a^{2}cd+12a{d}^2+4b^{3}c+24bcd+4c^3)+$

$+3u^2(4a{c}^3+6a^2{d}^2+24abcd+6b^2{c}^2+4b^{3}d+12b{d}^2+12c^{2}d)+$

$+5w(12ab{d}^2+12a{c}^{2}d+4b{c}^3+12b^{2}cd+12c{d}^2)+$

$-4v(12ac{d}^2+6b^2{d}^2+12b{c}^{2}d+c^4+4d^3)+3u(4a{d}^3+12bc{d}^2+4c^{3}d)+$

$+12u^{5}a-180u^{2}vwa-60u{v}^{3}a+20w^{3}a-16u^{4}v+24u^2{w}^2+48u{v}^{2}w+4v^4+5d^4=$

Now all these terms shall be separated along the c powers accurately!

By doing that and by adding the clues for the key coefficients we got following result for the linear term coefficient of the Bring Jerrard final form:

$m=v{c}^4+(2ud-5wb+3uv)c^3+$ $+(-12uw{a}^2-6v^2{a}^2+15uvab-15wad+12vbd-9u^{3}a+21vwa+12uwb-9u^{2}d+15u^{2}v)c^2+$ $+[-(3a^2+3b)(11u^{2}w+11u{v}^2)-(a^3+6ab)(-10u^{2}v+5w^2)-(3a^{2}b+3b^2+3d)(3u^3-9vw)-(3a{b}^2+6ad)(8uw+4v^2)+$ $+21uv{a}^{2}d-18u^{2}abd+12va{d}^2+7uv{b}^3-9ub{d}^2-15w{b}^{2}d+42uvbd-15w{d}^2-$ $-9u^{4}a+72uvwa+12v^{3}a+13u^{3}v-13u{w}^2-13v^{2}w]c+$ $+(6a^2+4b)(14u^{3}w+21u^2{v}^2-14v{w}^2)+(4a^3+12ab)(-13u^{3}v+13u{w}^2+13v^{2}w)+$ $+(a^4+12a^{2}b+6b^2+4d)(3u^4-24uvw-4v^3)+(4a^{3}b+12a{b}^2+12ad)(12u^{2}w+11u{v}^2)+$ $+(6a^2{b}^2+12a^{2}d+4b^3+12bd)(-10u^{2}v+5w^2)+(4a{b}^3+4a^{3}d+24abd)(3u^3-9vw)+$ $+(12a^{2}bd+b^4+12b^{2}d+6d^2)(8uw+4v^2)-84uva{b}^{2}d-84uva{d}^2+18u^2{a}^2{d}^2+12u^2{b}^{3}d+36u^{2}b{d}^2+60wab{d}^2-$ $-24v{b}^2{d}^2-16v{d}^3+12ua{d}^3+12u^{5}a-180u^{2}vwa-60u{v}^{3}a+20w^{3}a-16u^{4}v+24u^2{w}^2+48u{v}^{2}w+4v^4+5d^4$

Result of the transformation

$=-y^5+(-4s-3ar+5d)y^4+$

+ (4*r*t - 6*s^2 - 5*a*r*s - 3*a^2*r^2 + 3*b*r^2 - 5*a*b*t - 5*c*t + 16*d*s - 2*b^2*s - 4*a*c*s + 12*a*d*r - 3*b*c*r - 10*d^2)*y^3 +

+ (r^4 - 4*s^3 + 8*r*s*t + a*r^2*t - a*r*s^2 + 5*a^2*t^2 + 5*b*t^2 - a^2*r^2*s + 2*b*r^2*s + 3*a^3*s*t - 2*a*b*s*t - 11*c*s*t - a^3*r^3 + 3*a*b*r^3 - 3*c*r^3 - 7*a^2*b*r*t + a*c*r*t + 8*b^2*r*t - 12*d*r*t + 4*a^2*b*s^2 - 8*a*c*s^2 - 4*b^2*s^2 + 18*d*s^2 - 5*a^2*c*r*s + a*b^2*r*s + 15*a*d*r*s + 2*b*c*r*s + 9*a^2*d*r^2 - 3*a*b*c*r^2 + b^3*r^2 - 9*b*d*r^2 + 3*c^2*r^2 + 15*a*b*d*t - 5*a*c^2*t - 5*b^2*c*t + 15*c*d*t + 12*a*c*d*s + 6*b^2*d*s - 4*b*c^2*s - 24*d^2*s - 18*a*d^2*r + 9*b*c*d*r - c^3*r + 10*d^3)*y^2 +

