One problem with the proliferation of "open access" journals is the decrease in quality. A good example is this "proof" of Fermat's Last Theorem by a guy who seems to specialize in rather eccentric papers. This paper was passed around to great laughter at the van der Poorten memorial conference in Australia. (The list of keywords alone is funny to a professional mathematician.)
This journal - the Journal of Mathematical and Computational Science - and its editorial board should be ashamed of publishing this junk.
Sunday, March 18, 2012
Subscribe to:
Post Comments (Atom)
200 comments:
I had a long correspondence with one of these folks. A fine retired gentleman who made one mistake after another in an attempt to use basic high-school math to prove FLT. I kept telling him that although I would never say never, the tools that he was using were simply of too feeble mathematical content to be likely to prove anything. He persisted till I gave up answering his e-mails.
Wasn't this proved by a mathematician some time in the 1990s?
Should I infer that you are opposed to the basic idea of an open-access journal, or are you simply pointing out that open-access can lead to a proliferation of bad ones? (I happen to believe that open-access journals are a great idea, but they place ever more responsibility for integrity on the editors.)
SLC:
Yes, Fermat's last theorem was proved by Andrew Wiles (with some help from Richard Taylor).
That's the problem with false proofs of true theorems: it's not easy to produce a counterexample.
isohedral:
I edit a true open-access journal - that is, free to both authors and readers - so evidently no, I am not opposed to them.
There are a number of problems with many open access journals as they are currently set up. One is that most such journals have no provisions for waiving the author fees for those without a grant or those in 3rd world countries where the fee to publish might represent an average year's salary.
Thanks for the amusing link. (Surely he is credible, look at how diligent his research review was, why, he asks us to "see [1] or [3] or [4] or [5] or [6] or [7] or [8] or [9] or [10] or [11] or [12] or [13] or [15] or [18] or [19] or [20] or [21]"!)
However, I don't know why you focused on the openness of the journal's access. Closed journals are just as capable of publishing junk. Just look at Chaos, Solitons & Fractals.
Journals that make their money through author charges, rather than through institutional subscriptions, have less reason to insure quality, and more reason to accept nearly everything submitted.
Indeed, one might properly regard the Chaos, Solitons and Fractals issue as also being caused by a departure from sane journal-funding strategies.
If Elsevier sold journals individually, loads of people would want Advances in Mathematics (for example) and pretty much nobody would want Chaos, Solitons and Fractals, and it would not seem like such a big deal at all.
Apparently the author was charged $100 for publishing this paper. Forty papers have been published so far, and $4000 is a decent profit for putting up a web site. (It's clear from this paper that the level of copy-editing provided is minimal, if it's nonzero at all.)
Sadly, I recognize several names on the editorial board, probably people who have not closely looked into the standards of this journal:
http://scik.org/index.php/jmcs/announcement/view/5
Strangely, I know about one third of the editors... I am not sure they are aware of what is going on.
I didn't realize your journal was open access. I'd be interested in chatting with you about that at some point, particularly since I just took over as editor of a non open access journal in January. Do you have a web page up that explains the mechanics (policies, payments, copyediting, etc) for the journal?
Iso:
There are no payments! As I explained, it is completely free to both authors and readers.
You can see information about copyediting on the Journal's home page: http://www.cs.uwaterloo.ca/journals/JIS/ .
The paper is ridiculous. Thanks for the laugh. I shared it with my colleagues in the math department at Uppsala today and also linked it on my blog.
Incidentally, please be informed that the Goldbach conjecture has now been proven to hold:
Ikorong Anouk Gilbert Nemron. A proof of the Goldbach conjecture and the strong attachment to the Fermat's last conjecture. South Asian Journal of Mathematics, Vol. 1(2), 68- 80, 2011.
I note that Ikorong Anouk Gilbert Nemron is part of the editorial board of the South Asian Journal of Mathematics.
Nemron is not the only editor of the South Asian Journal of Mathematics who has published a "proof" of the Goldbach conjecture in the South Asian Journal of Mathematics:
http://www.sajm.com.nu/sajm2012_2_2_16gandhi.pdf
A Google search for "Ikorong Anouk Gilbert Nemron" yields several papers more or less identical to his FLT paper "proving" other famous results.
