Re: Fermat's Last Theorem

On Apr 15, 2:22 pm, Basti...@xxxxxxx wrote:
Dear newsgroup:

Since the Google group does not have attachments please see the
following, the text is as follows.

http://mathforum.org/kb/message.jspa?messageID=6178645&tstart=0

Only a few of my methods carry free variable types, in which case they
are used for solving integrals.

Almost others have a C as general constant of integration of Riccati.

The following is a degree 13 polynomial, generated by a separable
Riccati differential equation DE (as you can see).

Maple will solve it and upon substitution into the polynomial you get
its solution.

I have a program, which is designed for calculating them; it has many
combinations depending on degree.

This one has 56 combinations; the polynomial you see is the first one
on the list.

It is one of the beautiful characteristic of Riccati and polynomial
connections, that if you change t=T(t), m=m(t),n=n(t),r=r(t) in the
polynomial, the resulting differential equation is mainly Riccati as
well.

Since we have the general solution of the polynomial, we can get to
know the solution of the corresponding Riccati.

We need to solve Riccati differential equations as a class relative to
class polynomials and keep them in our library for reference.

Certainly there are eventually developments in non separable cases;
most of them are not solvable by Maple.

But there are procedures on my methods which solve them including its
integrals (It is part of the package of developments).

Generally as far as my experience shows, maple solution of polynomials
is similar to solving polynomials with higher order linear
differential equations.

This means Maple solutions of polynomials are in expanded forms.

It is not possible to handle even with computers many differentiations
of polynomial functions.

That is why after 100 years of research they could not develop much on
Galois theory (Thus the Galois theory is out of business).

But Riccati equations are of first order differential equations and we
have much hope to handle them nicely within higher order polynomials.

I have ample classes of polynomials and Riccati with applications for
the next 15 years.

Look at the AMS Mathjobs.org, no one was interested to even consider
me for a junior jobs.

Someone should question their committee and ask what they are doing
there?

They see only themselves and their lost math (they are friends of
Wiles!).

Dr.Mehran Basti

Format's last theorem for non zero integers (x, y, z, n), (n>2)

x^n + y^n = z^n

Can simply be reduced to the case where (n) is odd positive integer,
and (n > 1)

In fact it can be reduced further to the case where (n) is odd prime
number, and (x, y, z) are coprime in pairs, but I will consider (n) as
odd positive integer, where (n > 1)

What I think that Fermat realized that the following equation holds
true always, then his last theorem becomes so obvious to conclude.

The equation is the following, where a counter example simply doesn't
exist
..
"If (x) and (y) are two distinct integers belong to Z*, where (x) and
(y) are prime to each other, and (n) is odd positive integer grater
than one, then there exist three integers (you're a fucking moron
basti, just like tommy1729) (m, k, p), where (m) belongs to Z*, (k)
belongs to N, (p) is prime number, where (m) and (p) are prime to each
other, such that:

x^ n + y^n + m*(p^(2^k)) = 0 "

Of course, one simply can prove that the later term
m*(p^(2^k)) can't be equal to (z^n), where (z) belongs to (Z*), other
wise assume


z^n = m*(p^(2^k))

Then, (p) must be a prime factor of (z),
Let, z = t*(p^s), where (t) is integer belongs to (Z*), (s) is
positive integer, where (t) and (p) are prime to each other.
This implies:


z^n = m*(p^(2^k)) = (t^n)*(p^(n*s))

From which, we conclude that:
n*s = 2^k,

Since, the left hand side of the above equation has odd prime factor
and the right hand side of the same equation doesn't have any odd
prime factor, we conclude that (n*s) is not equal to (2^k), and
similarly
z^n is not equal to m*(p^(2^k)), a contradiction to our
assumption of equality,

Hence, Fermat equation as defined earlier
(x^n + y^n +z^n = 0), is impossible

"Copyright (c) 2006, Bassam Karzeddin. All rights reserved"

Bassam Karzeddin
Al-Hussein Bin Talal University
JORDAN
.

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