I would like to have a little discussion on the meaning of the word "short proof" which appears here and there without being defined. It is a general "experimental" fact that if hard problems or conjectures cannot be solved at a moment t in time in spite of the efforts of many great (and less great) mathematicians, it is mainly because at this instant t adequate tools are missing. Examples are plentiful. Let us cite the problems of constructing  the cubic root of 2 by ruler and compass, or computing the roots of the quintic polynomial by radicals, which resisted for many centuries before being solved in one easy stroke by Galois theory. And FLT of course. I agree with C. Landauer when he reminds us "that we waited 340 years for the first proof [of FLT], so patience [for a short proof] is warranted " . But again, what is a short proof ? If we consider that Wiles' work on  the modularity of elliptic curves defined over Q  is not merely a long proof of the Shimura-Taniyama-Weil conjecture, but actually  a fundamental development in the theory of modular forms and elliptic curves, then the short proof that V. Christianto waits for is already there.: the modular machinery constructed by Wiles and his predecessors is a powerful multi-tool, and one manifestation of its power is a proof of FLT (using the elliptic curve of Hellegouarch-Frey and the "level descent" of Ribet) which takes only a few pages. Even better, in the recent developments of the theory in the wake of Wiles' achievement, Khare and Wintenberger have proved a conjecture of Serre on modular (in both mathematical senses of this word) representations, and this reduces the proof of FLT to a few lines ! To summarize: one must not confuse the construction of the general machinery (which is a natural step in the development of science) with its application to the proof of a particular (and sometimes not significant) small problem. Galois theory was not designed only to get rid of the quintic. Differential and integral calculus were not invented only to show that the orbits of planets are ellipses. I think this should close any discussion on a "short proof" of FLT. Unless "short" is meant as "elementary", but this is a different matter    ¤

There are no simple proofs at this point.

Thank you, John, for your answer. Best wishes

In my opinion, the description on fermatproof.com is not a proof - it depends on pictures of intuitive properties of geometry that are not justified and likely not true in the generality needed - this is one of the two most common erroneous proof methods we have seen over the last 350 years

also, computer studies have taken the possibilities up to very large numbers without a solution (as the proof by Wiles shows there cannot be)

on the other hand, it is widely (if not universally) surmised that the proof fermat thought he had relied on a property (unique factorization) of integers that does not hold in the algebraic number systems used in the proofs (there is a very short pseudo-proof of the theorem that depends on the unique factorization, reachable at the college number theory level) - this is the other common incorrect proof

none of this is trying to argue that there cannot be a short proof, but the number of smart people who have tried to find one leads me to think that we need some more basic theory first

a similar example holds in finite group theory - the odd order proof (no finite non-commutative simple groups have odd order) was 255 pages long when it was first published, but now, with the systematization of the methods invented in that paper, and the widespread use of them for other problems in group theory, the proof can be reduced to 20 or 30 pages (and made simple enough to be checked by a computer proof assistant) - perhaps something analogous will happen with fermat's theorem

Thank you, Christopher, for your answer. Does it mean that someday we can expect that there will be shorter proofs on Fermat last theorem? Best wishes

I do expect one, though you have to remember that we waited 340 years for the first proof, so patience is warranted

"Beside rigorous proofs of Fermat's last theorem": can a proof be not rigorous? A proof is rigorous, or it is not a proof at all. I wrote my PhD thesis on a topic related to FLT, and I haven't yet read any elementary proof of it. Which doesn't mean, of course, that such a proof doesn't exist.

Dear Christopher and Fabio, thank you for your answers. best wishes

Dear all, thank you so much for all answers. Allow me to upload a summary of this discussion in pdf file. Hopefully it will stimulate further study on this issue. best wishes

The only credible attempts were made by Kummer , Wieferich...but these are not a complete proof .

@Zekiri, thanks for your answer. Best wishes

just a remark, Fermat was not the first who worked on this problem, the FLT for n =3 and 4 was known by previous mathematicians, even some attempt to resolve the above cases.

