# Solvability of algebraic equations

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## Polynomials

Expression of form f(x)=[a,b,c,...]x = a∙xk+ b∙xk−1+ c∙xk−2+ ...+p∙x +q is called a polynomial of k-th degree of one variable (indeterminate) x. Numbers a, b, c .. are coefficients of the polynomial. Assuming that the values of a, b, c .. and x are from any numeric field, considered (numerical) expressions (see chapter 1) expands by other expressions. We can write besides [2,4,3] x = 2² + 4x + 3 also eg. [2,−4,3]x = 2x²−4x+3. (This method is not used now; partly it was included in the Roman enrollment of numbers, for example IV = 5-1 = 4. Roman numeration method had been used already in Egypt).

To obtain the product of f(x) ∙ g(x) is necessary to multiply each member f(x) with each member of g(x) (numbers in the numeral system with basis x are multiplied the same way).
Two polynomials f(x) and g(y) with different variables (as well as two numbers recorded at different bases x and y) can be also multiplied.
For f(x)= 2x²+4x+3 a g(y) = y²+3y+2 is:

``` f(x)∙g(y)│  y²    3y    2
─────────┼──────────────────
2x²    │ 2x²y²  6x²y  4x²
4x     │ 4xy²  12xy  8x
3      │ 3y²    9y   6

h(x,y) = f(x)∙g(y) =
= 2x²y²+ 6x²y+ 4x²+ 4xy²+ 12xy+ 8x+ 3y²+ 9y+ 6
```

The term h (x, y) is called a polynomial of two variables x and y.

### Equations of polynomials

 Abel, Niels Henrik , 1802-1829, Norwegian mathematician, author of the first evidence that the general equation of 5th degree is not algebraically solvable using the finite formulas with roots (y.1824). He was interested in algebra, worked on the theory of elliptic and transcendental functions. Discovered a method for acceleration of convergence of infinite series, developed the theory of integral calculus, proved binomial theorem.

Consider two polynomials P(x) and R(x). Polynomials are algebraically equal, when all their quotients are equal. When their values are equal, we say that the polynomials functionally equal.
In fields with a finite number of elements may be cases where two polynomials with different coefficients (for all possible x) give the same values: Two different algebraic polynomials P(x) = x4 -1 and Q(x) = x4 + x³-x-1 are functionally equal in the field Z3 (x = 0,1,2), as they provide for all x equal value.

### Primitive polynomials

Polynomial, whose coefficients have no common divisor greater than 1, ie. (an,an−1,....a1,a0)=1, is called primitive polynomial.

Když f a g are primitive polynomials, pak f∙g is also primitive polynomial.
For example (3x²+4x+5)(2x+3) = 6x³+14x²+22x+15, where (6,14,22,15) = ((6,14),(22,15)) = (2,1) = 1.

## Algebraic equations

Expression of the form F(x)=0, where F(x)=[a,b,c,...]x is polynomial of k-th degree of one variable x, is called algebraic equation of k-th degree. To solve algebraic equation means to find such values x to given coefficients a,b,c,.. that F(x)=0. Equation may have generally more solutions.

Solution x1,x2,.. are called roots of algebraic equation. Solution of equation F(x)=0 is equivalent to finding divisors of polynomial F(x).

The equations that are not algebraic (e.g. exponential equation, logarithmic, trigonometric, ...) are called transcendent.

 Cantor, Georg Ferdinand [kantor], 1845-1918, German mathematician. He was interested in the theory of numbers and theory of functions. Research of convergence of series and representation of the numbers led him to the sets. He defined the basic concepts of set theory (1874), divided infinite sets into countable (whole, rational, algebraic numbers) and countless (real numbers), assigned so called cardinal numbers to the sets. Mathematically formulated the concept of infinity, and proved the existence of so-called. transcendental numbers. He created the first comprehensive overview of transcendental sets and numbers. Expressed hypothesis of continuum.

Algebraic solution of equation requires finding procedure with using of a finite number of 5-operations (+, -, ∙, /, √) respectively of 4 basic operations (+, -, ∙, /) and a solution of the binomial equation of shape xk=a.

Algebraic solution of algebraic equations (how was proved by N.H.Abel and E.Galois) for some equations of 5-th degree (e.g. x5 -10x +2 = 0) and higher levels does not exist.

### Algebraic elements

Consider an expression P(x) = a(n)xn+...+a(2)x²+a(1)x1+a(0), where x is from some (definition) field D. If value P(x) does not belong - neither for one x - to D, element x is called transcendental.
Otherwise (ie if at least one result of D belongs to the field), we say that the element x is algebraic.

By supplementing m transcendental elements (x, y, ...) we obtain polynomial of m unknowns.

If there is any relation between unknown (e.g. y=x²,..), we say that the transcendent elements are algebraically dependent. Algebraically independent transcendental elements are called indefinite.