+ (-2*r^2*t^2 + 4*r*s^2*t - s^4 - 3*a*r^2*s*t + a*r*s^3 - 5*a*t^3 - a^2*s*t^2 + 2*b*r^3*t - b*r^2*s^2 + 6*b*s*t^2 + 2*a^3*r*t^2 - 4*a*b*r*t^2 + 7*c*r*t^2 + -a^3*s^2*t + 3*a*b*s^2*t + c*r^3*s - 7*c*s^2*t + 3*a^4*r*s*t - 11*a^2*b*r*s*t + 10*a*c*r*s*t + 4*b^2*r*s*t - 16*d*r*s*t - a^4*s^3 + 4*a^2*b*s^3 - 4*a*c*s^3 - 2*b^2*s^3 - 2*d*r^4 + 8*d*s^3 - 2*a^3*b*r^2*t + 6*a*b^2*r^2*t - 2*a*d*r^2*t - 6*b*c*r^2*t + a^3*b*r*s^2 - a^2*c*r*s^2 - 3*a*b^2*r*s^2 + 2*a*d*r*s^2 + 5*b*c*r*s^2 + 5*a^3*c*t^2 - 5*a^2*b^2*t^2 - 10*a^2*d*t^2 + 5*a*b*c*t^2 + 5*b^3*t^2 - 10*b*d*t^2 - 5*c^2*t^2 - a^3*c*r^2*s + 2*a^2*d*r^2*s + 3*a*b*c*r^2*s - 4*b*d*r^2*s - 3*c^2*r^2*s - 6*a^3*d*s*t + 7*a^2*b*c*s*t - a*b^3*s*t + 4*a*b*d*s*t - 13*a*c^2*s*t - 3*b^2*c*s*t + 22*c*d*s*t + 2*a^3*d*r^3 - 6*a*b*d*r^3 + 6*c*d*r^3 + 14*a^2*b*d*r*t - 3*a^2*c^2*r*t - 6*a*b^2*c*r*t - 2*a*c*d*r*t + 2*b^4*r*t - 16*b^2*d*r*t + 9*b*c^2*r*t + 12*d^2*r*t - 8*a^2*b*d*s^2 - 2*a^2*c^2*s^2 + 4*a*b^2*c*s^2 + 16*a*c*d*s^2 - b^4*s^2 + 8*b^2*d*s^2 - 4*b*c^2*s^2 - 18*d^2*s^2 + 10*a^2*c*d*r*s - 2*a*b^2*d*r*s - 3*a*b*c^2*r*s - 15*a*d^2*r*s + b^3*c*r*s - 4*b*c*d*r*s + 3*c^3*r*s - 9*a^2*d^2*r^2 + 6*a*b*c*d*r^2 - 2*b^3*d*r^2 + 9*b*d^2*r^2 - 6*c^2*d*r^2 - 15*a*b*d^2*t + 10*a*c^2*d*t + 10*b^2*c*d*t - 5*b*c^3*t - 15*c*d^2*t - 12*a*c*d^2*s - 6*b^2*d^2*s + 8*b*c^2*d*s - c^4*s + 16*d^3*s + 12*a*d^3*r - 9*b*c*d^2*r + 2*c^3*d*r - 5*d^4)*y -