EDIT:

In the last page of this manuscrit , you will find the statement of "fermat last theorem" for n=3 , there is seven opens problems (for him) which the author state , the statements is as following :عدد مكعب قسم بقسمين مكعبين

"a cube divided to two cubes"

this is the image of the page, and the link of the manuscrit

https://www.qdl.qa/%D8%A7%D9%84%D8%B9%D8%B1%D8%A8%D9%8A%D8%A9/archive/81055/vdc_100044787806.0x000056

@Ali, thank you for your answer. Best wishes

Can u please read a proof submitted by me as well.

Thanks, Vinay. Ok i will try to read your proof, but please send your paper to my email: [email protected]. Best wishes

FERMAT IS CERTAINLY LINKED WITH RIEMANN'S HYPOTHESIS !

Found simple but not elementary proof for your numerically tested hypothesis: k=1 iff n=2. The hypothesis and the proof of it can be easily generalized.

@Valdemars: thank you for your answer, yes it is very interesting if you can find proof for our numerical test showing that FLT is unique, i.e. k equals 1 only iff n equals 2. Do you find proof to generalize that finding? Please kindly send your paper to my email: [email protected]. Thanks

The proof is attached to the post.

@Valdemars: thank you for your paper, it seems interesting enough to publish. If you wish, you can write up a complete paper then send it to Editor in Chief of IFSL journal. His email is : [email protected]. But it is okay if you do not wish. Best wishes

I think that it is possible to take again the Kummer's approach via the Bernoulli Numbers .

Please visit my page & see my article related to Fermat's in relation to trigonometry & polar coordinate form.

Modified Pythagoras and Fermat Formula ( MPFF)

I would like to have a little discussion on the meaning of the word "short proof" which appears here and there without being defined. It is a general "experimental" fact that if hard problems or conjectures cannot be solved at a moment t in time in spite of the efforts of many great (and less great) mathematicians, it is mainly because at this instant t adequate tools are missing. Examples are plentiful. Let us cite the problems of constructing  the cubic root of 2 by ruler and compass, or computing the roots of the quintic polynomial by radicals, which resisted for many centuries before being solved in one easy stroke by Galois theory. And FLT of course. I agree with C. Landauer when he reminds us "that we waited 340 years for the first proof [of FLT], so patience [for a short proof] is warranted " . But again, what is a short proof ? If we consider that Wiles' work on  the modularity of elliptic curves defined over Q  is not merely a long proof of the Shimura-Taniyama-Weil conjecture, but actually  a fundamental development in the theory of modular forms and elliptic curves, then the short proof that V. Christianto waits for is already there.: the modular machinery constructed by Wiles and his predecessors is a powerful multi-tool, and one manifestation of its power is a proof of FLT (using the elliptic curve of Hellegouarch-Frey and the "level descent" of Ribet) which takes only a few pages. Even better, in the recent developments of the theory in the wake of Wiles' achievement, Khare and Wintenberger have proved a conjecture of Serre on modular (in both mathematical senses of this word) representations, and this reduces the proof of FLT to a few lines ! To summarize: one must not confuse the construction of the general machinery (which is a natural step in the development of science) with its application to the proof of a particular (and sometimes not significant) small problem. Galois theory was not designed only to get rid of the quintic. Differential and integral calculus were not invented only to show that the orbits of planets are ellipses. I think this should close any discussion on a "short proof" of FLT. Unless "short" is meant as "elementary", but this is a different matter    ¤

Enjoy my short but no simple proof

https://www.researchgate.net/publication/301674377_Fermat%27s_Last_Theorem_Proof_fixed_edition?ev=prf_pub

Data Fermat's Last Theorem (Proof) (fixed edition)

Didn't have much time to fully comprehend the attempted proof however below are my thoughts on it. The reading of text was interesting, nevertheless there are several parts which need revision.

The grammar could be improved.

The equations and figures need to be numbered.

The attempt to prove that  []+(y/x)^n]^(1/n) is a rational number wrong - the correct way is to note that x and y are integers and see that an integer z is a product of integer x and a number of yet unknown property. That's why ]+(y/x)^n]^(1/n) is a rational number.

