Arbitrary n degree irreducible algebraic equation formula solution

WuZhong-xiang (Institute of Mechanics, AcademiaSinica)

Abstract

The formula solution and radical solution of arbitrary 1 to 4 degreealgebraic equations have long been successively obtained. But that of more than4 degree although the experience algebraist nearly 500 years of effort, it hasnot yet been. In particular, in 1830, Galois, E. was given criterion foralgebraic equation radical solution, the academic community seems to have beenaccepted on n> 4 irreducible algebraic equation has no radical solutions.

My 2011 article: "Arbitraryn degree irreducible algebraic equations radical solution"

http://blog.sciencenet.cn/blog-226-510331.html

Detailed analysis has been: Galois’s theory that, in fact, only" throughout the process of solving irreducible algebraic equations, theadded radical index, n *, should be less than 4", not the equation degree,n, should be less than 4, and not the equation degree n greater than 4 can nothave radical solution.

And, specifically, the radical solution of arbitrary 5 and 6 degree irreduciblealgebraic equations have been given. Also extended to m successive increases, arbitraryn = 2m and 2m+1 radical solution of algebraicequations corresponding solution. Wherein the radicals added are all less than4, and thus, are specifically indicated: The above understanding of Galois’stheory is correct with the actual.

This in turn gives that any n irreducible algebraic equations can besolved with variety formula and radical solutions.

The added radical, are also less than 4, are also more powerfulshow: Correct "the usual misinterpreted Galois theory" is correct andnecessary.

Keywords: irreducible algebraic equationsolution formula radical solution Galois’s theory

1. The specific expressions of arbitrary n degreeirreducible algebraic equation

Arbitrary n degree irreducible algebraic equation can be expressedas:

x ^ n + a(n-1) x ^ (n-1) + a (n-2) x ^ (n-2) + ... + a (1) x + a(0) = 0,         ( 1.1)

     Wherein, a (j); j = 0,1,2, ..., n-1, n of theequation coefficients are real rational number (can be extended to thecorresponding real and imaginary),

The first term (n degree term)coefficient is 1, if not for 1, each terms of this equation will divide by it.

Transformation: y = x + a (n-1) / (n-1) may also be used,substituting into the original equation, the original equation simplifies to y^ (n-1) coefficients b(n-2) = 0, of the form:

y ^ n + b(n-2) y ^ (n-2) + b (n-3) y ^ (n-3) + ... + b (1) y +b0 = 0,           (1.2)

Thus, the equation (1.2), can be expressed only with n-1 arbitrary coefficients.

But take note:

Only the coefficients by corresponding parameters algebraicexpression, equation is arbitrary, if it is a value, the equation is not arbitrary.

All numerical solution of equations can not be the formula, evenwith radical, because coefficients are not algebraic parameters, it can not becalled radical solution.

Only having no rational factor, the equations just be irreducible,otherwise, it can be solved resepacatively due to its each much lower degree factortypes.

2. Solving algebraic equations using the relationshipequations of their each coefficient and each root

The n degree equation (1.1), (1.2), have n roots, x[1] + x[2] + x[3]+ ... + x[n],  y[1] + y[2] + y[3] + ... +y[n]. The roots and the coefficients are respectively following relationships:

For the equation (1.1) as:

-(x[1]+x[2]+x[3]+...+x[n])=a(n-1),                              (2,1)

x[1](x[2]+x[3]+...+x[n])+x[2](x[3]+x[4]+...+x[n])+...+x[n-2](x[n-1]+x[n])+x[n-1]x[n]

=a(n-2),                                                 (2.2)

-(x[1](x[2](x[3]+x[4]+...+x[n])+x[3](x[4](x[5]+x[6]+...+x[n])+...+x[n-1]x[n])

+x[2](x[3](x[4]+x[5]+...+x[n])+x[4](x[5](x[6]+x[7]+...+x[n])+...+x[n-1]x[n])+...

+x[n-2]x[n-1]x[n])=a(n-3),                                    (2.3)

  .........

-x[1]x[2]x[3]...x[n]=a(0),                                      (2.n)

(When n is odd, the left is negative, when n is even, theleft-hand side are positive)

They are n un-dependent equations.

For equation (1.2), only need to: replace x by y; change a by b; andnote b(n-1)=0.

From equation (2.1), ie:

x[1] = - (x[2]+x[3]+...+x[n]+a(n-1)),                              (1 *)

Takng the n above equations only x[1] as the variable, the others ofx[j]; j = 2,3, ..., n, and the equation coefficients as parameters, thus the n notmutual dependent algebraic equations can be eliminated x [1] with each two ofthese equations, the new equations of number as combination number c (n, 2) =n(n-1) / 2.,with x[2] as variable, the others x[j]; j = 3,4, ..., n, and the equationcoefficients of these equations as parameters.

In these n(n-1)/2 equations, the degree of variable x[2] will besome increased.

When n> 3, elimination x[1], the number of combinations c(n, 2) =n(n-1)/2 is greater than 6, elimination of x [j]; j> 1, will obtain morecombination number.

Elimination x [1], we can choose the equations that are mutually un-dependentyor each other and containing all the coefficients of this equation. The simultaneousequations, successive descending degree until to 1 degree of x[2], we get the x[2]that expressed by x[j]; j = 3, 4, ..., n, and the coefficients of equationparameters.

Similarly, successiveelimination x [j]; j =, 3, ..., n-1, successive descending until to 1degree of x[j+1] , we get x [j +1].that is, too: expressed by x[k]; = j +2, j +3,, ..., n,and the coefficients of equation parameters.

   Where, the degreeof x[j +1] may be high, however, because the roots are all able to meet theoriginal equation, i.e., there must be:

x [k] ^ n + {a [j] ^ jx [k] ^ j; j = 0 ton-1 summation} = 0, k = 1,2, ... n, that is:

x [k] ^ n =- {a [j] ^ jx [k] ^ j; j = 0 to n-1 summation}, k = 1,2, ... n,

Thus, all the high-order x [k] are only up to n-1degree of theexpression.