Is Nemron actually a member of the Centre de Calcul de Enseignement et de Recherche of the Université de Paris VI (Pierre et Marie Curie)?
The South Asian Journal of Mathematics seems to be full of weird papers, such as http://www.sajm.com.nu/sajm2012_2_2_11gueye.pdf. Aside from what must be translation issues (switching between "prime" and "first"), some things just seem inexplicable. For example, n^2 is used to denote squaring, but n^* is used for cubing.
Well, his address is ...@ccr.jussieu.fr. Replacing this by www.ccr.jussieu.fr brings us to web page of a department which looks like computing support division. In other words, he's probably a computing officer (centre du calcul, in french, probably means computing [support]). So, my guess is, he's a computer hacker with lots of free time and/or good copy/paste skills.
"A good mathematical joke is better, and better mathematics, than a dozen mediocre papers." - J. E. Littlewood [taken from the quotes page of UC Berkeley Prof. Emeritus Paul Chernoff]
In this case the joke and the (less than less of a) mediocre paper are identical.
See the note on http://people.cs.uchicago.edu/~razborov/.
See Wile's Proof of Fermat's Last Theorem at http://www.coolissdues.com/mathematics/Wile'sproofofFLT.html
If you find Andrew Wiles proof is difficult, Pl. read on the intenet(open access pdf)CMNSEM,Vol.2 ,No.3,March 2011(General proof),CMNSEM,Vol.1,No.6,September2010(correct and simple use of method of infinite desscent of Fermat, in Case.N=3)CMNSEM,Vol.Vol.1,No.3,April 2010(except trivial typos,wonderful proofs seem OK)Extremely short simple proof for n=7 in CMNSEM,VOl.1,No.9,December 2011,However, a mistake may be seen in n=5 case published in the same journal.
Simple proofs for specific exponents have long been known and aren't of that much interest.
I had never heard of CMNSEM before. It is apparently the Canadian Journal on Computing in Mathematics, Natural Sciences, Engineering and Medicine. It does not appear to have very much of interest in it.
The journal lists no editorial board and the accompanying web pages are written in ungrammatical English. Not very impressive.
Thanks a lot for valuable information on open acesss journals and Fermat's last theorem.The Author,Ikrong Anouk Gilbert Nemrong has already proved Fermat's last theorem seperately-A Complete Simple Proof of the Fermat's last theorem,International Mathematical Porum,Vol.7,2012,np.20,953-971. The paer (wonderful one,if it is OK) in the Joiral of mathematical and computer science had been reviewed within ten days!
Ikorong Anouk Gilbert Nemron has already proved Dermat's last theorem seperately.(A Complete Simple Proof of Fermat's Last theorem,International mathematical Forum,Vol.7,2012,no.20,953-971),This is not Mathematics,I guess,but Logic,etc.. I don't understand.Another one.A Simple Proof and Short Proof of Fermat's last Theirem,Fayez Fok Al Adel,Advances in Applied Mathmatical Analysiss,Vol.3,No.1 (2008)7-15,I never seen any comment on this simple proof,Even though author asked to comment on this proof seperately.This year I did not see it on the internet,I don't know why. Journal is open or not , the correct proofs must be appreciated and recimmended by people like you since mathematically sound Andrew Wiles-Taylor proof is extremely difficult and lengthy. Fermat's marvelous proof(?) is also available noW. Author is a professor.Daniele De Pedis Anyway, your information is useful to Honest Researchers. Thank you.
The Author,Ikrong Anouk Gilbert Nemrong has already proved Fermat's last theorem seperately
No, he hasn't.
It is, A proof of the Goldbach conjecture and the strong attachment to the Fermat's last conjecture by Ikrong....NEMRON ,the paper on the South Asian Journal of Mathematics, 2011,Vol.1,No.2,68-80 had been reviwed within ten daye. I am soory, it is not the journal of mathematical and computational science.Thanks to you , anyway, we know that all big problems in mathematics have solved!!.Thank you also for your cooment on the same person's proof of Fermat's last theorem.
An Anonymous (not me) has reported the notice about the pubblication of my demostration of Fermat's Last Theorem.