## Roots of algebraic equation

### Solution of the quadratic equation

```    x1+x2 = e1
x1-x2 = √(e12 - 4e2) = √D
```

For solution of quadratic equation is used the relationship: (x1+x2)² − 4x1x2 = (x1−x2)². Z (x1+x2)² = e1² a (x1−x2)² = e1² − 4e2 = D, where D is so called discriminant, two linear equation arise and from them follows directly the solutions x1 and x2.

In a statement with the coefficients of equation, discriminant D = b² -4ac depends equally on the coefficient a and the coefficient c.
Thus e.g. equation x² -5x + 6 = 0 and 6x² -5x + 1 = 0 has the same discriminant.

### Inner substitution

Substitution uses for simplification of the given expression generally another expression, e.g. equation x4+x²+1 = 0 is solved using function y=x². It is also possible to substitute the the same expression into itself. E.g. equation x²−x−1 = 0 is possible to rewrite to x=1+1/x and then gradually substitute for x on the right side. In this way, the solution converts to the so-called continued fraction.   x = 1+ 1/x  =>  x = (1+1/(1+1/x))

=> x = (1+1/(1+1/(1+1/x)))  => ... x=(1+√5)/2

 Viéte, Francois , 1540-1603, French mathematician. He dealt with the problems of proportions, equivalent to the solution of equations. He began use letters instead of the usual numerical coefficients. Distinguished known and unknown quantities is the equations and required homogeneity of dimensions. Solved some cases of cubic equations, noting that the cubic equation is associated with problem of trisection of angle. Using of basic trigonometric function he solved the plane and also spherical triangles. Introduced the trigonometric form of Cardan's relations. He expressed number π by a sum of an infinite series.

### Viete's formulas

During solution of quadratic equation ax²+bx+c=0  (resp. x²+px+q=0) it appears that the coefficients of equation can be written using roots x1 a x2.

It holds:   a(x−x1)∙(x−x2) = ax²−a(x1+x2)x+a(x1∙x2)= 0  resp. (x−x1)∙(x−x2) = x²−(x1+x2)x+(x1∙x2)   = 0

And Viete's formulas:   (x1+x2) = −b/a = −p,      (x1∙x2) = c/a = q.

Position of roots x1 and x2 is symetrical. Swapping of roots will not change anything on the given coefficients, ie coefficients are symmetric functions of roots. Left side of formulas consists of orbital polynomials:  o2(x1)   = x1+x2=−p = e1      o2(x1∙x2)= x1∙x2= q = e2

Here, it is important to note that if the coefficients p, q are of the domain T                and roots of the domain U, U must contain T (i.e. U is a superset of T, including possibility U = T).                This applies in general - for equations of all degrees.

 D'Alembert, Jean-Baptiste Le Rond [dalambér], 1717-1783, French mathematician, physicist, naturalist and enlightened philosopher. He was interested in mechanics, theory of gravitation, the movement of the planets, music theory. He got the European reputation by his treatise on dynamics. Later he wrote mainly philosophical works (in the spirit of atheism and determinism). Drafted an outline of the history of the origin and development of knowledge, he was co-publisher of Encyclopedia. He sought to prove the fundamental theorem of algebra.

Equation a0xk + a1xk−1+...+ak−1x+ak = 0 is called inverted equation with regard to basic equation akxk + ak−1xk−1+...+a1x+a0 = 0. If xj is one solution of the basic equation, then corresponding solution of inverted equation is 1/xj.
E.g. equation x²−5x+6=0 has roots x1=2, x2=3 and 6x²−5x+1=0 has roots x1=1/2, x2=1/3.
Or e.g. in linear equation 2x−3=0 is x=3/2, meanwhile −3x+2=0 for x=2/3.

Hence, the equation that is symmetrical (see Inverse equations) must have with the root also the root 1/x.

Reciprocal equation is equation formed by a polynomial, in which the symmetric coefficients have equal values, e.g.

5x4 +7x³ +4x² +7x +5 = 0  or  x4 −7x³ +4x² −7x +5 = 0

Equation is negatively reciprocal in case when symmetric coefficients have opposite values, e.g.

5x4 −7x³ +7x −5 = 0

The intermediate member is in the reciprocal equation arbitrary, in the negative reciprocal equation must be zero.

If the reciprocal of equation has root x1, has also root 1/x1.
Reciprocal equation are algebraically solvable up to degree k = 9 (inclusive). Negatively reciprocal equations have always the root (x-1), so that are solvable up to degree k = 10.

### Fundamental theorem of algebra

Every algebraic equation P(x) = 0, where P(x) is a polynomial of degree k, has in the field of complex numbers, at least one root. If this root is x*, then P (x) is divisible by the formula (x-x*). Hence, the algebraic equation P(x) = 0 of k-th degree has in the field of complex numbers always just k roots (together with the multiplicity). Fundamental theorem of algebra was proved by K.F.Gauss. The proof has contributed to the recognition of complex numbers. The theorem guarantees the existence of a solution, but does not specify how to find a solution.