- c s^3 t - a b r s t^2 + 3 c r s t^2 + b s^2 t^2 + a^2 r t^3 - 2 b r t^3 - a s t^3 + t^4 - 2*a*c*r^2*t^2 + a*c*r*s^2*t + b^2*r^2*t^2 + 2*d*r^2*t^2 - 4*d*r*s^2*t + d*s^4 - a^5*t^3 + 5*a^3*b*t^3 - 5*a^2*c*t^3 - 5*a*b^2*t^3 + 3*a*d*r^2*s*t - a*d*r*s^3 + 5*a*d*t^3 - b*c*r^2*s*t + 5*b*c*t^3 + a^4*b*s*t^2 - a^3*c*s*t^2 - 4*a^2*b^2*s*t^2 + a^2*d*s*t^2 + 7*a*b*c*s*t^2 + 2*b^3*s*t^2 - 2*b*d*r^3*t + b*d*r^2*s^2 - 6*b*d*s*t^2 + c^2*r^3*t - 3*c^2*s*t^2 + 2*a^4*c*r*t^2 - a^3*b^2*r*t^2 - 3*a^3*d*r*t^2 - 6*a^2*b*c*r*t^2 + 3*a*b^3*r*t^2 + 4*a*b*d*r*t^2 + 7*a*c^2*r*t^2 - 3*b^2*c*r*t^2 - 7*c*d*r*t^2 - a^4*c*s^2*t + a^3*d*s^2*t + 4*a^2*b*c*s^2*t - 3*a*b*d*s^2*t - 4*a*c^2*s^2*t - 2*b^2*c*s^2*t - c*d*r^3*s + 7*c*d*s^2*t - 3*a^4*d*r*s*t + a^3*b*c*r*s*t + 11*a^2*b*d*r*s*t - a^2*c^2*r*s*t - 3*a*b^2*c*r*s*t - 10*c*a*d*r*s*t - 4*b^2*d*r*s*t + 5*b*c^2*r*s*t + 8*d^2*r*s*t + a^4*d*s^3 - 4*a^2*b*d*s^3 + 4*a*c*d*s^3 + 2*b^2*d*s^3 + d^2*r^4 - 4*d^2*s^3 + 2*a^3*b*d*r^2*t - a^3*c^2*r^2*t - 6*a*b^2*d*r^2*t + 3*a*b*c^2*r^2*t + a*d^2*r^2*t + 6*b*c*d*r^2*t - 3*c^3*r^2*t - a^3*b*d*r*s^2 + a^2*c*d*r*s^2 + 3*a*b^2*d*r*s^2 - a*d^2*r*s^2 - 5*b*c*d*r*s^2 - 5*a^3*c*d*t^2 + 5*a^2*b^2*d*t^2 + 5*a^2*b*c^2*t^2 + 5*a^2*d^2*t^2 - 5*a*b^3*c*t^2 - 5*a*b*c*d*t^2 - 5*a*c^3*t^2 + b^5*t^2 - 5*b^3*d*t^2 + 5*b^2*c^2*t^2 + 5*b*d^2*t^2 + 5*c^2*d*t^2 + a^3*c*d*r^2*s - a^2*d^2*r^2*s - 3*a*b*c*d*r^2*s + 2*b*d^2*r^2*s + 3*c^2*d*r^2*s + 3*a^3*d^2*s*t - 7*a^2*b*c*d*s*t - 2*a^2*c^3*s*t + a*b^3*d*s*t + 4*a*b^2*c^2*s*t - 2*a*b*d^2*s*t + 13*a*c^2*d*s*t - b^4*c*s*t + 3*b^2*c*d*s*t - 4*b*c^3*s*t - 11*c*d^2*s*t - a^3*d^2*r^3 + 3*a*b*d^2*r^3 - 3*c*d^2*r^3 - 7*a^2*b*d^2*r*t + 3*a^2*c^2*d*r*t + 6*a*b^2*c*d*r*t - 3*a*b*c^3*r*t + a*c*d^2*r*t - 2*b^4*d*r*t + b^3*c^2*r*t + 8*b^2*d^2*r*t - 9*b*c^2*d*r*t + 3*c^4*r*t - 4*d^3*r*t + 4*a^2*b*d^2*s^2 + 2*a^2*c^2*d*s^2 - 4*a*b^2*c*d*s^2 - 8*a*c*d^2*s^2 + b^4*d*s^2 - 4*b^2*d^2*s^2 + 4*b*c^2*d*s^2 + 6*d^3*s^2 - 5*a^2*c*d^2*r*s + a*b^2*d^2*r*s + 3*a*b*c^2*d*r*s + 5*a*d^3*r*s - b^3*c*d*r*s + 2*b*c*d^2*r*s - 3*c^3*d*r*s + 3*a^2*d^3*r^2 - 3*a*b*c*d^2*r^2 + b^3*d^2*r^2 - 3*b*d^3*r^2 + 3*c^2*d^2*r^2 + 5*a*b*d^3*t - 5*a*c^2*d^2*t - 5*b^2*c*d^2*t + 5*b*c^3*d*t - c^5*t + 5*c*d^3*t + 4*a*c*d^3*s + 2*b^2*d^3*s - 4*b*c^2*d^2*s + c^4*d*s - 4*d^4*s - 3*a*d^4*r + 3*b*c*d^3*r - c^3*d^2*r + d^5