And the equations of variable x [k], when n> 3, they all get alot of combination equations, and which are all only up to the n-1 degree andthe corresponding simultaneous equations, and have at least two mutualun-depandent, and will be able to successive descending degree to1 degree ofx[k-1] and can be successive eliminated until to get x [n] that is only expressedby the parameters of equation coefficients.

Then putting x[n] successively into x[k]; = n-1,n-2, ..., j, we getthe x[j-1]; j = n, n-1,. .., 2, that all expressed by only the rational functionof equation coefficients.

That is: The rational solution of arbitrary n degree irreducible arithmeticequation expressed only by resonal function of equation coefficients.

Moreover, these solutions are simply not introduce any radical, itsonly rational arithmetic expression of the coefficient formula solution, butalso are not contradictory with the correct understanding of Galois’s theory.

2.1. Arbitrary 2 degree irreduciblealgebraic equation solution

When n = 2, ie equation:

x ^ 2 + a1x + a0 = 0,                                          (2.1.1)

Its two roots x1, x2, and coefficients, have only following twoequations:

x1 + x2 =-a1,                                              (1)

x1x2 = a0,                                                 (2)

Because only two equations, it is not enough to use as beforesuccessive descending, elimination approach solving.

But only by (1) the square:

(x1 + x2) ^ 2 = x1 ^ 2 +2 x1x2 + x2 ^ 2 =a1 ^ 2,           (3)

From (3), (2):

(x1-x2) ^ 2 = x1 ^ 2-2x1x2 + x2 ^ 2 = a1 ^2-4a0, get:

x1-x2 = (a1 ^ 2-4a0) ^ (1/2)                               (4)

Introducting 2 degree radical contained parametric equationcoefficients, then by (1) and (4), that the solution was radical solution:

x =-a1 / 2 + ((a1 / 2) ^ 2-a0) ^ (1/2),

= -a1/2- ((A1 / 2) ^ 2-a0) ^ (1/2),                                            (2.1.2)

     Typically, this solution could be plural.

If ((a1 / 2) ^ 2-a0) ^ (1/2) is a rational number, then the equationis reducible.

The solving method of the equation, in fact, is changing theequation to move and rotate its roots in the complex plane, to be able to solvethe derived solution.

Because only introduction 2 degree radical, of course, it does notviolate properly understood Galois’s theory.

When a1 = a0 = 1, (2.1,1) as follows:

x ^ 2 + x +1 = 0,                                                                (2.1.1')

Solutions are:

x = -1/2- or + ((1/2) ^ 2-1) ^ (1/2)) = (-1- or + i3 ^ (1/2)) / 2 = w1, w2,      (2.1.2')

2.2. Radical solution of arbitrary 3 degreeirreducible algebraic equations

When n = 3, equation:

x ^ 3 + a2x ^ 2 + a1x + a0 = 0,                                      (2.2.1)

Seasonal: y = x + a2 / 3, x = y-a2 / 3 into the original equation, aboutthe original equation simplifies to the coefficients of y ^ 2 = 0, the form:

y ^ 3 + b1y + b0 = 0,                                                (2.2.2)

Where, b1 = a1-a2 ^ 2/3, b0 = a0-a2a1 / 3 +2 a2^ 3/27,

Equation has 3 roots, y1, y2, y3, and are:

- (Y1 + y2 + y3)= 0,  y1 (y2 + y3) + y2y3 = b1,  -y1y2y3 = b0,

Although successive descending degree as before, and seek andeliminate y1, y2, but get the three y3 equations are the same, and can not asbefore successive descending degree solving.

Which may use the two roots of x ^ 2 + x +1 = 0, w1=(-1-i3^(1/2))/2,w2=(-1+i3^(1/2))/2, the three roots of y respectively expressed as w1, w2, andthe two parameters z1, z2,:

y0 =z1+z2, y1=w1z1+w2z2, y2=w2z1+w1z2,while according to equation between roots and coefficients, there:

-(z1+z2+w1z1+w2z2+w2z1+w1z2)=0,

b1=(z1+z2)(w1z1+w2z2)+(w1z1+w2z2)(w2z1+w1z2)+(w2z1+w1z2)(z1+ z2)

=-3z1z2, namely:

z2=-b1/(3z1),                                   (1)

b0=-(z1+z2)(w1z1+w2z2)(w2z1+w1z2)

=-(Z1 ^ 3+z2 ^ 3), namely:

z1^ 3+z2^3=-b0,                                 (2)

Simultaneous (1)、(2), that solution was:

z1^6+b0z1^3-(b1/3)^3 = 0,

z1^ 3=(-b0+(b0^2+4(b1/ 3)^ 3)^(1/2))/2

=-b0/2 +((b0/2)^2+(b1/3)^3)^(1/2),

z1=(-b0/2+((b0/2)^2+(b1/3)^3)^(1/2))^(1/3),

z2=(-b0/2-((b0 2)^2+(b1/3)^3)^(1/2))^(1/3),

    Namely introducting 3degree radicalcontained the equation coefficients, and the solution is:

y0=(-b0/2+((b0/2)^2+(b1/3)^3)^(1/2))^(1/3)+(-b0/2-((b0/2)^2+(b1/3)^3)^(1/2))^(1/3),

y1=w1(-b0/2+((b0/2)^2+(b1/3)^3)^(1/2))^(1/3)+w2(-b0/2-((b0/2)^2+(b1/3)^3)^(1/2))^(1/3),

y2=w2(-b0/2+((b0/2)^2+(b1/3)^3)^(1/2))^(1/3)+w1(-b0/2-((b0/2)^2+(b1/3)^3)^(1/2))^(1/3),

Then, the solution is arbitrary 3 degree radical solution of arbitrary3 degree irreducible algebraic y equation.

The arbitrary 3 degree radical solution of x equation:

xj = yj-a2 / 3; j = 0,1,2,

   Solution of the equation, in fact, is the root of the equation changesto make it move and rotate in the complex plane, to be able to solve thederived solution.