1+R1n=R2n is not equivalent to R1n+R2n=1. The correct equivalence is R2n-R1n=1. 

R12+R22=1 is not equivalent to R2-R1=1/(R1+R2). Incidentally, the last equation is equivalent to 1+R1n=R2n in case of n=2.

How R2-R1=R3 is true? R3 is not previously mentioned. Did you mean denote R3=R2-R1? Seeing that the same sentence appears several times I think it needs to be put above the text "For n=2".

There's quite a large number of "Equation 1" and "Equation 2".

It would be better to state that function 1 is R1=R2-R3, function 2 is R1=1/R3-R2 , etc.

I think it would be better to replace Figure 1 with the case of for, example, R3=2/5 because it now gives impression that 1/R3 is always an integer.

What do you mean by "Equation X is an axis of symmetry of Equation Y"? Is it meant as "Plug Equation X into Equation Y, and Equation Y won't change"? Or did you mean "Graph of Function Y is symmetrical against graph of line X". Or maybe "Graph of line X serves as an axis of symmetry for graph of function Y"?

Didn't verify numerical calculations.

For me it's not obvious how property: "Graph of Function Y is symmetrical against graph of line X" is equivalent to "Intersection of graphs Y and line X has rational coordinates),  In backward direction it's false.

The math for factorizing expressions R2n-R1n=1 is probably is correct(didn't check), however, the signs of monomes in the original factorable expression seem to be wrong.

I don't see how verification of cases n=2,3,4,5 implicates anything about n>=6.

I’ve taken a quick glance at your « proof », which is longer than Fermat’s and shorter than Wiles’, but I must agree with  the flaws pointed out by Valdemars and could point out many others. Besides, you just stop at the exponent 5,  implying some obvious (?) induction process. Historically, the cases of exponents 3 and 5 had been proved – definitely -  by Euler and others, but certainly no one dared to consider that this could be extended just like that to higher exponents.

Now, why did I only take a quick glance, and not a thorough examination ? The reason is not with the difficulties -  because of your grammar and the numerous errors - to really see what you were after. The reason is a fundamental  one.Your argument only appeals to elementary operations, more precisely operations derived from addition and multiplication, and their inverse .Your goal is to manipulate these so called « algebraic » operations to show that a certain expression cannot stay in the field Q (the rationals). It does not matter whether your calculations are correct or not, because you CANNOT conclude in this way. Let me explain : if you replace Q by the field Q_p of p-adic numbers Q_p (p a prime), the same algebraic operations can be performed, but there exist values of x and y in Q_p such that your expression remains in Q_p. Even if you don’t know what Q_p is, its existence and simple logic show that algebraic arguments alone (in the precise sense above) can certainly not prove FLT.

Sorry Valdemārs Plociņš, when my assistant transcribed leave an editing mistake

right is R2n-R1n=1

Thanks for your comment

https://www.researchgate.net/publication/302602715_Fermat%27s_Last_Theorem_Proof_fixed_edition_2?ev=prf_pub

Data Fermat's Last Theorem (Proof) (fixed edition) 2

I know the grammar can be better.

only transcribe my notebook. Regards

if you want to know what I mean about simetry of curves please read Taniyama - Shimura,  Weierstrass , Mordell - Weil

Just fixing in few places a sign of monomes does not make the text fixed. The critical flaws are yet to be addressed - 1) formulation and proof (or referencing to it) of lemma which makes  use of symmetry and rational points; 2) validation of inductive step; 3) improvement of the readability of text.

Correct me if I'm wrong but the only common field I'm aware of, in which those mentioned great mathematicians worked are elliptic curves. There's a known procedure of obtaining all rational points of elliptic curve through application of symmetry. However, 1) the function (R2-R1)/(R2^n-R1^n)=const is not elliptic curve; 2) the intersecting curve through which the rational points are obtained is not the axis of symmetry for the elliptic curve; 3) consider equation x*y/(x+y)=z=const. Through simple manipulations one can obtain that y=z*x/(x-z). The graph is a hyperbole which has an axis of symmetry y=x-z. All other lines are not axis of symmetry. Despite it, one can draw infinitely many lines which intersect that hyperbole and the intersection will have rational coordinates

I think that part not present in the best way, and even think that my job requires adjustments, and this is the right place to present this type of work and to correct errors, I never said that my conclusion is the only one, but I firmly believe that this close to it. instead of feeling upset the contrary I appreciate your comments in the best way. Regards

Valdemars

please considers that it is not any line , only R2 -R1 = Z . my references are not because I copy them , but to give another point of view. 