$\mathrm{S}\mathrm{u}\mathrm{m}\mathrm{m}\mathrm{a}\mathrm{n}\mathrm{d}\mathrm{i}\mathrm{z}\mathrm{e}\mathrm{d}\mathrm{F}\mathrm{o}\mathrm{r}\mathrm{m}$

Exactly this summandized form directly leads to the clues that connect the coefficients of the Principal form with the coefficient of the quartic key and also with the coefficients of the Bring Jerrard form that is strived for. By setting the summandization negative the coefficients of the Quintic result can be read in its complete form.

Clues of the Transformation

In the now following step the shown summandized quintic form must be set equal with the Bring Jerrard Form. The coefficients of the quartic, the cubic and the quadratic term must be set to zero. Therefore these are the first two clues for the determination of the coefficients of the quartic key:

Equation for the quartic term of the resulting quintic:

$\mathrm{I})4s+3ar-5d=0$

Equation for the cubic term of the resulting quintic:

$\mathrm{I}\mathrm{I})-4rt+6s^2+5ars+3a^2{r}^2-3b{r}^2+5abt+5ct-16ds+2b^{2}s+4acs-12adr+3bcr+10d^2=0$

Equation for the quadratic term of the resulting quintic:

$\mathrm{I}\mathrm{I}\mathrm{I})$ r*c^3 + (-3*r^2 + 5*a*t + 4*b*s)*c^2 + (11*s*t + 3*r^3 - a*r*t + 8*a*s^2 + 5*a^2*r*s - 2*b*r*s + 3*a*b*r^2 + 5*b^2*t - 15*d*t - 12*a*d*s - 9*b*d*r)*c - (r^4 - 4*s^3 + 8*r*s*t + a*r^2*t - a*r*s^2 + 5*a^2*t^2 + 5*b*t^2 - a^2*r^2*s + 2*b*r^2*s + 3*a^3*s*t - 2*a*b*s*t - a^3*r^3 + 3*a*b*r^3 - 7*a^2*b*r*t + 8*b^2*r*t - 12*d*r*t + 4*a^2*b*s^2 - 4*b^2*s^2 + 18*d*s^2 + a*b^2*r*s + 15*a*d*r*s + 9*a^2*d*r^2 + b^3*r^2 - 9*b*d*r^2 + 15*a*b*d*t + 6*b^2*d*s - 24*d^2*s - 18*a*d^2*r + 10*d^3) = 0

This equation system of three equations but four unknowns a, b, c, d has infinite many solutions for those four unknowns. But only a few solutions of that equation stand in an elementary relation to the coefficients r and s and t of the given Principal Quintic. However one solution of that system of three equation clues does always have an elementary relation to the coefficients of the Principal Quintic and can be generated by adding a fourth condition. In the coefficient of the cubic term of the final form there is only one unknown variable that appears linear, and that is variable c accurately. And this variable, the unknown c appears in the coefficient of the final quadratic term in a cubic equation. So it is useful to take the independence of the cubic term from the variable c as the further condition. To make the coefficient of the cubic term of the final equation independent in this way, that cubic coefficient term in relation to c must have a horizontal slope. Therefore the slope must be set equal to zero. This looks exactly like that:

$\mathrm{I}\mathrm{I})-4rt+6s^2+5ars+3a^2{r}^2-3b{r}^2+5abt-16ds+2b^{2}s-12adr+10d^2+(5t+4as+3br)c=0$

New condition:

$5t+4as+3br=0$

The expression in the round brackets is set equal to zero because this is the mentioned slope. So the further condition is formulated. In this way a new equation system can be built for the unknowns a, b and d that is succeeded by the already mentioned equation from the quadratic term for solving the left unknown c too. The determination of c requires the solving of a cubic equation. And so the quadratic key for the Tschirnhaus Transformation is synthesized.