## Pitfalls of solution

 Ferro, Scipione Del , -1526, Italian mathematician, the first known solver of the cubic equation.

### Cubic equation

In cubic equation x³+px²+qx+r=0 platí for roots x1,x2,x3:  o3(x1) = x1+x2+x3  = −p  = e1  o3(x1∙x2) = x1∙x2+x1∙x3+x2∙x3 =  q  = e2  o3(x1∙x2∙x3) = x1∙x2∙x3   = −r  = e3

Equation is usually first converted to a reduced form in which p=0. According to the values of q and r is possible to distinguish three types of equations: 1/ q>0,r<0, 2/ q<0,r>0 3/ q<0,r<0.

### Example

Let us consider a cubic equation x³−7x+6 = 0, so p=0, q=−7, r=6.

 Tartaglia, Niccolo , 1500-1557, Italian mathematician, he solved (y.1535) independently of S.de Ferro general case of cubic equation. He dealt with the translation of Euclid's works.

It applies:

```    x1+x2+x3   =  0
x1∙x2+x1∙x3+x2∙x3 = −7
x1∙x2∙x3   = −6
```

For solution it offers:

to estimate one of the roots x1=1, to divide polynomial x³−7x+6 by x−1 and compute the other twoo roots from the quadratic equation. knowing that the roots are integer, to use the decomposition of number −6 (i.e.x1x2x3) and to choose experimentally suitable numbers from possibilities ±1,±2 a ±3. draw cubic parabola f(x)=x³ and find its intersection with the line g(x)=7x−6.

The equation has solution x1=1, x2=2, x3=−3. But to get this solution in a general case is not easy.

### Symmetry

From equation x1+x2+x3 = 0 we express one of the roots and substitute it to the remaining two equations. We always get the same shapes of the equation: Expression x3  Expression x2   Expression x1     x1² + x1x2 + x2²= 7   x1² + x1x3 + x3²= 7    x2² + x2x3 + x3²= 7 x1²x2 + x1x2²   = 6   x1²x3 + x1x3²   = 6    x2²x3 + x2x3²   = 6

But how to get a solution from this equation is still not clear. Substitution of orbital polynomials does not help x1+x2=u, x1x2=v, It returns us through relations u²−v = 7, uv= 6 to the original equation.

### Cardano's formulas

Italian mathematicians (S.Ferro, N.Tartaglia ..) managed to discover a general formula for solution of cubic equation. Solution (reduced in the case of p=0) is a combination of values u1 and u2, where:

#### u1,2 = 3√(−r/2±√D), where D=(r/2)² +(q/3)³

The simplest solution has the form: x1=u1 + u2.

 Cardano Girolomo [kardano], 1501-1576, Italian mathematician, physician and astrologer. Published (y.1545) as the first (Tartaglia's) solution of cubic equation and (Ferrari's) solution of biquadratic equation.

But when the equation has three real roots, the result is hardly applicable. In our example is D = 3² +(−7/3)³ = −100/27 and u1,2 = 3√(−3±10i/3√3).

This case was called "casus irreducibilis" and was not solved. Neither the theory of imaginary numbers drawn by mathematician R.Bombelli (y.1572) helped to its transparency.

In this case, equation with real roots solves trigonometrically, looking for the angle φ, for which cosφ=−9√3/7√7.

### Casus irreducibilis

Let's look in more detail at x³ -15x -4 = 0. From the Cardan formulas, the solution will be in the form:

•   x= ³√(2 + 11i) + ³√(2 - 11i)

But it is simply not possible to root out complex numbers.
In our example we know that: (2 + i)³ = 2 + 11i   and generally for solution (f, g) = f + gi of expression (r, s) = r + si it holds: (f + gi) ³ = r + si, so:

#### g³ - 3gf² = -s.    (2)

However, substitution of one equation into another - after several necessary adjustments - leads again to the (original) unsolvable form...

There is another option. Let's look at examples of some solutions (f, g) ← (r, s):

```
(1,1) ← (-2,-2),  (1,2) ← (-11,2),   (1,3) ← (-26,18), (1,4) ← (-47,52), (1,5) ← (-74,110), ...
(2,1) ← (2,-11),  (2,2) ← (-16,-16), (2,3) ← (-46,9),  (2,4) ← (-88,16), (2,5) ← (-142,65), ...
...
```

The norm |f + gi| = f² + g² is always the cubic root of the norm |r + si| = r² + s² .
For example N(1,3) = 1 + 9 = 10 a N(-26,18) = 676 + 324 = 1000 = 10² So we get one more relationship::

#### N(f,g) = f² + g² = ³√(r² + s²) = ³√N(r,s)   (3)

Not even the third equation (which places the roots of the solution on an ellipse with an equation f² + g² = C (=³√N(r,s)) helps.
From substituting of equations (1) and (2) we get 4f³ - 15f - 2 =0; and 4g³ - 15g + 11 =0; and these are again the (original) unsolvable forms of equations ...!?