Because only introducting 3 degree radical containing equationcoefficients, it does not violate properly understood of course, Galois’stheory.

2.3. Radical solution of arbitrary 4 degreeirreducible algebraic equations (solution presented in this paper is benefit forextended to n = 2m)

When m = 2; n = 4, the (2.1) equation:

x^ 4+a3x^3+a2x^2+a1x+a0 = 0,                                         (2.3.1)

Now, consider the introduction of the corresponding radicalsimplification of solution method. Introducting a function y, and take:

(x^2+a1x/2+y/2)^2 =x^4+a3x^3+((a3/2)^2+y)x^2+(a3y/2)x+(y/2)^2

Original equation can be rewritten as:

(x^2+a3x/2+y/2)^2=((a3/2)^2-a2+y)x^ 2+(a3y/2-a1)x+(y ^ 2/4-a0) ,

When the right-hand side set to become entirely a function of xsquared:

((A3 / 2) ^ 2-a2+ y) x ^ 2 + (a3y/2-a1) x + (y ^ 2/4-a0) = (c1x + c0) ^ 2

By the (c1x + c0) ^ 2 = c1 ^ 2x ^ 2 +2 c1c0x+ c0 ^ 2, are:

c1^2 =(a3/2)^2-a2+y, 2c1c0 =a3y/2-a1, c0^2 = y^ 2/4-a0,c1=((a3/2)^2-a2+y)^(1/2),

c0=(y^2/4-a0)^(1/2),2c1c0 = 2((a3/2)^2-a2+ y)^(1/2) (y^2/4-a0)^(1/2)=a3y/2-a1,

((A3/2)^2-a2+y)(y^2/4-a0)=(a3y/2-a1)^ 2/4,which is obtained:

y^3+(3(a3/2)^2/4-a2)y^2+(a3a1-4a0)y-4((a3/2)^2-a2)a0-a1^2= 0,       (2.3.2)

Seasonal: s =y+((a3/2)^ 2/4-a2/3), y=s-((a3/2)^ 2/4-a2/3),          (2.3.3)

To (2.3.3) into equation (2.3.2),ie, it is reduced to the coefficients of the form s^2=0,.

Namely:

s ^ 3 + b1s+ b0 = 0,                                                (2.3.4)

b1 =-3a3 ^ 4/256 + a2a3 ^ 2/8 + a2 ^ 2 + a3a1-4a0,

b0=2a3^6/16^3-a2a3^4/128-a1a3^3/16+(a2^2/8-3a0/4)a3^2+a1a2a3/3-2a2^3/27+8a0a2/3-a1^2,

The solusions are:

s0=(-b0/2+((b0/2)^2+(b1/3)^3)^(1/2))^(1/3)+(-b0/2-((b0/2)^2+(b1/3)^3)^(1/2))^(1/3),

s1=w1(-b0/2+((b0/2)^2+(b1/3)^ 3)^(1/2))^(1/3)+w2(-b0/2-((b0/2)^2+(b1/3)^3)^(1/2))^(1/3),

s2 =w2(-b0/2+((b0/2)^2+(b1/3)^3)^(1/2))^(1/3)+w1(-b0/2-((b0/2)^2+(b1/3)^3)^(1/2))^(1/3),

(2.3.5)

And for 3 degree y equation (2.3.2),the solutions are:

yj = sj + ((a3 / 2) ^ 2 + a2) / 3; j =0,1,2,                            (2.3.6)

The original equation can be expressed as two x equations:

(x^ 2 + a3x / 2 + y / 2) ^ 2 = (c1x + c0) ^2, that is:

x ^ 2 + a3x / 2 + y / 2 + or - (c1x + c0) =0, namely:

x ^ 2 + (a3 / 2 +(a3 ^ 2/4-a2 + y) ^ (1/2)) x + y / 2 + (y ^ 2/4-a0) ^ (1/2)) = 0,

x ^ 2 + (a3/2- (a3 ^ 2/4-a2 + y) ^ (1/2)) x+ y/2- (y ^ 2/4-a0) ^ (1/2)) = 0,        (2.3.7)

Substituting solution obtained in (2.3.6),y roots, into equation (2.3.7), respectively, to solve these two x equations(due to the results of the roots are the same, so only to y0 into), namely

We get the four roots of 4 degree equation(2.3.1).

Which, in s equations, introducing 2, 3 degree radical; in 2 xequations, introducing 2 degree, of course, there no higher degree radical, it doesnot violate properly understood Galois theory.

2.4. Radical solutions of arbitrary 6 degreeirreducible algebraic equations (innovation of this paper)

When m = 3; n = 6, (2,1) equation:

x^6+a5x^5+a4x^4+a3x^3+a2x^2+a1x+a0=0，                        (2.4.1)

Introduce two functions y1, y2, and take:

(x^3+a5x^2/2+(a4+y1)x/2+y0/2)^2=x^6+a5x^5+(a5^ 2/4 +(a4+y1))x^4

+(a5(a4+y1)/2+y0)x^3+(a5y0/2+(a4+y1)^ 2/4)x^2+(a4+y1)y0x/2+y0^2/4,

Original equation can be rewritten as:

(x^3+a5x^2/2+(a4+y1)x/2+y0/2)^2=(a5^2/4+y1)x^4+(a5(a4+y1)/2-a3+y0)x^3

+((a4+y1)^2/4-a2+a5y0/2)x^2+((a4+y1)y0/2-a1)x+(y0 ^ 2/4-a0),

When the right-hand side set to become entirely a function of xsquared:

The original equation (x^3+a5x^2/2+(a4+y1)x/2+y0/2)^2=(c2x^2+c1x+c0)^2, can be broken down into the following two equations:

x^3+(a5/2+c2)x^2+((a4+y1)/2+c1)x+(y0/2+c0)=0;

x^3+(a5/2-c2)x^2+((a4+y1)/2-c1)x+(y0/2-c0)=0,                (2.4.2)

And has:

(a5^2/4+y1)x^4+(a5(a4+y1)/2-a3+y0)x^3+((a4+y1)^2/4-a2+a5y0/2)x^2+((a4+y1)y0/2-a1)x

+(y0^2/4-a0)=(c2x^2+c1x+c0)^2=c2^2x^4+2c2c1x^3+(c1^2+2c2c0)x^2+2c1c0x+c0^2,

As a result, each solution can be obtained:

c2 = a5 / 2, 2c2c1= a5 (a4 + y1) / 2-a3 + y0, c1 ^ 2 +2 c2c0 = (a4 + y1) ^ 2/4-a2 + a5y0 / 2,

2c1c0 = (a4 + y1) y0/2-a1, c0 ^ 2 = y0 ^ 2/4-a0,

From the 1st, we have: c2 =a5/2, puttng c2into the 2nd, we obtain: c1=(y0-a3)/a5+(a4+y1)/2,

Puttng c1 into 4, we obtain: c0 = ((a4 +y1) y0/2-a1) / (2 (y0-a3) / a5 + (a4 + y1)), that is:

c2=a5/2,c1=(y0-a3)/a5+(a4+y1)/2, c0=((a4+y1)y0/2-a1)/(2(y0-a3)/a5+(a4+y1)),   (2.4.3)

Pottng c2, c1, c0 into 3, we obtain:

(y0-a3)^2/a5^2+(y0-a3)(a4+y1)/(2a5)+(a4+y1)^2/4+a5((a4+y1)y0/2-a1)/(2(y0-a3)/a5+(a4+y1))

-(a4+y1)^2/4+a2-a5y0/2 = 0, namely:

2y0^3/a5^3+(-6a3/a5^3+2a4/(a5^2)-1+2y1/(a5^2))y0^2

+(2a2/a5+4a3^2/a5^3+a3^2/a5^3-4a3a4/(a5^2)+a3+(-4a3/(a5^2))y1)y0

-a1a5-2a2a3/a5+a2a4-2a3^3/a5^3+2a3^2a4/(a5^2)-a4^3/4+(a2+2a3^2/(a5^2)-3a4^2/4)y1

-3a3y1^2/(4a5)-3a4y1^2/4-y1^3/4=0,                                    (1 ')

Puttng c0 into 5, we obtain:((a4+y1)y0/2-a1)^2/(2(y0-a3)/a5+(a4+y1))^2-y0^2/4+a0=0 ,   namely:

-4y0^4/a5^2+(-4a4/a5+8a3/a5^2-4y1/a5)y0 ^3

+(a4^2/4+4a0/a5^2+(-3a4/2+4a3/a5)y1-3y1^2/4))y0^ 2

+(4a0a4/a5-8a0a3/a5^ 2-a1a4 + (4a0/a5-a1) y1) y0

+a1^2+4a0a3^2/a5^2-4a0a3a4/a5+a0a4^2+ 2a0a4-4a0a3/a5)y1+a0y1^ 2)=0,    (2 ')

By (1 '), (2'), the first successivedescending y0 to 1degree, that get y0 expressed by y1 and ai; i = 0 to 5, thus,eliminating y0, to get the 5 degree y1 algebraic equation expressed by ai ; i =0 to 5. The whole operation process according to the second and third sectionsis carried out similarly.

Solving this 5 degree y1 algebraic equation, we get y1 expressed byai; i = 0 to 5.

Then putting1y1 into, we get y0 expressed by ai; i = 0 to 5.

Thus, they are substituted into (2.4.3),we obtain c2, c1, c0, expressed by ai; i = 0 to 5.

Then they are substituted into (2.4.2),the two 3 x equation are gained, and to solve them, ie: radical solution of 6 degreeirreducible algebraic equation.

Of course, this 6 degree solution is much more convenient than thedirect solution.

Among them, in the 2 solutions of m degree x equations, introducing 2and 3 degree radicals but no higher order radicals. Of course, it does not violatethe right understand Galois’s theory..

2.5. Radicalsolutions of 2m(m> 2) degree irreducible algebraic equations (innovation of this paper)

Analog 4, 6 degree irreducible radical solution of algebraic equationsthe solution, for all the even higher-order equation we can introduce functons yi;i = 0,1, ..., m-2, and constants cj; j = 0,1, ..., m-1, to rewrite the original2m (m> 2) equation as:two m degree x equations with y.

And solving toget the expressions of m+1 lower degree yi; i=0,1,…,m-2，equations and constants cj; j = 0,1, ..., m-1, expressed only by thethe equation coefficients. Substituting these into two m degree x equations andsolved separately, ie: 2m rootsof original 2m degreeequation. And solve the 2m(m> 2) irreducible radical solution of algebraic equations. Of course, thissolusion is much simpler than direct solution of 2m irreducible algebraic equation.

Moreover, since the solution of m> 4 irreducible algebraicequation does not contain any radical, except to the solution of 2m irreducible algebraic equations, allintroduced 2 and 3 degree, radicals but no higher order radical. Of course, itdoes not violate the right to understand Galois’s theory.

3. Galois’s theory and its correctunderstanding

As early as the third century BC, had drawn two irreduciblealgebraic equations radical solution. However, only up to the 16th century AD, Ithas been 3 and 4 degree irreducible radical solution of algebraic equations.They all have introduced 2, 3 degree radical solutions containing parametricequation coefficients.

Although since nearly more than four centuries, many people seek toget n> 4 irreducible radical solution of algebraic equations, but all failedto success.

Furthermore after Galois possible from characteristics of equationgroups introducted radical in the process of solving equations, discussed andgiven the criterion of algebra equations radical solutions can be obtained, Abel(Abel, NN 1830) Accordingly, first proposed n> 4 irreducible algebraicequation can not be solved radical, it seems to have been accepted in academia thatn> 4 irreducible algebraic equation has no radical solutions [1].

For n> 4 irreducible algebraic equations, the only on the basisof detailed analysis of ther "solution" positions on the number field,numerically approximation, or introduce some special function to solve.