Synthesis of the Transformation

Synthesis of the quartic key

Along with the already given explanation a system of three equations and three unknowns can be created. By adding the further demanded condition and simplifying the equation system, the following system for the determination of a, b and d is generated:

${\mathrm{I}}^{\mathrm{*}})4s+3ar-5d=0$

$\mathrm{I}{\mathrm{I}}^{\mathrm{*}})5t+4as+3br=0$

$\mathrm{I}\mathrm{I}{\mathrm{I}}^{\mathrm{*}})3rt-2s^2+15abt+6b^{2}s-3adr+ds=0$

In a successive way, the unknown c can be solved by the mentioned cubic equation:

$\mathrm{I}{\mathrm{V}}^{\mathrm{*}})$ r*c^3 + (-3*r^2 + 5*a*t + 4*b*s)*c^2 + (11*s*t + 3*r^3 - a*r*t + 8*a*s^2 + 5*a^2*r*s - 2*b*r*s + 3*a*b*r^2 + 5*b^2*t - 15*d*t - 12*a*d*s - 9*b*d*r)*c - (r^4 - 4*s^3 + 8*r*s*t + a*r^2*t - a*r*s^2 + 5*a^2*t^2 + 5*b*t^2 - a^2*r^2*s + 2*b*r^2*s + 3*a^3*s*t - 2*a*b*s*t - a^3*r^3 + 3*a*b*r^3 - 7*a^2*b*r*t + 8*b^2*r*t - 12*d*r*t + 4*a^2*b*s^2 - 4*b^2*s^2 + 18*d*s^2 + a*b^2*r*s + 15*a*d*r*s + 9*a^2*d*r^2 + b^3*r^2 - 9*b*d*r^2 + 15*a*b*d*t + 6*b^2*d*s - 24*d^2*s - 18*a*d^2*r + 10*d^3) = 0

These equation clues lead to the complete quartic key for the Adamchik Transformation as a special case of the Tschirnhaus Transformation.

Synthesis of the Bring Jerrard form

The last two mentioned formulas of the old equation system lead to the coefficient of the linear term and to the coefficient of the absolute term of the aimed Bring Jerrard form:

$y^5+my+n=0$

$-m=$ -2*r^2*t^2 + 4*r*s^2*t - s^4 - 3*a*r^2*s*t + a*r*s^3 - 5*a*t^3 - a^2*s*t^2 + 2*b*r^3*t - b*r^2*s^2 + 6*b*s*t^2 + 2*a^3*r*t^2 - 4*a*b*r*t^2 + 7*c*r*t^2 + -a^3*s^2*t + 3*a*b*s^2*t + c*r^3*s - 7*c*s^2*t + 3*a^4*r*s*t - 11*a^2*b*r*s*t + 10*a*c*r*s*t + 4*b^2*r*s*t - 16*d*r*s*t - a^4*s^3 + 4*a^2*b*s^3 - 4*a*c*s^3 - 2*b^2*s^3 - 2*d*r^4 + 8*d*s^3 - 2*a^3*b*r^2*t + 6*a*b^2*r^2*t - 2*a*d*r^2*t - 6*b*c*r^2*t + a^3*b*r*s^2 - a^2*c*r*s^2 - 3*a*b^2*r*s^2 + 2*a*d*r*s^2 + 5*b*c*r*s^2 + 5*a^3*c*t^2 - 5*a^2*b^2*t^2 - 10*a^2*d*t^2 + 5*a*b*c*t^2 + 5*b^3*t^2 - 10*b*d*t^2 - 5*c^2*t^2 - a^3*c*r^2*s + 2*a^2*d*r^2*s + 3*a*b*c*r^2*s - 4*b*d*r^2*s - 3*c^2*r^2*s - 6*a^3*d*s*t + 7*a^2*b*c*s*t - a*b^3*s*t + 4*a*b*d*s*t - 13*a*c^2*s*t - 3*b^2*c*s*t + 22*c*d*s*t + 2*a^3*d*r^3 - 6*a*b*d*r^3 + 6*c*d*r^3 + 14*a^2*b*d*r*t - 3*a^2*c^2*r*t - 6*a*b^2*c*r*t - 2*a*c*d*r*t + 2*b^4*r*t - 16*b^2*d*r*t + 9*b*c^2*r*t + 12*d^2*r*t - 8*a^2*b*d*s^2 - 2*a^2*c^2*s^2 + 4*a*b^2*c*s^2 + 16*a*c*d*s^2 - b^4*s^2 + 8*b^2*d*s^2 - 4*b*c^2*s^2 - 18*d^2*s^2 + 10*a^2*c*d*r*s - 2*a*b^2*d*r*s - 3*a*b*c^2*r*s - 15*a*d^2*r*s + b^3*c*r*s - 4*b*c*d*r*s + 3*c^3*r*s - 9*a^2*d^2*r^2 + 6*a*b*c*d*r^2 - 2*b^3*d*r^2 + 9*b*d^2*r^2 - 6*c^2*d*r^2 - 15*a*b*d^2*t + 10*a*c^2*d*t + 10*b^2*c*d*t - 5*b*c^3*t - 15*c*d^2*t - 12*a*c*d^2*s - 6*b^2*d^2*s + 8*b*c^2*d*s - c^4*s + 16*d^3*s + 12*a*d^3*r - 9*b*c*d^2*r + 2*c^3*d*r - 5*d^4