If you are interested in, you can found it on
http://arxiv.org/ftp/arxiv/papers/1105/1105.0669.pdf
Any comment/suggestion are welcome
daniele.depedis@roma1.infn.it
It is true that empty vessels make most noise.
Publishing of nonsense in the open access journals seem to be possible then and has been done as well.This acctually kill the moral of honest reserachers who do not want fame or anything.There are many valuable results in open access journals which we can study freely.This must be apprected while hackneying nonsense,
See BealFermatPythagorasTriplets.htm
I'm sorry to come rather late to this discussion, but the "South Asian Journal of Mathematics" has just been drawn to my attention via another route, and a web search for that journal led here.
Over the last few years various vested interests have promoted the "author pays" model of publication under the weaselly phrase "open access" (or even "gold open access"). There have been warnings that the adoption of such a model would lead to a rash of poor papers being published in fly-by-night journals. What I have only now realised is that this is already happening.
As regards I. A. G. Nemron, looking at his abstracts for Concerning the Goldbach Conjecture and Fermat Last Assertion in Commmunications in Mathematics and Applications and A Curious Strong Resemblance between the Goldbach Conjecture and Fermat Last Assertion in Journal of informatics and Mathematical Sciences reveals them (the abstracts) to be virtually identical. Whether the papers are identical, I cannot comment, as both journals adhere to both the doctrines "author pays" and "readers pay".
The elementary "proof" posted on arxiv by Daniele De Pedis is flawed.
At a certain point he claims that gcd(A, B, C) = 1 implies that gcd(B+A, C-A)=1.
This is clearly false. A counterexample is:
A = 2
B = 13
C = 37
Thanks a lot to the genteman's comment on Daniell De pedis elementary proof.
I think no researcher will be interested in commenting on the so called elementary proofs of Fermat;s last theorem not given in refereed journals.It is utter waste of time.Many of these authors do not know Fermat riples do not satisfy Pythagoras' equation.This is, by no means a prof of Fermat's last theorem.Unfortunately, some of these proofs have been given by Professors.
Prof.Fayez Fok Al Adeh's proof of Fermat's last theorem is not simple at all to me,a university teacher, which has been published in the Journal of Advances in Applied Mathematucal Analysis. However,I am trying to understand it.
Answer to Paolo Tassotti.
Your counterexample is not valid, of course the A, B and C variables must fulfill the 's equation!
On the other hand, the property gcd(B+A, C-A)=1 is reported also in the P.Ribbemboin book- "Fermat's last theorem for amateur".
Clearly the statement must be interpreted as: "if there were three integers that satisfy the Fermat's equation then ..."
In any case, recently I have posted on http://arxiv.org/pdf/1105.0669.pdf a new version of my proposed "proof" much more simpler and general, any comment and/or suggestion are welcome.
My request message to the moderator.
On 12:21 PM, August 13, 2012 I have posted my answer to Paolo Tassotti concerning my "proof" of the Last Theorem. In that post,due to my inexperience about this blog, appear two errors:
1) The author of post is Master instead of my name "Daniele De Pedis"
2)on the first row is missing the name, that is:"...must fulfill the 's equation!" instead of "...must fulfill the 's equation!"
Please could you fix these two points ??
Thank for your help
Daniele De Pedis
By reading this Blog(very very yseful), I understand that CMNSEM(Canadian Journal of Computing in mathematics,Natural Science,Engineering and Msrdicine) does not contain the general proof of Fermat's last theorem. This is rather unfortunate.Simple and short analytical proof of Fermat's lasttheorem is there.This will certainly help to reduce the calculation of Prof.Daniele De Pedis(Wrong!) proof drastically.Would you please point out a simple proof of Fermat's last theorem for n=3 case that I( University firstyear student of a third world country) can understand.Thank you very much regarding your comment on CMNSEM as well.
Take it easy, whatever you have heard or read about CMNSEM.It has already published 'A simple and snort Analytical proof of Fermat's last theorem'.CMNSEM is a refereed journal.
It has already published 'A simple and snort Analytical proof of Fermat's last theorem'
I rest my case.
I am going to killed.Long live CNMSEM!!
Thank you so much.Don't you like to comment on the validity of the proof in CMNSEM,Any way ,thank you so mych.Thanks to God ,you are a proffesor in Canada.