### Goniometric solution

Some relations of the trigonometric functions can be converted into algebraic equations (see Chebyshev polynomials). E.g. cos(3u) = 4cos³(u)−3cos(u), i.e. 4cos³(t/3)−3cos(t/3)−cos(t) = 0 for given c=cos(t) and unknown x=cos(t/3) gives 4x³−3x−c = 0. Hence (from the other side), we conclude that the cubic equation 4x³−3x−c = 0 can be easily solved trigonometrically. Equation of third degree has no representation in the plane, so trisection of angle is not possible.

 Descartes, René du Perron [dekart] (in Latin - Renatus Cartesius), 1596-1650, French philosopher and mathematician, founder of analytic geometry. He defined the term equation of curve and elevated arithmetic and algebra over geometry. Introduced the use of symbolism - still used - for writing polynomials, formulated fundamental theorem of algebra. He named square roots of negative numbers - "imaginary numbers." Proposed a method of knowledge based on doubt. He also dealt with physics, optics, meteorology and music theory.

### Estimate of the number of roots

According to Descartes' theorem algebraic equation has at most as many positive roots, how many sign changes it contains. i.e. how many times is fulfilled a(j)∙a(j+1)<0 in equation
a(k)∙xk+...+a(2)∙x²+a(1)∙x+a(0)=0.
Depending on the number of these changes is also the number of positive roots even or odd.

### Sign of equation coefficients

Algebraic equations is convenient to write with varying accents: (At the same time, we should also rephrase the Descartes theorem...)

```    Usual notation        Transcription
─────────────────────────────────────
x²+px+q = 0;         x²−e1x+e2 = 0
x³+px²+qx+r=0       x³−e1x²+e2x−e3 = 0
```
 Lobačevskij, Nikolaj Ivanovič, 1793-1856, Russian mathematician, one of the founders of non-Euclidean geometry. He derived solutions of some until unsolvable integrals. He sought to confirm his theories by astronomical measurements.

### Equations with real roots

If the second power of the equation has alternating signs, given equation has only real roots.
E.g. because (x³−7x)² = (−6)² a x6−14x4+49x²−36 = 0 has střídavá znaménka, has equation x³−7x+6 = 0 jen reálné roots.
The determination that all of the roots are real is necessary eg. in Graff-Lobachevsky method of solving algebraic equations.

In biquadratic equation x4+px³+qx²+rx+s=0 roots x1,x2,x3,x4 satisfy:  o4(x1)   = x1+x2+x3+x4       = −p = e1  o4(x1∙x2)   = x1∙x2+x1∙x3+x1∙x4+x2∙x3+x2∙x4+x3∙x4 = q  = e2

 Ferrari Ludovico , 1522-1565, Italian mathematician solved the general case of biquadratic equation.

o4(x1∙x2∙x3)    = x1∙x2∙x3+x1∙x2∙x4+x2∙x3∙x4   = −r = e3  o4(x1∙x2∙x3∙x4) = x1∙x2∙x3∙x4       = s  = e4

Ferrari's solution consisted in transferring biquadratic equation to cubic equation.

### Powers of roots of equations

Solving equations of n-th degree is equivalent to n-equations for sums of powers of roots (see Waring formula coefficients)

```k=2:
x1+x2  =  e1  = −p
x1²+x2²  = e1²−2e2  =  p²−2q
────────────────────────────────────────────
k=3:
x1+x2+x3  =  e1  = −p
x1²+x2²+x3²  =  e1²−2e2  = p²−2q
x1³+x2³+x3³  =  e1³−3e1e2+3e3 =   −p³+3pq−3r
```

## Permutations

### Invariant polynomials

By renumbering of the indexes of variables in polynomial f=f(t1,t2,t3...) according to certain permutation p, we get new polynomial f'(t1,t2,t3...). Permutation with cycle [1,2] changes f(t1,t2)=t1³+t2 to f'(t1,t2)=t2³+t1. If polynomial has not more than 3 variables, we will - instead of designation t1,t2,t3 use the usual x,y,z: thus f(x,y)=x³+y changes to f'(x,y)=y³+x.

The polynomial f is invariant with respect to the permutation p, if f and f' are the same, eg. f(x,y)=f'(x,y)=x+y.

### Group of inertia of the polynomial

each polynomial is invariant relative to an identical permutation if it is invariant relative to permutations p1 and p2, then is invariant also to composite permutation p1∙p2 if it is invariant relative to permutation p, then it is invariant also to its inverse p−1.

All permutations that do not affect polynomial, create group so called group of inertia of the polynomial. Group of inertia of symmetrical polynomials is a group of all permutations Sn.