This of course gave many practical problems and theoretical work inconvenience.

In fact, the detailed analysis of Galois theory [2] [3], can indeedprove: equation solvability radical solution is corresponding to the transformationgroup solvability of each equation coefficient for rational calculation and addthe corresponding radical successively, and the order of this transformationgroup equal to the maximum index of added radical in its entire solving process,and when the order of this transformation group> 4 symmetry permutationgroup, and its subgroups, are all the non-abelian simple group, they are all un-solvable.

Therefore, Galois’s theory can prove, but: "When the wholeprocess of solving equations add radicals maximum degree n *> 4, the generalirreducible algebraic equation has no radical solutions."

Obviously, the entire process of solving the maximum index addedradical, n *, not the number of degree of the solution equation n, according toGalois theory, have not completely that the greatest degree radical, n * addedin whole solved process is equal to the solution equation degree n, or them both,under any relationship. Abel also failed to give any basis.of n> 4irreducible algebraic equation would be no radical solutions. Thus, so far,seems to have recognized "n> 4 irreducible algebraic equation has noradical solution," concluded only when n * is equal to n, can be obtained.

However, n * does not have to equal n, if n * always maintained lessthan 4, for example, a variety of solution used in this paper as before and followscan be able to seek radical solution of arbitrary n degree irreduciblealgebraic equations.and does not contradict properly understood Galois theory.

4. Solving 4 degree irreducible algebraicequations with the successive descending by the relationship of coefficientsand roots of equation.

x^4 + a3x^3 + a2x^2 + a1x + a0 = 0,                         (4.1)

According to of the relationships of the equationcoefficients and roots, there are:

x1 + x2 +x3 + x4 =-a3,                                   (1)

x1 = - (a3+ x2 + x3 + x4),                                    (a)

x1 (x2 + x3+ x4) + x2 (x3 + x4) + x3x4 = a2,             (2)

x1 (x2 (x3 +x4) + x3x4) + x2x3x4 =-a1,                   (3)

x1x2x3x4 = a0,                                           (4)

Thus, to solve and eliminate orderly and successively each root, anduse that all roots satisfy the original equation, the magnitudes of equal or higherthan fourth power roots can all be expressed by that less than fourth power .That is:

x2=-(x3^3x4^3-x3^2(x4^2a2+x4a1+a0)-x3(x4^2a1+x4a0)-x4^2a0)

/(x3^3x4^2+x3^2(x4^3+x4^2a3)-x3(x4a1+a0)-x4a0),          (b)

x3 = (x4 ^ 2a3 ^ 2 +2 a0) / (x4 ^ 2a3)                        (c)

x4 =-A(5,0) / A (5,1),                                        (d)

   Of which:

A (5,1) = A (4,2) A (3,1)-A (3,2) A (4,1),

A (5,0) = A(4,2) A (3,0)-A (3,2) A (4,0),

A (4,2) = A(3,2) A (2,2)-A (2,3) A (3,1),

A (4,1) = A(3,2) A (2,1)-A (2,3) A (3,0),

A (4,0) = A(3,2) A (2,0),

A (3,2) = A(2,3) A (1,2) - A (1,3) A (2,2),

A (3,1) = A(2,3) A (1,1)-A (1,3) A (2,1),

A (3,0) = A(2,3) A (1,0)-A (1,3) A (2,0),

A (2,3) = A(1,3) a3-A (1,2),

A (2,2) = A(1,3) a2-A (1,1),

A (2,1) = A(1,3) a1-A (1,0),

A (2,0) = A(1,3) a0,

A(1,3)=-a3^9-a3^7a2-3a3^6a1+a3^5(a2^2+3a0)+a3^4a2a1-2a3^3a2a0-3a3^2a1a0+4a3a0^2,

A(1,2)=-a3^8a2+a3^7a1-a3^6(2a2^2+a0)-a3^5a2a1+a3^4(4a2a0+a1^2)-3a3^3a1a0

+14a3 ^2a0^2,

A (1,1)=-a3 ^ 8a1 + a3 ^ 7a0-2a3^ 6a2a1 + a3 ^ 5 (a2a0-2a1 ^ 2) +5 a3 ^ 4a1a0-3a3^ 3a0 ^ 2,

A (1,0)=-a3 ^ 8a0-2a3 ^ 6a2a0-2a3^ 5a1a0 +4 a3^ 4a0 ^ 2 +8 a0 ^ 3,

They are all the functions of the equation coefficients.

Then putting (d) sequentially into (c), (b), (a), we have the solutions.

     Specific show: For more than 4 degreeirreducible algebraic equations have not introduced any radical and directly successivedescending to derive solutions by.the relationship of equation coefficients androots.

5. Solving 5 degree irreducible algebraicequations with the successive descending by the relationship of coefficientsand roots of equation.

Arbtrary 5 degree irreducible algebraic equation:

x^5+a4x^4+a3x^3+a2x^2+a1x+a0 = 0,                        (5 .1)

It can be converted to: 4th degree coefficient = 0, by theconversion y = x + a4 / 5 as the following form:

y^5 + b3y^3 + b2y^2 + b1y + b0 = 0,                      (5.1 ")

Its 5 roots and coefficients have the following relationship:

y0 + y1 + y2 + y3+ y4 = 0,                                                         (1)

y0 = - (y1 + y2 +y3 + y4),                                        (a)

     Namely: y0 can be expressed by y1, y2, y3,y4.

y0 (y1 + y2 + y3+ y4) + y1 (y2 + y3 + y4) + y2 (y3 + y4) + y3y4-b3 = 0,            (2)

y0 (y1(y2+y3+y4)+y2(y3+y4)+y3y4)+y1(y2(y3+y4)+y3y4)+y2y3y4+b2=0,                   (3)

y0 (y1y2 (y3 +y4) + y1y3y4 + y2y3y4) + y1y2y3y4-b1 = 0,                            (4)

y0y1y2y3y4 + b0 = 0,                                                               (5)

(2), (3), (4), (5) are all algebraic equations un-depanded eachother and only expressed of y1, y2, y3, y4, and equation coefficients.