Let me join this discussion since I'm also engaged in despicable business of proving FLT. Here is an address of the abstract of the offered proof. Just in case if it may interest somebody
http://www.fileden.com/files/2012/5/4/3300464/Abstract%20of%20short%20Proof%20PDF%281%29.pdf
I am the euthor of A simple and short analytical proof of Fermat's last theorem in CNMSEM and some other as well.
Thanks to all.
I don't argue with the people. Howwever,proving Pythagorean triples do not satisdy the Fermat equation(General)eqyation is not the proof of Fermat's last theorem.
No one can nullify the papers I have published in CNMSEM.
Answer to Anonymous on 2:28 PM, August 13, 2012
If you have found any error in my proof posted on arXiv.org please report it to me, also by private email, if you wish, I will be grateful a lot!
In any case, on the first version of my paper: http://arxiv.org/vc/arxiv/papers/1105/1105.0669v1.pdf you can find an explicit proof for cases n=3 and n=5 based on my proposed technic.
What I did?
First, i heard and read that no simple prrof for Fermat's last theorem for n=3.At the same time ,i understood that correcting Euler's proof (n=3 case)is very difficult.
Secondly ,my aim was to give a simple proof for the general case.
Finally , I wanted to know how peiple look at Andrew Wiles Proof.
Taking these into account I did work on tme theorem for about three years.I think the proofe given in CMNSEM are sufficient at present,I thank you very much for the following,
(1) Anonymous can ask you queations
(2) Your commnet on CMNSEM is fare while others 'Killing' it.
(3) Above all you are a proffesor in Canada I thank you again.
For proof of FLT see http://www.coolissues.com/mathematics/Fermat/fermat.htm
Wile's proof of FLT violates the Pythagorean Theorem. see
http://www.coolissues.com/mathematics/Wile'sproofofFLT.html
For short proof of FLT see http://www.coolissues.com/mathematics/BealFermatPythagorasTriplets.html
Fermat's last theorem has been the most difficult and the most famous theorem in mathemtics.I have read the proofs of some people.All I read that unpublished in a refereed journal are incorrect.It is nonsense to argue with the authors of these proofs to my mind, since I had experienced once .Proof developed in CMNSEM is OK and based on simple mathematics.Any one can challenge against the validity of the proof.
I don't want to be too much annoying in debunking De Pedis's "proof", probably just a waste of time.
However, he tries to construct a univariate polynomial from the original trivariate statement saying "for each choice of the two other variables".
It could be legal unless, some paragraphs below, he mixes up the FLT's trivial solution with this polynomial having a single integer root.
This is clearly false because of the "for each choice of the other two variables" statement.
In any case he never use the property of the exponent "p" of being prime, and doesn't split the problem in the classic two cases (p | abc of not).
He neither takes into account the divisibility property of the binomial coefficient.
All symptoms of an hopeless attempt.
I don't know Paolo mrans Paulo Rebenboim.I have a great respect to the book of him,Fermat's Last
Theorem for Anateurs.Thanks a lot for your comment.
Answer to Paolo Tassotti post.
I don't understand the objection on first part of your post. Please could you "waste a little of your time" to explain better?
About the second part, the answers at all your objections are reported (as said in footnote 2) in the Rimbenboin book.
If you have found any error in my proof posted on arXiv.org please report it to me in clear statments, also by private email, if you wish, I will be grateful a lot!
"All symptoms of an hopeless attempt" are not proof of fault !!
Best regards
'A Simple and Short Analytical Proof of Fermar's Last Theorem' in CMNSEM is actually simple and correct and it must be re-written so that a high-scool student can understand.
I'm again asking to look at concise (1 page) description of 4 steps of offered proof of FLT. Any basic flaw will be seen from it immediately. Its URL
http://www.fileden.com/files/2012/5/4/3300464/Abstract%20of%20short%20Proof%20PDF%281%29.pdf
Regarding the South Asian Journal
of Mathematics, a number of mathematicians, including A. Razborov,
A. Abebe and N. Billor were
listed as editors by the journal
without their knowledge or consent. (Their names have now been removed).
U have received an e-mail from a man sayung that CMNSEM is not a legitmate journal.What one means by this?