An elementary proof of the Last Theorem of Fermat is available on site

http://arxiv.org/abs/1604.03753

good Reading

Carlos Mauricio Santacruz,

Seems like I wrote wrong equation. I did mean to write 3/2*x*y/(x+y)=z=const. Solving system of equations

3/2*x*y/(x+y)=z

y-x=z

for x and y gives (-2z/3;z/3) and (z;2z) as coordinates for intersection. Assuming z is ratiional number, x and y are also rational.

Andrea Ossicini,

I see that you made an effort to write your paper in plausible way, therefore catching an error will take time. However, the "fundamental theorem" on page 8 involving bi-quadrates I see as wrong.

You have to show what is the error present in the "Fundamental Theorem" on page 8, I unfortunately do not see it !!!

Best regards.

The necessary and sufficient condition that the equation X^n + Y^n = Z^n, where X, Y, Z are pairwise relatively prime, and n is odd prime, does not provide integer solutions is that all the three indeterminate X,Y and Z are equal to a perfect biquadrate.

How I interpret that statement: If GCD(X,Y,Z)=1 and X, Y, Z are not biquadrats, then equation X^n + Y^n = Z^n has  a solution for n as an odd prime. And that's false.

Edit. Replace "GCD(X,Y,Z)=1" with "GCD(X,Y)=1, GCD(X,Z)=1 and GCD(Y,Z)=1". 

Read the additional notes 2 and maybe everything will be clearer

Also your statement  "How I interpret that statement: If GCD(X,Y,Z)=1 and X, Y, Z are not biquadrats, then equation X^n + Y^n = Z^n has a solution for n as an odd prime" must be proven.

For example, the Pythagorean equation admits of entire solutions as I determine the parametric solutions that meet.

A careful reading of Appendices 1, 2 and 3 shall be obligatory !!!

All conditions for theorem should be explicitly stated, or referred to within that theorem.

I don't have to prove that statement. I simply used part of so-called "Fundamental Theorem" and applied formula (A => B) (not B => not A).  Or did you mean I need to provide counterexample to that statement?

Your "Fundamental Theorem" has two theorems within it due to "necessary and sufficient condition".  Therefore it needs to be written carefully, in order to reduce ambiguity. 

Pythagorean equation has nothing to do with the "Fundamental Theorem"  -  in case of Pythagorean equation n=2, which is not an odd prime.

If reading of Appendices 1, 2 and 3 is obligatory, why they are appendices? I view appendix as something that is not required to get the idea of text. Appendices could be used, for example, to contain a long proof of lemma.

If the indeterminate X,Y and Z are biquadrates, thanks to the case n = 4, the Fundamental Theorem is true.

In the opposite case however it is the chapter 2 to ensure that the nature of the indeterminate X, Y and Z  can not be different from that of a square, and if this added to the nature of the obvious solutions of Pythagoras equation arrivals to the condition of biquadrates.

The link between the Diophantine system with three homogeneous equations of the second degree and the equation of the elliptic curve Frey assures me that the Fundamental Theorem is correct.

I agree with you that perhaps it would be better to write a more direct proof of the Fundamental Theorem, I unfortunately have not been able to find a more effective exposure, but I'm sure of its validity.

In fact that would explain why  Fermat is limited to the case n = 4 !!!

He did not need another to generalize his Last Theorem.