When taking y1as variable, y2, y3, y4, and the equation coefficientsas parameters, whereby by these 4 algebraic equations, successive descendinguntil:

y1=(2y2(y3^3y4^2b0+y3^2y4^3b0)+2y3^3y4^3b0)

/(y2^2y3^4y4^4-y2(y3^3(y4^3(b2+b0)+y4^2b1+y4b0)+y3^2(y4^3b1+3y4^2b0)+y3y4^3b0)

-(y3^3(y4^3b1+3y4^2b0)+y3^2y4^3b0)),                      (b)

That is: y1 expressed only by the y2, y3, y4, and the equationcoefficients.

Thus we get 4 algebraic equations.only 3 variables of y2, y3, y4.

When taken y2 as variable, y3, y4, and the equation coefficients asparameters, whereby selectng 2 mutual un-depandent algebraic equations,successive descending until:

y2=(y3^4+y3^3y4+y3^2(y4^2+b3)+y3(y4^3+y4b3+b2)+y4^4+y4^2b3+y4b2+b1)

/(y3y4^2),                                                (c)

That is, too: y2 expressed only by the y3, y4, and each equation coefficient.

Thus we get 3 algebraic equations only 2-variables of y3, y4.

When taken y3 as variable, y4 and the equation coefficients as parameters,thereby selecting two mutual un-depandent algebraic equations, successivedescending until:

y3=-(y4^8+2y4^6b3+2y4^5b2+y4^4(b3^2+b1)+y4^3(2b3b2+b0)+y4^2(b2^2+b1b3)

+y4(b1b2+b3b0)+b2 b0)

/(y4^7+2y4^5b3+y4^4b2+y4^3(b3^2+b1)+y4^2(b2b3+b0)+y4b1b3+b3b0)

And in this process, has been: y4^5+y4^3b3+y4^2b2+y4b1+b0=0,        (1 ")

(In fact, that shows: The equation root y4 satisfy the originalequation)

Leaving the formula simplifies to:

y3=-(y4 ^ 3+y4b3+b2)/(y4^2+b3),                                     (d)

(Specifically indicate:  Thedegree of y4 higher than 3 can all be expressed by the terms not more than 3)

That is, too: y3 expressed only by the y4, and the equationcoefficients.

By the (1 "), (d), we get two un-depangent algebraic equations thatonly y4 is variable.

When taking y4 is variable, equation coefficients for theparameters, the two mutual un-depandent algebraic equations will inevitably besimultaneous successively descending until 1-degree of y4, derived solution:

y4B (1,1) + B (1,0) = 0,                                                 (2")

Wherein,

B (1,1) = B '(2,2) B (2,1) - B (2,2) B'(2,1),

B (1,0) = B '(2,2) B (2,0) - B (2,2) B'(2,0),

B '(2,2) = (B (2,2) B (3,2)-B (3,3) B(2,1)),

B '(2,1) = (B (2,2) B (3,1)-B (3,3) B(2,0)),

B '(2,0) = B (2,2) B (3,0),

B (2,2) = (B '(3,3) B (3,2) - B (3,3) B'(3,2)),

B (2,1) = (B '(3,3) B (3,1) - B (3,3) B'(3,1)),

B (2,0) = (B '(3,3) B (3,0) - B (3,3) B'(3,0)),

B '(3,3)=(27b3^11-18b3^9b1+39b3^8b2^2+b3^7(-39b2b0-15b1^2)+18B3^6b2^2b1

+b3^5(+6b1^3+10b2^4)+b3^4(-25b2^3b0+b2^2b1^2)

+B3^3(6b2^3b2b1+12b2^2b0^2+3b1^4+8b1b2^4)+B3^2(2b2^2b1^3+4b2^6)

-4b3b2^5b0+2b3b2^4b1^2+2B2^5b2b1+4b2^4b0^2+b2^2b1^4)

B '(3,2)=(27b3^10b2-9b3^9b0+b3^7(30b2^3-15b1b0)+B3^6(-24b2^2b0+9b2b1^2)

+b3^5(3b1^2b0+12b1b2^3+3b2b0^2)+B3^4(10b2^5-5b2^2b1b0)

-b3^3(6b2b1b0^2+24b2^3b1^2+7b2^4b0+6b1^3b0)

+ B3^2(4b2^5b1+3b2^2b1^2b0+b2^3b0^2)+b3(b2^7-2b2^4b1b0)-2b2^6b0+5b2^5b1^2

+2b2^3b1b0^2-2b2^3b1b0^2+2b2^2b1^3b0),

B '(3,1)=(27b3^10b1+b3^8(9b2b0-b1^2)+b3^7(30b2^2b1-9b0^2)+B3^6(-9b1^3)

+b3^5(6b1 ^ 2b2 ^ 2 +3 b1b0 ^ 2 +6 b2 ^ 3b0)+B3^4(10b2^4b1-6 ^2b0^2)

+b3^3(+12b2^2b1^3-3b1^2b0^2+6b2b0^3)

+B3^2(b2^5b0+3b2^4b1^2+b2^2b1b0^2)+b3(b2^6b1-b2^4b0^2)

+3B2^4b1^3+2b2^3b0^3+b2^2b1^2b0 ^ 2).