### Orbital polynomials

Polynomial, which is the sum of all permutations of given polynomial f is called orbital and denoted by on(f). Each orbital polynomial is invariant with respect to any permutation.   o2(xy²) = xy² + yx²

For polynomial f(t1,t2,t3,..) = t1k, acquires orbital polynomial a form of power sum: on(t1k) = t1k+t2k+t3k+.... = sn(k).   s3(x5) = x5 + y5 + z5

### Breakdown of orbital polynomials

Any orbital polynomial on can be written using of power polynomials sn. E.g. from the breakdown of product s2(u)∙s2(v) we get:    s2(u)∙ s2(v) =(xu+yu)(xv+yv) = xu+v+ xu∙yv+ xv∙yu   = s2(u+v) + o2(xu,yv)

So: o2(xu,yv) = s2(u)∙ s2(v) − s2(u+v).

### Elementary symmetric polynomials

Orbital polynomials on(t1),on(t1∙t2),on(t1∙t2∙t3),... are called elementary symetric polynomials .   o1(x) = x   o2(x) = x+y    o2(x∙y) = x∙y.   o3(x) = x+y+z  o3(x∙y) = x∙y+x∙z+y∙z  o3(x∙y∙z) = x∙y∙z

Every symmetric polynomial may be assembled from elementary symmetric polynomials.

### Preservation of symmetry

Polynomial f(t1,t2,..) is called symetric, když is invariant with regard to any permutation. Similarly the function f(t1,t2,..) is symetric, if its value does not depend on the rearrangement of arguments t1,t2,...

Sum and also product of two symetric polynomials is also symetric. Therefore, when are into any polynomial f(t1,t2,t3,..) substituted for t1,t2,t3,... symetric polynomials, resulting polynomial is also symetric.
E.g. into polynomial g(t1,t2)= 2t1²+3t2 we substitute t1=x+y and t2=x∙y. We get:
g(x+y,x∙y) = 2(x+y)²+3xy = 2x²+2y²+7xy

i.e. symetric polynomial.

### Even-symmetric polynomials

Polynomial is even-symmetric, if it is invariant with regard to even permutations. Even-symmetric polynomials are either symmetric or antisymmetric.
E.g. x+y is symetric (so also even-symmetric), x−y is antisymetric (is even-symmetric and is not symetric).

Products of symmetric (S) and antisymmetric (A) polynomials respect - for multiplying - the following table:

```      S   A        S∙S = S    (x+1)(x+1) = x²+2x+1
─────────        S∙A = A    (x+1)(x−1) = x²−1
S   S   A        A∙S = A    (x−1)(x+1) = x²−1
A   A   S        A∙A = S    (x−1)(x−1) = x²−2x+1
```

### Copy of the structure

 Poincaré, Jules Henri , 1854-1912, French mathematician and physicist, has contributed significantly to the solution of problems in celestial mechanics, in particular to the problem of motion of three or more bodies. In mathematics he is famous for its research of so-called automorphic functions. He dealt with a wide range of physic fields (relativity, thermodynamics, rotational motion, ...) and mathematic fields (sets, probability, topology, series ..).

We call (in general) deformation (morphism) - the representation (depiction), that assigns to a given structure some other structure. The structures may have different number of elements and also different internal composition.

A special case of projection is copy (isomorphic representation, isomorphism) when both algebraic structures have the same number of elements and the same internal composition.

E.g. group made by values f(x) is copy of group of values x:

```    x        f(x)
────────────────────
0 1 2    I R S
1 2 0    R S I
2 0 1    S I R
```

In that case:

#### f(a∙b) = f(a)∙f(b)

Composition of copies is a copy

Table T(r,∙), rεP can be converted to T(φ(r),+). Let us rewrite e.g. T(5,∙) according to rule:  1→0,2→1,4→2,3→3 (i.e. each one will be replaced by zero, a deuce by one, four by deuce, with number three leaved).

```T(5,*):            T(4,+):
* │ 1 2 3 4        + │ 0 1 3 2
──┼────────        ──┼────────
1 │ 1 2 3 4        0 │ 0 1 3 2
2 │ 2 4 1 3        1 │ 1 2 0 3
3 │ 3 1 4 2        3 │ 3 0 2 1
4 │ 4 3 2 1        2 │ 2 3 1 0
```

The new table is (except shuffle of rows and columns) table for counting numbers (according to module 4). Because T(5 ∙) is group, T(4 +) is also group.

All the tables for rεP behave the same way. Multiplicative group of order r and additive group of order φ(r) are two copies of the same structure.

### Projection of the structure

The deformation that do not disrupt operations introduced in these structures, will be called projection (homomorphic mapping, homomorphism). The projection of the structure may have a different number of elements than the default structure. Projection is e.g transformation of the whole numbers into the set of congruence classes according to module r.   Z ───> Zr

Field Z is infinite, while the field Z7 consists of 7 elements (0,..,6). For r= 7 is:

In the field of Z   In the field Z7    3+9= 12  ───> (3 mod 7)+(9 mod 7)= (12 mod 7)  i.e. 3 + 2 = 5    3∙9= 27  ───> (3 mod 7)∙(9 mod 7)= (27 mod 7)  i.e. 3 ∙ 2 = 6

Composition of the projections is the projection.