@Luis C. L. Botelho

What do you think of the recent attribution of the Abel 2016 prize to A. Wiles "for his stunning proof of Fermat's Last Theorem by way of the modularity conjecture for semistable elliptic curves, opening a new era in number theory", to quote the Abel Committee ? I don't think that the Committee fears "anti-establishment" accusations such as a few we may have heard here and there, but note that on this occasion (as is usual), a small conference around the laureate's work has been held, with an "expert" talk by Henri Darmon and a "non expert" report by Alex Bellos (which can surely be found on the web). But I would like above all to draw your attention to the motivation of the Committee :

" The Abel Prize recognizes contributions of extraordinary depth and influence to the mathematical sciences(...)Proof of the Taniyama-Shimura conjecture, a result now known as the modularity theorem, meant that Wiles had also proved Fermat’s Last Theorem, thus bringing to a close a chapter of mathematical history that began 350 years previously. Yet more than settling an old and famous riddle, the impact of the modularity theorem onmathematics has been immense. Wiles demonstrated a fundamental structural connection between elliptic curves and modular forms, a rich and important result within number theory with many deep consequences. He also devised a powerful conceptual toolkit that has been used over the past two decades by other mathematicians in spectacular ways ". You will not be surprised if I insist on the second part of te acknowledgement.

A. Wiles is deserved the Abel Prize, but that does not mean it can not be an elementary proof of the Last Theorem of Fermat.

In my work I have shown that there is a close link between the elliptic curve Frey and so-called '' double equations "Fermat and that from this we can derive an elementary proof of his Last Theorem.

@ Valdemars Plocins

Logically you are right to claim that  (A => B) (not B => not A), and I don't think that Andrea contests that. What he asks you, I think, is to prove your claim (in your own words) : " if GCD(X,Y,Z)=1 and  [one of] X, Y, Z is not a biquadrat, then equation X^n + Y^n = Z^n has  a solution for n an odd prime. And that's false"  . I'm convinced that this would be as difficult as showing Andrea's original assertion. Of course, what I say does not mean that HIS proof is correct.

@Thong Nguyen Quang Do:

In light of how ambiguous the original "necessary and sufficient condition" theorem, from which my statement is derived, will you forgive me some forgotten words? Also, I didn't mean to prove the general statement that there are no solutions. All that I intended was to find a counterexample to the statement that there should be a solution with certain properties.

Let's consider following example: X=1, Y=1, Z=8. X, Y and Z are pairwise relatively prime, Z is not even a square. Let's find a value of  n. According to statement it should be an odd prime.

1^n+1^n=8^n

2=8^n

n=-3 - is not a prime.

It means that a counterexample for that theorem is found.

I don't understand your counter-example. But anyway, there is a systematic process to get counter-examples to FLT in degree p (p an odd prime) with variables in the ring Z_p of p-adic integers or the field Q_p of p-adic numbers . In particular, you can find p-adic integers x, y, z which are not all biquadrates and yet are solutions of  (F_p) , the Fermat equation of degree p. Since all ALGEBRAIC calculations (addition, multiplication,and the inverse operations) which are performed in Z and Q can be repeated word for word in the p-adics, this shows in advance, I believe,  that the proof given by  Andrea (that solutions of (F_p)  must be biquadrates) cannot work. Same situation for the previous proof of Martin Cruz.

Consider the following

 Theorem:  There are not  integers X,Y, Z   pairwise relatively prime such that X^n+Y^n=Z^4 for n prime  >2.

 This result may be obtained by the application of a theorem of  H. Darmon and

L. Merel, who have shown that for n greather than 4   the equation  a^n+b^n=c^2 does not admit integer solutions with a,b,c  relatively prime.  [ l"Winding quotients and some variants of Fermat's Last Theorem."    Journal für die reine und   angewandte Mathematik  490 (1997), 81-100]  and  a theorem proved by N. Bruin, who claims that  the equation    X^3+Y^3=Z^4   has integer solutions if G.C.D.  (X,Y;Z)>1   [“On powers as sums of two cubes”, Algorithmic number theory (edited. by W. Bosma), Lecture Notes in Computer. Science. 1838, 169–184, Springer, Berlin, 2000].

 Therefore it suffices to show that Z is a biquadrate  to prove the Fermat's Last Theorem, with X, Y, Z are pairwise relatively prime and n prime >2.