B '(3,0)=27b3^10b0-9b3^8b1b0+30b3^7b2^2b0+B3^6(-9b1^2b0-18b2b0^2)+6b3^5b1b2^2b0

+10B3^4b2^4b0+b3^3(-7b2^3b0^2+6b2^2b1^2b0)+3B3^2b2^4b1b0+b3b2^6b0

-2b2^5b0^2+3b2^4b1^2b0,

B(3,3)=(9b3^7-3b3^5b1+7b3^4b2^2-b3^3(6b2b0+3b1^2)+3b3^2b2^2b1+b3b2^4

-2b2^3b0+3b2^2b1^2),

B (3,2) = (9b3 ^ 6b2-3b3 ^ 5b0 + b3 ^ 4b2b1+ b3 ^ 3 (3b2 ^ 3-6b1b0 +3 b2 ^ 3)

+ B3 ^ 2 (b2 ^ 2b0 +3 b2b1 ^ 2) + b2 ^ 5 +2 b2 ^ 2b1b0 +2 b2b1 ^ 3),

B (3,1) = (9b3 ^ 6b1 + b3 ^ 4b2b0 + b3 ^ 3(6b2 ^ 2b1-3b0 ^ 2),

+4 B3 ^ 2b2b1b0 + b2 ^ 4b1-b2 ^ 2b0 ^ 2 +4 b2b1 ^ 2b0)

B (3,0) = 9b3 ^ 6b0 +6 b3 ^ 3b2 ^ 2b0 + b3^ 2b2b0 ^ 2 + b2 ^ 4b0 +2 b2b1b0 ^ 2,

They are just the rational function of the equation coefficients.

     While:

y4 = - B (1,0) / B (1,1),                                               (e)

That is, too: y4 expressed only by the equation coefficients.

Puttng (e) successively into (d), (c), (b), (a), we get all fiveequation solutions of y4, y3, y2, y1, y0 those are only expressed by the rationaloperation of the equation coefficients.

     For any arbitrary 5 degree irreduciblealgebraic equations, only needed puttng their each coefficient into (e), (d),(c), (b), (a), all forms, namely obtaining their solutions.

This specificly gives arbitrary 5 degree irreducible algebraicequation formula solutions. Moreover, these solutions are simply not introducedany radical, but only the formula

solutions expressed by rational arithmeticof equation coefficients, also are not contradictory with the correctunderstanding of Galois theory.

6. The solvng equation methods of introductingappropriate number parameters to make it expressed as the product of two factors(innovation of this paper)

When n> 1, by transformation: y = x + a[n-1]/n, to make (2.1) as:

(y-a[n-1]/n)^n+{a[j]^j(y-a[n-1]/n)^j;j = 0 to n-1 sumation} = 0,              (6.1 ')

That is, too:

y ^n+{b[j]y^j; j = 0 to n-2 summation } =0,                                   (6.1")

Where, b[n-1] = 0,

When n is even, may be introduced n parameters: s[0], s[1], ...,s[n/2-1], t[0], t[1], ...,

t[n/2-1], the equation is rewritten as:

(y^(n/2)+s[n/2-1]y^(n/2-1)+... +s[1]y+s[0])(y^(n/2)+t[n/2-1]y^(n/2-1)+ ... +t[1]y+t[0])=0,

And there are:

s [n/2-1] t[0] + t [n/2-1] s [0] = 0,

s [n/2-2] t[0] + t [n/2-2] s [0] + s [n/2-1] t [1] + t [n/2-1] s [1 ] = b [n/2-2],

s[n/2-3]t[0]+t[n/2-3]s[0]+s[n/2-2]t[1]+t[n/2-2]s[1]+s[n/2-]t[2]+t[n/2-1]s[2]=b[n/2-2],

.........

s [0] t [0] = b [0],

When n is odd, can be introduced (n +1) parameters: s[0], s[1], ...,s[(n+1)/2], t[0], t[1], ...,

t[(n+1)/2-1],, the equation is rewritten as:

(y^((n+1)/2)+s[(n+1)/2-1]y^((n+1)/2-1)+... +s[1]y +s[0])(y^(n(n+1)/2)+t[(n+1)/2-1]y^((n+1)/2-1) + ... +t[1]y+t[0])=0,  And there are:

s[(n+1)/2-1]t [0] = 0,

s[(n+1)/2-2]t[0]+t[(n+1)/2-2]s[0]+s[(n+1)/2-1]t[1]+t[(n+1)/2-1]s[1]= b[(n+1)/2-1],

s[(n+1)/2-3]t[0]+t[(n+1)/2-3]s[0]+s[(n+1)/2-2]t[1]+t[(n+1)/2-2]s[1]+s[(n+1)/2-]t[2]

+t[(n+1)/2-1]s[2]=b[(n+1)/2-2],

.........

s[0]t[0]=b[0],

   That each have thesame number of equations and parameters, and each parameter can be in order to obtainsolution expressed by the after parameters and the two equations of the finalparameter can be obtained the solution, for higher degree, the equations can bealternately descending derived to get solution, thus obtaining the solution ofeach parameter. For example:

6,1, 2 degree quation: x^2 +a1x +a0 = 0,

Introducting s, t, two parameters to express the equation as:

(x + s) (x + t) = 0, There are:

s + t = a1, st = a0,

(s + t) ^ 2 = s ^ 2 +2 st + t ^ 2 = a1 ^ 2,

a1 ^ 2-4a0= s ^ 2-2st + t ^ 2 = (s-t) ^ 2,

s-t = (+, -) (a1 ^ 2-4a0) ^ (1/2), which have solutions:

x1 =-s =(a1 (+, -) (a1 ^ 2-4a0) ^(1/2)) / 2,

x2 =-t =(a1 (-, +) (a1 ^ 2-4a0) ^(1/2)) / 2,

   Still, but also just introduced two times radical. Also it does notcontradict properly understood Galois’s theory.

As a special case:

Equation: x ^ 2 + x +1 = 0, the solutionis:

w1 = (-1 (+, -) i 3 ^ (1/2)) / 2,

w2 = (-1 (-, +) i 3 ^ (1/2)) / 2,

6.2. 3 degree equation: x ^ 3 + a1x + a0 = 0,

Introducting s1, s0, t0, 3 parameters to express the equation as:

(x^ 2 + s1x + s0) (x + t0) = 0, There are:

s1 + t0 = 0, s1 =-t0, s0t0 = a0,

s0 = a0/t0, s1t0 + s0 = a1, s1 = (a1-s0) /t0 = (a1-a0/t0) / t0, t0 ^ 3 + a1t0-a0 = 0,

Adopt (2.2.2), similarly,solution of 3 t0 roots was:

t00=(a0/2+((a0/2)^2+(a1/3)^3)^(1/2))^(1/3)+(a0/2-((a0/2)^2+(a1/3)^3)^(1/2))^(1/3),

t01=w1(a0/2+((a0/2)^2+(a1/3)^3)^(1/2))^(1/3)+w2(a0/2-((a0/2)^2+(a1/3)^3)^(1/2))^(1/3),

t02= w2(a0/2+((a0/2)^2+(a1/3)^3)^(1/2))^(1/3)+w1(a0/2-((a0/2)^2+(a1/3)^3)^(1/2))^(1/3),

Then, the solution is arbitrary 3 degree t0 equation radicalsolution.