## Structure of equations

### Lagrange's resolvents

For solution of any algebraic equation is necessary to use symmetry roots. But it is difficult to imitate relation (x1+x2)² − (x1−x2)² = 4x1x2, which makes it easy to solve a quadratic equation (see Quadratic equation). E.g. difference of (x1+x2+x3)³ and some expression (x1±x2±x3)³ still contains too many members.

Lagrange studied the expressions of the form:

#### L(x1,x2,x3) = x1 + αx2 + α²x3 ..

and showed, that (specially in the case k=3) their k-th power gives for all the possible permutations of roots only 2 distinct values. Expressions L() are called Lagrange resolvents. Jejich koeficienty are roots binomické equation a are tedy rovnoměrně rozestoupené na jednotkové kružnici.
In the case of cubic equation is α = (−1+i√3)/2 (see Binomial equation/Basic root) and for each permutation of the equation x³−7x+6 = 0 we get:

```  x1  x2  x3  L(x1,x2,x3)  L³(x1,x2,x3)
──────────────────────────────────────────
1   2  −3   ( 3+5i√3)/2    3(−27+10i√3)
1  −3   2   ( 3−5i√3)/2    3(−27+10i√3)
2   1  −3   ( 6+4i√3)/2    3(−27+10i√3)
2  −3   1   ( 6−4i√3)/2    3(−27+10i√3)
−3   1   2   (−9− i√3)/2    3(−27+10i√3)
−3   2   1   (−9+ i√3)/2    3(−27+10i√3)
```
 Ruffini, Paolo , 1765-1822, Italian mathematician. Accepted imaginary numbers as roots of solution of equation (quadratic, cubic and also biquadratic) and questioned the possibility of a solution of algebraic equation of any grade through basic operations.

The norm (see Complex numbers) of each of these numbers is 21. Lagrange works on equations inspired P.Ruffini, Abel and also Galois.

### Function with a limited number of values

Some functions of roots does not change value under any permutation roots or returns only a limited number of values.

### Alternating polynomials

Polynomial, which after renumbering according to any permutation just changes sign is called alternating (sign-changing) polynomial. For these polynomials only odd permutations changes the sign, even permutations leave the polynomial unchanged.

E.g. the following functions give only two values differing by sign:

n=2: (x1−x2),(x2−x1) n=3: (x1−x2)(x1−x3)(x2−x3),(x1−x3)(x1−x2)(x3−x2),...

Polynomial (x1−x2) has 2!=2 permutations: (x2−x1)=−(x1−x2), polynomial (x1−x2)(x1−x3)(x2−x3) 3!=6 permutations:

```    pi  u   (123)   Parity  Alternating polynomial
──────────────────────────────────────────────────
p0  1   (123)   S    (x1−x2)(x1−x3)(x2−x3)  =  A
p1  2   (312)   S    (x3−x1)(x3−x2)(x1−x2)  =  A
p2  4   (231)   S    (x2−x3)(x2−x1)(x3−x1)  =  A
p3  3   (132)   L    (x1−x3)(x1−x2)(x3−x2)  = −A
p4  6   (321)   L    (x3−x2)(x3−x1)(x2−x1)  = −A
p5  5   (213)   L    (x2−x1)(x2−x3)(x1−x3)  = −A
```

Alternating is generally each polynomial of the form:

### Alternating groups

All the elements of alternate groups are even permutations.
``` A(3)
Permutation        Cycles and order     Length   Parity      Type and indicator
──────────────────────────────────────────────────────────
p0     123  (1)(2)(3)   1 1 1   even (+)    (3,0,0)    123   1   x1³
───────────────────────────────────────────────────────────
p2     123   (1,2,3)    3       even (+)    (0,0,1)    231   3   x31
───────────────────────────────────────────────────────────
p4     123   (1,3,2)    3       even (+)    (0,0,1)    312   3   x31
```

Alternating group An is subgroup of permutation groups Pn, has a half-order -in comparison with the permutation group.

``` Values u    Permutation
1 2 4      p0 p2 p4
2 4 1      p2 p4 p0
4 1 2      p4 p0 p2
```

### Functions unchanging value

For the solution of quadratic equation we use the so-called discriminant. Discriminant D is value shared by all the roots. It is defined:

#### D = (δ)2

(see Alternating polynomials).

### Klein's group

Let's go back to the first from Latin squares of the 4-th order R4(a):

 Klein, Christian Felix [klain], 1849-1925, German mathematician. He was engaged in theory of group, theory of functions and by topology. He promoted the uniform classification of geometry using theory of groups, proved that different geometries (Euclidean, hyperbolic, elliptical) are consistent with each other.
```    1  2  3  4       1  i  j  k
2  1  4  3       i  1  k  j
3  4  1  2       j  k  1  i
4  3  2  1       k  j  i  1
```

Corresponding group is called Klein's group. This is a subgroup of the permutation group P4 and also of alternate group A4 (which allows the solvability of algebraic equation 4-th degree).