The point is HOW to show that z  in the equation (F_p) is a biquadrate . The situation is the same as previously : if your proof consists only in algebraic manipulations (in the sense above), then again it can be applied verbatim to the p-adics ... and in there you can find, as I said, a triple (x, y, z) which satisfies (F_p) , without z being a biquadrate. This means logically that such an "algebraic proof" is certainly wrong.

z, in the equation (F_p), is a biquadrate  and this is  proved without doubt (see page 14 of my paper http://arxiv.org/abs/1604.03753v2)  in Appendix 3, entitled :

“the arquebus by Euler”.

What is wrong ???

What is wrong ? I don't know, because I was not able to follow thoroughly the calculations (and  it seems that nobody could, up to now, so it's somewhat premature to say that it's "proved beyond doubt") . But the proof is certainly wrong if it uses only algebraic manipulations, as I said.

I have not used only algebraic manipulations, but I have used many remarkable developments of Algebraic geometry with the most refined techniques Diophantine Analysis of the second degree, including the fact that if exists an obvious or trivial solution to the problem, all the others solutions must be obtained from the obvious or trivial solution.

This last statement is present in the pages 39-44 of the book of L. E. Dickson and E. Bodewig – Einführung in die Zahlentheorie, B. G. Teubner, Leipzig & Berlin, 1931.

The Appendix A1 is also more than sufficient to understand this last statement, just study it and understand it !!!

Dear Andrea,

From what I have tried to understand of your calculations, i can't agree that you used "  many remarkable developments of Algebraic geometry with the most refined techniques Diophantine Analysis ", especially when you cite a book which dates from 1931. And I maintain that throughout all your paper, including the appendices, you only manipulate algebraic expressions, polynomials and rational fractions. So let us stop here a discussion which leads nowhere.  Since your preprint has been posted on Arxiv, the next natural step is to submit it for publication in an international journal specialized in number theory. It's  the only way out. No bad feeling, I hope.

@David Cole

Your deduction (with a "smiley") of (4) and (5)  from (3) reminds me of a funny cartoon: here we have this famous (or not) mathematician at the blackboard writing "Here a miracle occurs...", and going on. Then some one in the audience raises a hand: "Could you please explain the miracle?"

FYI:  A Simpler Proof of Fermat's Last Theorem

Data Proved a statement about a conditional Diophantine equation

Dear David, thank you for your file. Btw i just uploaded a new paper on simpler graphical method to solve FLT. You can find it in my new project.

Yours

Victor

Thanks, Victor!

Here's another attempt:

http://vixra.org/pdf/1701.0630v1.pdf

An elementary proof of the Last Theorem of Fermat Is in

https://arxiv.org/abs/1704.06335

But it's in Italian

Appendix 3 is a tribute to Euler

The proof is much clearer than the one contained in the paper:

The origin of the Frey elliptic curve in a too narrow margin.

Good reading  !!!!!!

Dear Pietro

thanks for the reference, I ll try to read it. btw is there anyway to find out the english version of this file?

yours

vic

Oh! It's  Prof. Orsini's contribution. It has been discussed at length right here in this section.

Dear Prof Thong Nguyen Quang Do is true what he says.

Please read Appendix 3 is can now verify that the proof really is full !!!

I think the Italian language is not a difficult language.

Best regards

All right, I'll try honestly to read the appendix 3. But why did the author write his paper in Italian ? If his purpose was to get a maximal exposure, then (like it or not) English would have been a better choice.

You are right, I wrote to A. Ossicini, asking for a English version of his paper.

Did he answer me with a question?

Do you know the story of Grigorij Jakovlevič Perel'man?

He also informed me that he will make available, scheduled for May 7, an even simpler version than the current one.

Without Appendices and with a direct Proof !!!

I apologize, but the paper of A. Ossicini  is scheduled to be announced at Tue, 9 May 2017 00:00:00 GMT in ArXiv :

Title: The lost proof of Fermat's last theorem

Authors: Andrea Ossicini

Categories: math.GM

Comments: 10 pages, in Italian - The following version is simpler and contains

a direct proof

MSC-class: 11D41, 11G05

Greetings to all.