The extreme breadth of the journal's coverage; the small editorial board with tiny geographical coverage and unimpressive reputations; the fact that reviews are done in two weeks; the use of Microsoft Word as the text preparation software; -- all these are red flags that something is rather odd about this journal.
I have published all of my papers on Fermat's last theorem in CMNSEM. I actually did not know any bad thing about CMNSEM.
In n=5 (with one of my students) we have made a mistake.Apart from that any one can challenge the validity of the proofs.
As far as I know CNMSEM is better tha the journal 'Advaces in pure mathematics' since ut has not published noncence.
We have corrected the case n=5 and now we have two proofs of Fermat's last theorem (FLT)for n=5 using the Identity of Fermat equation, x+y-z.
A fox, i have no any suitable word to call this man, has sent an E-mail to the authorities of my University saying that CNMSEM is not legtmate. Would you please tell us the open access journal that you edit.We are looking for the second simple proof of FLT similar to n=7 case in CMNSEM with one of my colleagues.R.A.D.P
Even CNMSEM saya 2 weeks review, they take moure than one month.
It is not nice to make fun of the intelectual capacities of someone. And he is not a system engineer, because the math he presented is on a post-graduate level , even if it has mistakes.
It is not clear who has been aimed at by the saying of anonymous of October 28.If it is Ikorong Anouk Gilbert Nemron,many will be against you because his work on Fermat's last theorem is so hilarious.
I must indorm everyone who is interested in Fermat's last theorem that,mainly, the Dean of the university of Kelaniya has decided that CNMSEM is not a refereed journal.I know that the proof of FLT in CNMSEM is correct.However, anyones comment is appreciated.
It is the Dean of the science faculty, professor in mathematics,(I dont know what he knows) as far as I know that
wanted to nullify all the papers related to FLT publisshed in CMNSEM by a lecturer of the department of mathematics of the University of Kelaniya.This is nasty and disgusting practice of this man.Let us see what will happen to the journal and proofs of FLT for many exponents.
CMNSEM is a refereed journal.
I Can prove.
Fermat,s last theorem(FLT) is the most notorious,difficult and famous theorem.If someone make(in finding a simple proof) a mistake in proving it there are a lot to laugh at it.Do these people appreciate the good work of it?.No.This is the truth.
The so called experts look at the marvelous proofs in CNMSEM with closed eyes. I an sure , the other (are being done ar present by the )proofs by the same author will make FLT very easy for all.
Harvey Friedman grand conjecture says simple mathemtics can
prove Fermat's last theorem. Also ,Colin Mactarty (philosopher)has reached the same conclusion.
One has proved Fermat's last theorem for all exponents using elementary mathematics and published in CMNSEM.Anyone can challenge the world against the validity of the proofs on behalf of the author of the papers.
I try my best to find (one is already OK)three simple proofs of Fermat's last theorem.
R.A.D.
Look at
http://www.fileden.com/files/2012/5/4/3300464/Abstract%20of%20short%20Proof%20PDF%281%29.pdf
To prove Fermat's last theorem one needs Fundamental theorem of Arithmetic and Remainder theorem (A special form) only.R.A.D.P.
Structure of Fermat triples in Mcpgr proof and proof in CMNSEM Vol.1,No.2,March 2011,pp57-63 are ninety percent coincide.One set in step.1 is wrong in Mcpgr ,therefore.I am reading the full proof however.
It is utter waste of time. I don't understand Mcpgr proof.
Take a look at:
http://www.mymathforum.com/viewtopic.php?f=40&t=40857
By now,I have read four proofs of Fermat's last theorem.Unfortunately, all are wrong.I have given up reading such proofs for sometime.
"A simple and short analytical proof of Fermat;s last theorem 'is available now.CMNSEM,Vol.2,No3. March 2011,p.57-63.
Using the fundamental theorem of Arithmetic the theorem can be proved justifying Harvey Friedman conjecture and Colin McLarty prediction.However, by no means this underestimate Andrew Wiles work,to my mind.
I thank you very much for giving me a very good chance to tell the above to the world.
R.A.D.P
Investigate one paper, and you find a network of publishers and authors working together, http://www.journalshub.com/all-journal.php, example: American Journal of Math .. http://www.journalshub.com/journal-detail.php?journals_id=127
then google for Diploma Mills, Vanity Press, and cross reference the references.