Grupa is non-cyclical and commutative (applies ij=ji=k, i²=j²=k² = ijk= 1).

Klein's group is made by e.g. eclipsing operations on the rectangle.

(1-identity, 2-vertical axis, 3-horizontal axis, 4- rotation by 180°).

```Permutation  Cycles and order Lengths Parity  Type and indicator
──────────────────────────────────────────────────────────
p1    1234    (1)(2)(3)(4)  1 1 1 1  even (+)    (4,0,0)
1234     1     x14
──────────────────────────────────────────────────────────
p2    1234    (1,3)(2,4)    2 2      even (+)    (0,2,0)
3412     2     x2²
──────────────────────────────────────────────────────────
p3    1234    (1,2)(3,4)    2 2      even (+)    (0,2,0)
2143     2     x2²
──────────────────────────────────────────────────────────
p4    1234    (1,4)(2,3)    2 2      even (+)    (0,2,0)
4321     2     x2²
```

### Copy of the complex number

Consider the function f in the field of real numbers, for which each number rεR returns its identical copy: f(r) = r. Then the relation (isomorphism), which we have defined for copies, will apply:

#### f(a∙b) = f(a)∙f(b)

What happens if we attempt to create a similar function in the field of complex numbers? Since f(-1) = -1 must be also f(i²)=f(i∙i)=f(i)∙f(i)=−1 and therefore f(i) = ±√−1 = ±i.
In this sense, then the copy of number a+bi is not only a+bi but also a−bi, i.e. complex conjugated number.

(Relation of complex association belongs to the automorphisms studied in the Galois theory of solvability of algebraic equations.)

 Galois, Evariste [galoa], 1811-1832, French mathematician. Dealt with the solvability of algebraic equations, confirmed the impossibility of the solution of certain types of equations of higher degrees by using roots. Defined precisely the circumstances in which traditional methods of solution of equations work and when are failing. Introduced the term 'group', discovered normal subgroups. He noted that the solution of some equations of higher degrees fails when there are no other normal subgroups than simple. His works published J.Liouville y.1846.

### Galois' group of the equation

Galois, inspired by the works of Lagrange and Gauss, assigned to each equation all permutations of the roots, which do not change the values of certain polynomials.

Equation x4−10x²+1= 0, i.e. (x²−5)²−24 = 0, has roots:
x1=+√2+√3,   x2=+√2−√3,   x3=−√2+√3,   x4=−√2−√3.

Build the table for products of roots:

```     │   x1       x2      x3      x4          │ x1 x2 x3 x4
───┼───────────────────────────────      ───┼───────────
x1 │  +5+2√6     −1      +1  −5−2√6      x1 │ 1  2  3  4
x2 │  −1     +5−2√6  −5+2√6      +1      x2 │ 2  1  4  3
x3 │  +1     −5+2√6  +5−2√6      −1      x3 │ 3  4  1  2
x4 │  −5−2√6     +1      −1  +5+2√6      x4 │ 4  3  2  1
```

By designation 1:+5±2√6  2:−1   3:+1  4:−5±2√6  it corresponds to the table of the Klein's group.

### Decomposition of the group to classes

In the non-commutative permutation group of order 6:

``` 1 2 4  3 6 5
2 4 1  6 5 3
4 1 2  5 3 6
3 5 6  1 4 2
6 3 5  2 1 4
5 6 3  4 2 1
```

four subgroups exists:

```    1 3   1 6   1 5   1 2 4
3 1   6 1   5 1   2 4 1
4 1 2
```
```  Left classes   Right classes
────────────────────────────
1*{1,6}={1,6}  {1,6}={1,6}*1
6*{1,6}={1,6}  {1,6}={1,6}*6
2*{1,6}={2,5}  {2,3}={1,6}*2
5*{1,6}={2,5}  {4,5}={1,6}*5
3*{1,6}={3,4}  {2,3}={1,6}*3
4*{1,6}={3,4}  {4,5}={1,6}*4
```

Multiplication of each element of given group with elements of one its particular subgroup makes so called decomposition of group according to subgroup. According to multiplication from the left or the right we distinguish left and right classes. Elements that belong to the same class are called conjugated. E.g. according to subgroup {1,6} we get:  left classes {1,6},{2,5},{3,4} and right classes {1,6},{2,3},{4,5}:

 Jordan, Camille [žordan], 1838-1922, French mathematician. He published the first systematic exposition of group theory and Galois' ideas. He was also engaged in mathematical analysis.

The number of left classes is always the same as the number of right classes and number of (conjugated) elements in classes is a divisor of order of the final group (Lagrange's theorem).

### Normal subgroups

Subgroup is a normal (invariant) subgroup only if respective left and right classes of its decomposition are the same.