9 May 2017

Today I finally have been able to admire Fermat's wonderful Proof !!!!

https://arxiv.org/ftp/arxiv/papers/1704/1704.06335.pdf

An English version of this paper would be appreciated. !!!

W Fermat, but especially an immense Euler thanks.

Dear Professor Thong Nguyen Quang Do has a few things about the second chapter of the paper : https://arxiv.org/ftp/arxiv/papers/1704/1704.06335.pdf, that the alleged proof of Fermat?

He would have liked his comment, given his experience in the field of Number Theory.

Best regards

Ing. Pierre Lacroux

What do I have to think of this long silence?

Perhaps the proof of Ossicini is correct?

We are, perhaps, in the presence of an admirable demonstration of Fermat?

Please read my proof...

regards

Dear Pietro Lacroux,

I have been attending a conference abroad, which explains what you call my "long silence". But anyway, let me make my position clear :

1) I don't read Italian, which makes it difficult to follow possible subtleties (if any) in the proof. I repeat that Prof. Ossicini should have presented an English version in order to attract the widest attention possible

2) Even if my Italian is more than poor, I skipped through the calculations to get an idea of the argument. The writing is certainly better than in the first versions, but my conclusion remains the same as in my previous discussions with the author :     

- first, the allusions to elliptic curves are purely cosmetic, the specific theory of these curves does not intervene in the main part of the proof  

 - second, the central calculations are purely algebraic, more precisely  they only use the operations + and x, as well as their inverses, i.e. they take place in a field, sometimes extending that field by taking radicals. As I said before (see the numerous discussions in these columns) and repeat now, these calculations can be performed without any change in the field Qp of p-adic numbers, where the Fermat equation does admit a non trivial solution. This means that the "admirable demonstration" here is certainly wrong, without even bothering to check where things went astray

3) I'm afraid that I'll not be able to convince neither Prof. Ossicini  nor yourself (?), but now that this note is posted on ArXiv, many people have surely read it, and you should have other reactions than my own

Best regards

Dear Professor, that you do not know Italian, it seems to me a huge excuse.

The proof  seems to be contained in just 3 pages (see chapter 2: la prova).

Therefore, if it is not correct, it should not be difficult to verify it.

Best regards

See the many requests for a version in English

The following paper is scheduled to be announced at Thu, 22 Jun 2017 00:00:00 GMT in ArXiv  with  URL  https://arxiv.org/abs/1704.06335.

The lost proof of Fermat's last theorem

Andrea Ossicini

 It is shown that an appropriate use of so-called double equations by Diophantus provides the origin of the Frey elliptic curve and from it we can deduce an elementary proof of Fermat's Last Theorem

Comments:

The following version is simpler and contains a direct proof –

The paper is bilingual :10 pages in English and 10 pages in Italian

Best wishes to the professor Thong Nguyen Quang Do

See the many requests for a version in English

The following paper is also availabe in    http://vixra.org/abs/1706.0421

The lost proof of Fermat's last theorem

Andrea Ossicini

 It is shown that an appropriate use of so-called double equations by Diophantus provides the origin of the Frey elliptic curve and from it we can deduce an elementary proof of Fermat's Last Theorem

Comments:

The following version is simpler and contains a direct proof –

The paper is bilingual :10 pages in English and 10 pages in Italian

The proof is fantastic !!!

Finally it is also in English

A definitive version of my paper "THE LOST PROOF OF FERMAT?S LAT THEOREM will appear on July 18 in    https://arxiv.org/abs/1704.06335  .

Comments:

13 pages, 1 figure - The Fundamental Theorem of this publication is a theorem equivalent to the Fermat's Last Theorem. The main goal is to rediscover what Fermat had in mind (no square number can be a congruent number). The current version is English language, that is improved !!!.

Good reading at all.

Any comments or comments are welcome

Prof. ANDREA OSSICINI

Dear Professor Ossicini,

I can see that there are no more reports coming from the forum.

