Home #1 Mathematics When this quintic have solutions? ...
When this quintic have solutions? ... PDF Print E-mail
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Friday, 12 February 2010 00:00

In other words, what condition must be fulfilled, such that the following equation of fifth degree have solutions

x5-ax-b = 0?

Like probably you know only polynomials which degree is lower than five have solutions that can be found through formulae. Logically, these formulae will involve radicals.

Again, like probably you know this has been proved by Niels Heinrich Abel in 1856, as a special case and preparation for the full solution realized by Evaroits Galois

Now, there are some equations of fifth degree, called quintic that can be solved.

The question is: Would be possible to know when a quintic can or not be solved?

The Italian mathematician GianFrancesco Malfatti.

This question, transformed in exercise comes from the book: Heinrich Dorrie - 100 Great Problems of Elementary Mathematics (pag 116-127)

So, we have to prove the following Lema:

Lemma: The polynomial,

x5-ax-b = 0     (1)

cannot be solved when a and b are positive integers that are divisible by a prime p, b is indivisible by p2, and when 44a5>55b4.

To prove this lemma we can in part use the Schoenemann's Theorem, which states that if the integral coefficients C0, C1, C2, ... CN-1 of the polynomial

   f(x) = C0 + C1 x + C2 x2 + ... + CN-1 xN-1 + xN

are divisible by a prime number p, while the free term C0 is not divisible by p2. Then, f(x) is irreducible in the rational domain.

The proof of this theorem is available in the Dorrie's book.

What we need to fix the condition: 44a5>55b4 ?

Would be necessary to investigate with the resolvent?

No, is not necessary!

According to the Sturm's Theorem the roots are located between the sequence of polynomials, f0,f1, ... where

f0 is the original polynomial,

f1 is the derivative of f1 = f'0

f2=-rem (f0, f1)

f3=-rem (f1, f2)

...

0 = -rem (fm-1, fm)

You can check this on the same book ... at page 116.

In fact, we have that,

f0 = x5-ax-b, f1 = 5x4-a, f2 = 4ax-5b and f3= 44a5 - 55b4.

To get f3 we need to apply (logically) the Euclidean Algorithm for polynomials and the divisor is: -5/(4a) x3+52b/(4a)2x2 -53b2/(4a)3x+54b3/(4a)4

This complete the proof of the Lema! ...

Now, I see interesting to investigate the result, for someone that does not know the Sturm Theorem, and therefore was forced to navigate in the resolvent of the quintic!

To reach the condition, we need information about the Malfatti resolvent, which according to Leonard E. Dickson's paper published in the AMS Bulletin, in 1925, Leonard Eugene Dickson - Resolvent Sextics of Quintic Equations.pdf, is equivalent to,

The Jacobi-Cayley Resolvent which states that The discriminant Λ of,

    f(x,y) = a0 x5 + 5 a1 x4y + 10a2 x3 y2 + 10a3 x2y3 + 5 a4 x y4+ a5 y5        (2)

is defined to be the polynomial such that 55a0-8 Λ is equal to the product of the squares of the ten differences of the roots xi of f(x,1)=0.

Our (1) to have (2), we must have,

a0  = 1,  a1 = a2 + a3 = 0. a4 = -1/5 and a5 = -b.

f(x,1)=0 implies, x5 - a x - b = 0 and then 55a0-8 Λ

If you can to get the Malfatti resolvent you can read the book, Jörg Bewersdorff - Galois Theory for Beginners.pdf or directly study the original paper published by him, in Latin: "De natura radicum in aequationibus quarti gradus ad clarissimum virum P. Vincentium Riccatum Soc. Jesu epistola". Ferrariae, apud Iosephum Barberium (1758).

Can you prove that these resolvent are equivalent?

According to the book, Christian U. Jensen, Arne Ledet, and Noriko Yui - Generic Polynomials Constructive Aspects of the Inverse Galois Theory.pdf there are still another equivalent resolvent published by Weber in the Lerbuch der Algebra.

Mathematics is a very wonderful Science and very natural. Mathematics never will betray you, like God.

A good mathematician need to know not only the solution, but also the best solution. The Sturm Theorem, not very popular in High School books, offers a wonderfull solution and conditions, full equivalent to Galois Theory, but using simply the derivative of the original polynomial and the Euclid Algorithm.

In fact, Professor Heinrich Dorrie, note that the polynomial of degree seven, x7-ax-b=0, will be algebraically not soluble if: 66a7>77b6.

Thanks,

Giovanni A. Orlando.

 

 

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Last Updated on Friday, 12 February 2010 16:27