In commutative groups are all subgroups normal. E.g. -in commutative group- left and right classes of decomposition according to subgroup {1,6} do not differ:

```1 2 4   3 6 5
2 4 1   6 5 3
4 1 2   5 3 6
3 6 5   2 4 1
6 5 3   4 1 2
5 3 6   1 2 4
```
``` 1*{1,6} = {1,6} = {1,6}*1
6*{1,6} = {1,6} = {1,6}*6
2*{1,6} = {2,5} = {1,6}*2
5*{1,6} = {2,5} = {1,6}*5
3*{1,6} = {3,4} = {1,6}*3
4*{1,6} = {3,4} = {1,6}*4
```

Normal subgroups is not affected by the direction of operation with other elements. E.g. shift operation constitutes normal subgroup of motion group, while the turn operation not:

```        turn +  shift   − turn = shift
shift   +  turn − shift   = ?
```

The creation of normal subgroups is especially important for non-commutative groups. Any commutative (Abel) group is solvable.

Normal subgroup N of group G commutes with arbitrary subgroup A of groups G, i.e. A∙N = N∙A.
The analogy of normal subgroup in structures with multiple operations is called ideal.

### Simple group

The simple group has no nontrivial normal subgroup. Such is e.g each cyclic group of prime order.

## Equations of higher degrees

### Breakdown of power sums

Let us mark o2(x) = x+y = e1 o2(x∙y) = x∙y = e2. For F(k) = o2(xk) = xk+yk it holds:

```  k  o2(xk)    f(e1,e2)
──────────────────────────────────────
1   x1+y1 = e1
2   x²+y² = e1²−2e2
3   x³+y³ = e1³−3e1e2
4   x4+y4= e14−4e1²e2+2e2²
....
7   x7+y7= e17−7e15e2+14e1³e2²−7e1e2³
```

The following recurrent relation applies:

#### F(k) = e1∙F(k−1) − e2∙F(k−2)

e.g. F(3) = x³+y³ = e1∙(e1²−2e2) − q∙e1 = e1³−3e1e2

### Waring formula coefficients

E.Waring create a formula for the direct expression of power sums sn(k) using elementary function. We will notice only the coefficients of expressions that are the result of Waring's formula.

```     2    2    2    2    2    2
1    3    5    7    9    11    13
1    4    9   16   25    36    49
1    5   14   30    55    91 ...
1    6   20   50   105   196 ..
1    7   27    77   182   378
1    8   35   112   294 ...
1    9    44   156 ...
1   10    54   210 ...
1    11    65 ...
1    12    77 ...
```

Writing a constant sequence a0(i) = 2 and from it gradually, starting in every line by number one, we derive cumulative sequences (see sequences):

 Waring, Edward , 1734-1798, English mathematician. He was interested in the theory of numbers, symmetric functions, algebraic curves. Some of his thoughts anticipated the Galois theory. He was the first who published the so called Goldbach hypothesis.

In this schema we read numbers in the columns from the bottom. These are (except the signs, that we change in members on the even positions) coefficients of expressions of the Waring's formula.

E.g. in the column above the seventh number one - we read sequentially 1,7,14,7. Accordingly applies (relation of the previous paragraph):  x7+y7 = e17−7e15e2+14e1³e2²−7e1e2³

### Sums of more variables

 Hermite, Charles [], 1822-1901, French mathematician. He was interested in the theory of complex numbers, interpolation, elliptic functions and in solving equation. He explained the Steiner theorem about the impossibility of finding the center of a circle using a ruler. He proved that e is a transcendental number, i.e not a solution of any algebraic equation.

Using Waring's formula is possible to quantify also power sums of more variables (xk+yk+zk+..) [Kuffner].
Eg. for x1+x2+x3 = e1, x1∙x2+x1∙x3+x2∙x3 = e2 a x1∙x2∙x3 = e3 it applies: x1³+x2³+x3³ = e1³−3e1e2+3e3.

To verify this relation- after substraction    e1³−3e1e2 = (x1+x2+x3)³ − 3∙(x1+x2)∙x1∙x2 ...

these members: (x1+x2+x3)³ drop out of the listing    3(x1x2²+x1x3²+x2x3²+x1²x2+x1²x3+x²x3)

and remains: x1³+x2³+x3³−3x1x2x3, where 3x1x2x3 = 3e3.

### Non-algebraic solution of equations

Lack of algebraic solution does not say anything about functionality of other possible methods for solution of algebraic equations.

For quadratic and cubic equation is a trigonometric solution available, but it requires enumeration of trigonometric functions (cos, sin, ..). Approximate methods of solution of algebraic equations: Lagrangian method (expression of real roots by continued fraction) Lin's method (algorithmic decomposition of the polynomial to product), graphical methods, ... Iterative methods clarification of estimated roots: method of bowstrings (regula falsi), Newton's method of tangents, method of separation of roots, quadratic interpolation, and the like. Ch.Hermite demonstrated (y.1858) general solution of the equation of 5-th degree using elliptic function.

The question arises: is it not possible yet another, simpler algebraic solution than using of the elliptic functions?

Is there any other operation (besides the five identified - addition/subtraction, multiplication/division and extraction of the root), which could be considered as algebraic and that could be used to a solution?