$-n=$ - c s^3 t - a b r s t^2 + 3 c r s t^2 + b s^2 t^2 + a^2 r t^3 - 2 b r t^3 - a s t^3 + t^4 - 2*a*c*r^2*t^2 + a*c*r*s^2*t + b^2*r^2*t^2 + 2*d*r^2*t^2 - 4*d*r*s^2*t + d*s^4 - a^5*t^3 + 5*a^3*b*t^3 - 5*a^2*c*t^3 - 5*a*b^2*t^3 + 3*a*d*r^2*s*t - a*d*r*s^3 + 5*a*d*t^3 - b*c*r^2*s*t + 5*b*c*t^3 + a^4*b*s*t^2 - a^3*c*s*t^2 - 4*a^2*b^2*s*t^2 + a^2*d*s*t^2 + 7*a*b*c*s*t^2 + 2*b^3*s*t^2 - 2*b*d*r^3*t + b*d*r^2*s^2 - 6*b*d*s*t^2 + c^2*r^3*t - 3*c^2*s*t^2 + 2*a^4*c*r*t^2 - a^3*b^2*r*t^2 - 3*a^3*d*r*t^2 - 6*a^2*b*c*r*t^2 + 3*a*b^3*r*t^2 + 4*a*b*d*r*t^2 + 7*a*c^2*r*t^2 - 3*b^2*c*r*t^2 - 7*c*d*r*t^2 - a^4*c*s^2*t + a^3*d*s^2*t + 4*a^2*b*c*s^2*t - 3*a*b*d*s^2*t - 4*a*c^2*s^2*t - 2*b^2*c*s^2*t - c*d*r^3*s + 7*c*d*s^2*t - 3*a^4*d*r*s*t + a^3*b*c*r*s*t + 11*a^2*b*d*r*s*t - a^2*c^2*r*s*t - 3*a*b^2*c*r*s*t - 10*c*a*d*r*s*t - 4*b^2*d*r*s*t + 5*b*c^2*r*s*t + 8*d^2*r*s*t + a^4*d*s^3 - 4*a^2*b*d*s^3 + 4*a*c*d*s^3 + 2*b^2*d*s^3 + d^2*r^4 - 4*d^2*s^3 + 2*a^3*b*d*r^2*t - a^3*c^2*r^2*t - 6*a*b^2*d*r^2*t + 3*a*b*c^2*r^2*t + a*d^2*r^2*t + 6*b*c*d*r^2*t - 3*c^3*r^2*t - a^3*b*d*r*s^2 + a^2*c*d*r*s^2 + 3*a*b^2*d*r*s^2 - a*d^2*r*s^2 - 5*b*c*d*r*s^2 - 5*a^3*c*d*t^2 + 5*a^2*b^2*d*t^2 + 5*a^2*b*c^2*t^2 + 5*a^2*d^2*t^2 - 5*a*b^3*c*t^2 - 5*a*b*c*d*t^2 - 5*a*c^3*t^2 + b^5*t^2 - 5*b^3*d*t^2 + 5*b^2*c^2*t^2 + 5*b*d^2*t^2 + 5*c^2*d*t^2 + a^3*c*d*r^2*s - a^2*d^2*r^2*s - 3*a*b*c*d*r^2*s + 2*b*d^2*r^2*s + 3*c^2*d*r^2*s + 3*a^3*d^2*s*t - 7*a^2*b*c*d*s*t - 2*a^2*c^3*s*t + a*b^3*d*s*t + 4*a*b^2*c^2*s*t - 2*a*b*d^2*s*t + 13*a*c^2*d*s*t - b^4*c*s*t + 3*b^2*c*d*s*t - 4*b*c^3*s*t - 11*c*d^2*s*t - a^3*d^2*r^3 + 3*a*b*d^2*r^3 - 3*c*d^2*r^3 - 7*a^2*b*d^2*r*t + 3*a^2*c^2*d*r*t + 6*a*b^2*c*d*r*t - 3*a*b*c^3*r*t + a*c*d^2*r*t - 2*b^4*d*r*t + b^3*c^2*r*t + 8*b^2*d^2*r*t - 9*b*c^2*d*r*t + 3*c^4*r*t - 4*d^3*r*t + 4*a^2*b*d^2*s^2 + 2*a^2*c^2*d*s^2 - 4*a*b^2*c*d*s^2 - 8*a*c*d^2*s^2 + b^4*d*s^2 - 4*b^2*d^2*s^2 + 4*b*c^2*d*s^2 + 6*d^3*s^2 - 5*a^2*c*d^2*r*s + a*b^2*d^2*r*s + 3*a*b*c^2*d*r*s + 5*a*d^3*r*s - b^3*c*d*r*s + 2*b*c*d^2*r*s - 3*c^3*d*r*s + 3*a^2*d^3*r^2 - 3*a*b*c*d^2*r^2 + b^3*d^2*r^2 - 3*b*d^3*r^2 + 3*c^2*d^2*r^2 + 5*a*b*d^3*t - 5*a*c^2*d^2*t - 5*b^2*c*d^2*t + 5*b*c^3*d*t - c^5*t + 5*c*d^3*t + 4*a*c*d^3*s + 2*b^2*d^3*s - 4*b*c^2*d^2*s + c^4*d*s - 4*d^4*s - 3*a*d^4*r + 3*b*c*d^3*r - c^3*d^2*r + d^5

