Stefano Maruelli

Stream: new members

[Stefano Maruelli]

Hope you will be interested in writing FLT via this property:

$$$$ $$A^n= \sum_{X=1}^{A}M_n = \sum_{x=1/K}^{A}M_{n,K}$$

where:

$$$$ $$M_{n,K}= {n \choose 1} \frac{x^{n-1}}{K} - {n \choose 2}\frac{x^{n-2}}{K^2} + {n \choose 3}\frac{x^{n-3}}{K^3} -... +/- \frac{1}{K^n}$$

So IFF A is an intger we can write:

$$A^2=\sum_{X=1}^{A} 2X-1 = \sum_{x=1/K}^{A}\left(\frac{2x}{K}-\frac{1}{K^2}\right) = \lim_{K\to\infty} \sum_{x=1/K}^{A}\left(\frac{2x}{K}-\frac{1}{K^2}\right)= \int_{0}^{A} 2x dx = A^2$$

And then prove why Pythagora was right, since there is in some case a finite number of moves to let:

$$A^2=C^2-B^2$$

And this is trivial to be proven using such sums of 1x1 squares or 1*1/2 triangles.

But also that Fermat was since if A, B, C are integers it is clearly impossible to find an arrangement in a finite number of moves to:

$$$$ $$A^{2m+1}= \sum_{x=\frac{1}{\sqrt{A}}}^{A^m{\sqrt{A}}} \left(\frac{2x}{\sqrt{A}}-\frac{1}{A}\right) \neq \sum_{x=\frac{1}{\sqrt{C}}}^{C^m{\sqrt{C}}} \left(\frac{2x}{\sqrt{C}}-\frac{1}{{C}}\right)- \sum_{x=\frac{1}{\sqrt{B}}}^{B^m{\sqrt{B}}} \left(\frac{2x}{\sqrt{B}}-\frac{1}{{B}} \right)$$

While there will always be a solution if C (f.ex.) can be an irrational so if:

$$$$ $$A^{2m+1}= \sum_{x=\frac{1}{\sqrt{A}}}^{A^m{\sqrt{A}}} \left(\frac{2x}{\sqrt{A}}-\frac{1}{A}\right) = \int_{x=0}^{C^m{\sqrt{C}}} \frac{2x}{\sqrt{C}}dx - \sum_{x=\frac{1}{\sqrt{B}}}^{B^m{\sqrt{B}}} \left(\frac{2x}{\sqrt{B}}-\frac{1}{{B}} \right)$$

Because we can geometrically prove that any integer C and any Rational C do not allow to complete the area is C^n-B^n with n>3 is not the power of a rational (by this contruction has an irrational limit cannot be risen via a finite number of step, so moves in tessellation).

Hope you've patience enough to turn your mind in this direction do not needs the use of abstract concepts of rings etc...

[Notification Bot]

A message was moved here from #FLT > Functoriality of infinite completion for number fields by Johan Commelin.

[Johan Commelin]

@Stefano Maruelli I don't think your message had anything to do with the topic of the thread you posted in: "Functoriality of infinite completion of number fields"
So I have moved it to a new thread.

[Stefano Maruelli]

The only difficoult part of this proof is to understand how to show (geometrically onto the cartesian plane !) that the Real C that allow the equality in the case of the integral is Bigger than any Rational C we can use to tesselate the area represents C^n-B^n from n=3. And it is clearly done showing that such Real C is a Dedekind Cut, we can approach using rational, but we can reach it just via an infinite process is the integral. Here the first simple exchange of variable:

MARUELLI-FROM-INTEGER-TO-RATIONAL.jpg

and why Pythagora was right, still if the geometric construction depends on an irrational limit... but under a linear 1st derivative...

FLT-N2-NON-UNIQUE-FACTORIZATION-IN-R-Q.jpg

Was not easy to comes out from this problem... but was just question of good definitions on the right picture. (hope the preview and the edit button will ne added soon here !)

[Notification Bot]

A message was moved here from #FLT > Functoriality of infinite completion for number fields by Johan Commelin.

[Johan Commelin]

@Stefano Maruelli Please do not post in unrelated threads.

[Stefano Maruelli]

Pls check how the concept of a morphism can be easily seen as an exchange of variables in a Sum that describe the power of an integer. Sorry an administrator don't understand the subject is strictly related to this problem moves them here: #new members > Stefano Maruelli @ 💬

[Stefano Maruelli]

It is strictly related to FLT, morphism and is a more simple version of the Ring of Adeles property !

[Ruben Van de Velde]

We're looking forward to your sorry-free formalization

[Ruben Van de Velde]

Returning to this stream after @Johan Commelin has moved your message out of it does not seem like a sensible idea

[Stefano Maruelli]

I add several "sorry"16 years ago, but from that time no mathematician seriously take in count my results, nor help me in pubblication. We are saying the same things, by my complicate modulus algebra is several order of magnitude more direct and simple than Abstract (abstruse) Alegebra needs Lean to be checked.... because no human undestand his errors in bla, bla definitions. The relation to Adeles can be seen here and spot the problem of the abstract definition: both n=2 and n>2 shares irrational factorization, but just in n=2 thanks to the linear 1st derivative the product of 2 couples of irrationals produce the same integer value. (thanks for the edit button !)

[Notification Bot]

2 messages were moved here from #FLT > Functoriality of infinite completion for number fields by Kim Morrison.