Perhaps we can finally enjoy the wonderful proof of Fermat, which for over three centuries has been sought in vain.

My most sincere compliments.

Ing Pietro Lacroux

Post scriptum.

How Victor Christianto does not expose ?

What did Prof. Thong Nguyen Quang Do do?

I have an old one in the margin. This is truly short.

Let's think c^n as an 'equilateral' reserved space and inside it has

similar object of b^n in shared center point. Between their surfaces

must exist room for a^n. Now we assume that b, c & n are integers.

What are the terms of possible solution of a^n + b^n = c^n, so that

a is integer too?

c > b > a.

Set k = 2/(c-b) -> (ka)^n = (kb+2)^n - (kb)^n | sum - 2^n

=> (ka)^n - 2^n = (kb+2)^n - (kb)^n - 2^n

=> (ka)^n - 2^n = (2^n - 2)(kb)*sum(i=1...n-1){(n i)(kb)^(i-1)}

(n i) = n!/[(n-i)!i!] is binom factor.

Then we distribute all units into every 'side' (2n st) eliminating

halves at the same time:

2 * (2^n - 2) / 2n must be integer.

Why? Because of 'squeezing' directions from form of (kc)^n - (kb)^n

to form of (ka)^n.

Besides, the base ka is even-like and (ka)^n is build up with shells

similar to (kc)^n - (kb)^n and the core is symmetrical object like 2^n.

The number of (n-1)-dimensionally oriented units (on 'sides') divided

by 2n is integer and 2n is the smallest divider among binom factors of

power of n.

Primes seem to be solutions of this conditional statement.

We have 2^n units left. Let's do the same to them:

2^n / 2n must be integer.

Why? because when 'squeezing' we get 2^n units freed every step

and the new growing 'equilateral' n-object (for (ka)^n) get center

of every side filled up as last position!

These requirements have only solutions:

n=1, n=2.

:)

Today I downloaded the latest version of Ossicini work on Fermat's Last Theorem.

see:    https://arxiv.org/pdf/1704.06335.pdf

I was astonished,  in chapter 4  Ossicini wrote in the details the demonstration he had in mind Fermat.

A proof  with the method of complete Induction, which is really marvelous!!!

It begins with  the proof for n = 3 and n = 4, of which Fermat certainly possessed a demonstartion, and completes the whole with a theorem dealing with all even exponents greater than 2 and not multiples of 4.

For odd exponents, there is a "logical" step in the  method of Mathematical Induction.

The margin seems to be really unveiled, and since in all cases it uses the method of infinite descent, which is a kind of backward Induction, I feel that this is the most beautiful and at the same time simplest proof of all the history of mathematics.

I wish all the success to Ossicini and thank Ossicini you because in the face of some of my requests it has always been kind and courteous.

This email is in a sense a way to show him all my gratitude !!!

Ing. Pietro LaCroux

My proof is simple, but require you digest a new way to use the Sum Operator:

Extending the Telescoping  Sum property into the Rationals as a New Modular Algebra and taking it's Limit (the Integral) as reference.

Summarizing the trick: it start from the equation (pls see picture):

\Delta+=\Delta-

or:

\int_{A}^{Xb} 2nX^{n-1} = \int_{Xb}^{B} 2nX^{n-1}

Then I'm able to transform it in a "special" Sum, capable to work between Known, Fixed, Irrational Limits, to show that from n=3 the Left hand (is a genuine power of an integer) will be different from the Right one (for a clear simple math. reason, or that there are more terms into the Sum between the same limits).

While for n=2 the equality \Delta+=\Delta- will take back in few lines to A^2+B^2=C^2 so giving us the reply that there can be a solution to the problem.

Here all the story....

Deleted research item The research item mentioned here has been deleted

Deleted research item The research item mentioned here has been deleted

Here the new instrument I developed for Powers of Integer, Rational and Irrationals too, extending the Sum properties in case the Telecoping Sum Property can be used.

Hi Victor Christianto. We have proof in this link

Preprint N.Z. Orujov, Z.S. Orujov In the previous version of the manu...

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