And so the complete Adamchik Transformation is created.

Calculation example

Determination of the Quartic key

Given is the Principal Quintic equation that can not be solved by elementary roots:

$x^5-5x^2+5x-5=0$

In this example following pattern is valid:

$r=-5,s=5,t=-5$

The system of three equations and three unknowns a, b and d is set up:

${\mathrm{I}}^{\mathrm{*}})20-15a-5d=0$

$\mathrm{I}{\mathrm{I}}^{\mathrm{*}})-25+20a-15b=0$

$\mathrm{I}\mathrm{I}{\mathrm{I}}^{\mathrm{*}})25-75ab+30b^2+15ad+5d=0$

This equation system has following solution:

$a=-1,b=-3,d=7$

$a=7/5,b=1/5,d=-1/5$

Only the upper solution of that equation system leads to a cubic equation for c that has one real and two imaginary solution. In the other case there appear three real solutions. Therefore exactly the upper solution now will be entered into the mentioned cubic equation:

$-5c^3-110c^2-1100c-3080=0$

This is idential to that:

$c^3+22c^2+220c+616=0$

And in this way the real solution for c appears:

$c=\frac{8}{3}\sqrt{11}\sinh [\frac{1}{3}\mathrm{arsinh}(\frac{4}{11}\sqrt{11}\left)\right]-\frac{22}{3}$

$c\approx -4.2684949995680718959592849407954973796$

Determination of the Bring Jerrard form coefficients

This solution now is entered into the expressions for the coefficients of the Bring Jerrard form m and n to accurately find out these two coefficients:

The value of m is this:

$m=\frac{14080}{3}\{11-2\sqrt{22}\cosh [\frac{1}{3}\mathrm{arcosh}\left(\frac{7}{22}\sqrt{22}\right)\left]\right\}$

$m\approx 5346.7659799413342912673562726724783512142$

Corresponding equation:

$m^3-154880m^2+6542131200m-30704402432000=0$

And the value of n is that:

$n=\frac{11264}{3}\{11-110\sqrt{2}\sinh [\frac{1}{3}\mathrm{arsinh}\left(\frac{1}{22}\sqrt{2}\right)\left]\right\}$

$n\approx 28793.4255440915279846424972250875$

Corresponding equation:

$n^3-123904n^2+260987420672n-7435869363372032=0$

So the complete Adamchik Transformation for this calculation example is made:

$x^5-5x^2+5x-5=0$

$y=x^4-x^3-3x^2+\{\frac{8}{3}\sqrt{11}\sinh [\frac{1}{3}\mathrm{arsinh}\left(\frac{4}{11}\sqrt{11}\right)]-\frac{22}{3}\}x+7$

$y^5+\frac{14080}{3}\{11-2\sqrt{22}\cosh [\frac{1}{3}\mathrm{arcosh}\left(\frac{7}{22}\sqrt{22}\right)\left]\right\}y+\frac{11264}{3}\{11-110\sqrt{2}\sinh [\frac{1}{3}\mathrm{arsinh}\left(\frac{1}{22}\sqrt{2}\right)\left]\right\}=0$

$y\approx -4.87187987090341241739191116705958390845844658170795795268900739402026742$

$x\approx 1.56670895425072582758152133323240667646412076995364965189840377191745$

This described Adamchik Transformation gives the exact solution of the given Principal Quintic equation.

Shortcut to the Brioschi form

Description of the shortcut

For finding out the corresponding Brioschi form to the given Principal quintic the same equation system delivers a shortcut directly to the coefficients of the broken rational key of the Brioschi form that is described in the Titus-Piezas-III-essay. This is the equation system again. The coefficients a, b and d give the complete transformation of the Principal form into the Brioschi form. This is the given principal form:

$x^5+r{x}^2+sx+t=0$

And this is the given equation system:

${\mathrm{I}}^{\mathrm{*}})4s+3ar-5d=0$

$\mathrm{I}{\mathrm{I}}^{\mathrm{*}})5t+4as+3br=0$

$\mathrm{I}\mathrm{I}{\mathrm{I}}^{\mathrm{*}})3rt-2s^2+15abt+6b^{2}s-3adr+ds=0$

Now the coefficients of the broken rational key of the Brioschi form will be created by using the coefficent a in this way:

$u=\frac{(27r^4-160s^3+300rst)a+30r^{3}s-125s^{2}t}{3r^4-15s^3+75rst}$

$v=1728-\frac{5(15t-r{u}^2+3su{)}^3}{r^2(s^{2}u-5rtu+5st)}$

$w=\frac{5(15t-r{u}^2+3su{)}^3-(s^{2}u-5rtu+5st)(40r{u}^3+360s{u}^2+1800tu)}{5(s^{2}u-5rtu+5st)(r{u}^2+su+5t)}$

This is the broken rational key for the Brioschi form:

$x=\frac{wz+u}{v{z}^2-3}$

The resulting Brioschi equation looks this way:

$v^2{z}^5-10v{z}^3+45z-1=0$

Mentioned example

By taking the already mentioned calculation example following Brioschi form appears:

$x^5-5x^2+5x-5=0$

The value for a was already computed in the mentioned calculation example:

$a=-1$

This solution brings those values:

$u=-4\text{and}v=1849\text{and}w=129$

This is the broken rational key:

$x=\frac{129z-4}{1849z^2-3}$

It shows the Tschirnhaus Transformation in this way:

$[\frac{129z-4}{1849z^2-3}{]}^5-5[\frac{129z-4}{1849z^2-3}{]}^2+5\left[\frac{129z-4}{1849z^2-3}\right]-5=0$

And this is the resulting Brioschi equation:

$3418801z^5-18490z^3+45z-1=0$

Other sources

• Victor Adamchik, David Jeffrey: Polynomial Transformations of Tschirnhaus, Bring and Jerrard at the Wayback Machine (archived 2007-09-28), ACM Sigsam Bulletin, Band 37, 2003

• F. Brioschi: Sulla risoluzione delle equazioni del quinto grado: Hermite — Sur la résolution de l'Équation du cinquiéme degré Comptes rendus —. N. 11. Mars. 1858. 1. Dezember 1858, doi:10.1007/bf03197334

This article "Adamchik transformation" is from Wikipedia. The list of its authors can be seen in its historical and/or the page Edithistory:Adamchik transformation. Articles copied from Draft Namespace on Wikipedia could be seen on the Draft Namespace of Wikipedia and not main one.

📰 Article(s) of the same category(ies)

• Semiabelian group (Galois theory)
• Cyclotomic pre-polynomial
• Extended Langton's ant
• Bézout method
• Chaos communications
• AC Method
• RegularChains

---