The arbitrary 3 degree x equation radical solution is that of:

x =-t0,

     Visible, arbitrary x 2nd and 3rd equation arestill introducing their corresponding radical solution to obtain the formula.And their introduction is not more than three times radical. Also properlyunderstood Galois theory are not contradictory.

6.3 Arbitrary 4 degree irreducible algebraicequation:

x^4+a3x^3+a2x^2+a1x+a0= 0,                                  (6.3.1)

   By the conversion:y = x + a [3] / 4, the (2.3.1) is converted to:

y ^ 4 +b2y^2+b1x+b0= 0,                                      (6.3.1")

Where, b3 = 0, but also the introduction ofparameters s1, s0, t1, t0, (6.3.1") will be writtenas:

(y^2 + s1y + s0)(y^2 + t1y + t0) = 0, are:

s1 + t1 = 0, s1=-t1, s1 = b1 / (t0-b0/t0),

s0t0 = b0, s0 =b0/t0, s0 = b0 / (b2/t0 (+, -) ((b2 ^ 2-b0) ^ (1/2))

s1t0 + s0t1 =b1,-t1t0 + t1b0/t0 = b1, t1 = b1 / (-t0 + b0/t0)

s0 + t0 = b2,b0/t0 + t0 = b2, t0 ^ 2-b2t0 + b0 = 0, t0 = b2 / 2 (+, -) ((b2 ^ 2-b0) ^ (1/2)

Equation is obtained only a function of thecoefficients of the radical expression s1, s0, t1, t0, which, s0, t0 there weretwo different values, but were substituted into the back, the results should bethe same.

Each solution was:

y ^ 2 + s1y + s0 = 0,

y ^ 2 + t1y + t0 = 0,

Two solutions of the two equations are 4 degree equation 4 radicalsolutions only expressed by function of the coefficients.

   Among them, theradical index of only two. Also not contradict with properly understood Galoistheory.

6.4. y^ 5 + b3y ^ 3 + b2y ^ 2 + b1y + b0 =0,                   (6,4.1 ")

  Introduction ofparameters s1, s0, t1, t0, can be (6.4.1") be rewritten as: (are also four equations, four unknowns)

(y^ 3 + s1y + s0)(y ^ 2 + t1y + t0) = 0, there are:

t0 + s1 = b3, t0= b3-s1,

s0t0 = b0, s0 = b0/t0, s0 = b0 / (b3-s1),

s0 + s1t1 = b2, t1s1 = (b2-s0), t1s1 ^ 2 -(t1b3 + b2) s1 + b2b3-b0 = 0,          (1)

s1t0+s0t1=b1, s1t0=b1-s0t1, t1s1^3-2t1b3s1^2+t1(b3^2+b1)s1-(t1b1b3-b0)=0,        (2)

  By the (1), (2)successive descending on s1, we get:

s1=-(b1b3t1^2-(b2b3^2-b0b3+b0)t1-b2^2b3+b2b0)/(-b1t1^2+(b2b3-b0)t1-b2^2),        (3)

(b1b2^2b3^3+(b0b1b2+b0b1)b3^2+b0^2b1b3+b0b1^2b2-b0^2b1)t1^4

+(-b2^3b3^4+(3b1-2b0+b0)b2^2b3^3

-(2b1b2^3+b0b2^2+b0(1+2b1)b2-b0^2(2b2+1))b3^2+(b0^2(-2b2+b1-1)-b0b1b2^2)b3

-b0b1b2^2+b0^2b1(4b2+1))t1^3

+(-3b2^3b3^4+(b2^4+5b0b2^2)b3^3-(3b0b2^3+3b0^2b2+4b0b2^2)b3^2

-(2b1b2^4-b0b2^3-2b0^2b2-b0^3)b3+3b0b1b2^3-b0^2b2^2+2b0^3b2-b0^2b2)t1^2

+(-3b2^4b3^3+5b0b2^3b3^2-(4b0b2^4+2b0b2^3+3b0^2b2^2)b3

-b0b2^4+2b0^2b2^3+4b0^2b2^2+b0^3b2)t1

-(2b2^6b3-2b0b2^5)=0,                                                            (4)

By the (4) according to solving 4 degree t1 equation, to get t1, or(4) and (3), elimination s1, get another t1 equation, and wth (4) successivedescending, to get the solution of t1, and seek various other parameters, andpress 3 degree and 2 degree x equation, solve 5 degree x equation.

It only contains 2 degree and 3 degree radical. Also does notcontradict properly understood Galois’s theory.

6.5. Any high degree irreducible algebraicequation radical solutions (innovation of this paper)

In summary, the 2-4 degree irreducible algebraic equation radicalsolutions have been gained early, this article has used a variety of methodsspecifically: 5 degree and 6 degree irreducible algebraic equation formula solutionsand radical solutions, then by analogy to higher orders, we can solve successivelyany high irreducible algebraic equation formula solutions and radical solutions.

Moreover, they do not contain more than 3 degree radical solutions.Thus, the solution is also not violate proper understanding of Galois theory.

7. References:

[1] Encyclopedia of Mathematics Editorial(Consultant) Su Buqing etc. (Director) Wang Yuan Science Press, 1994, etc.

[2] Basic algebra 1-2 Jacobson, N. Freeman1974-1980

[3] Algebra 1-2BI Van Der Waerden Springer-Verleg 1955-1959

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