Thank you very much for your information,Anonymous.R.A.D.P.
I predicted "the death of the university" after being failed on a CELTA course in 2008.
A few weeks ago, I noticed that the rentacoder site was offering work for an assignment to be written. I e-mailed the lecturer to say that I did not want to make trouble for the person using rentacoder, but that this signalled to me the decline of civilisation as we know it.
I am very puzzled by blogs I have seen today on Fermat's Last Theorem.
There is Orwellian mention of an "open access journal". Does this mean that the journal is free to access on internet? I see now evidence that such a journal would be of poor quality.
50 years ago, we were told "publish or perish". We knew that 90% of the papers being published were rubbish.
Perhaps this is a blessing, as there is less to read. So it may be even more of a blessing if it turns out that at 99% of what is published today is rubbish.
Reading a blog on Fermats's Last Theorem. I can't get from one line to the next of Wiles's paper.But these blogs have at least the advantage of being easy to read. Maybe Wiles could republish his work in a way that is accessible to someone with only high school maths.
I will be very happy if Andrew Wiles publish
or republish his work in such a way that a High School Student can understand the proof.Then I can study something else.
R.A.D.Piyadasa
I will be very happy if Andrew Wiles publish
or republish his work in such a way that a High School Student can understand the proof.
You're going to be waiting a long time.
Thank you very much Prof Jeffrey.
I take your word.
Wish you a very happy New year as well.
R.A.D.Piyadasa
Second elementary proof of Fermat's last theorem is OK.Only high school mathematics is used.I am going to publish it.
R.A.D.Piyadasa
Here is one of some simple proofs of Fermat's last theorem as follows:
1.
X^p + Y^p ?= Z^p (X,Y,Z are integers, p: any prime >2) (1)
2. Let‘s divide (1) by (Z-X)^p, we shall get: (X/Z-X)^p +( Y/Z-X)^p ?= (Z/Z-X)^p (2)
3. That means we shall have:
X’^p + Y’^p ?= Z’^p and Z’ = X’+1 , with X’ =(X/Z-X), Y’ =(Y/Z-X), Z’ =(Z/Z-X) (3)
4. From (3), we shall have these equivalent forms (4) and (5): Y’^p ?= pX’^(p-1) + …+pX’ +1 (4)
Y’^p ?= p(-Z’)^(p-1) + …+p(-Z’) +1 (5)
5. Similarly, let’s divide (1) by (Z-Y)^p, we shall get: (X/Z-Y)^p +(Y/Z-Y)^p ?= (Z/Z-Y)^p (6)
That means we shall have these equivalent forms (7), (8) and (9):
X”^p + Y”^p ?= Z”^p and Z” = Y”+1 ,with X” =(X/Z-Y), Y” =(Y/Z-Y), Z” =(Z/Z-Y) (7)
From (7), we shall have:
X”^p ?= pY’’^(p-1) + …+pY’’ +1 (8)
X”^p ?= p(-Z”)^(p-1) + …+p(-Z”) +1 (9)
Since p is a prime that is greater than 2, p is an odd number. Then, in (4), for any X’ we should have only one Y’ (that corresponds with X’) as a solution of (1), (3), (4), (5), if X’ could generate any solution of Fermat’s last theorem in (4).
By the equivalence between X’^p + Y’^p ?= Z’^p (3) and X”^p + Y”^p ?= Z”^p (7), we can deduce a result, that for any X” in (8), we should have only one Y” (that corresponds with X’’ ) as a solution of (1),(7),(8),(9), if X” could generate any solution of Fermat’s last theorem.
X” cannot generate any solution of Fermat’s last theorem, because we have illogical mathematical deductions, for examples, as follows:
i) In (8), (9), if an X”1 could generate any solution of Fermat’s last theorem, there had to be at least two values Y”1 and Y”2 or at most (p-1) values Y”1, Y”2,…, Y”(p-1),
that were solutions generated by X”, of Fermat’s last theorem. (Please note the even number (p-1) of pY”^(p-1) in (8)). But we already have a condition stated above, that for any X” we should have only one Y” (that corresponds with X”) as a solution of (1),(7),(8),(9), if X” could generate any solution of Fermat’s last theorem.
Fermat’s last theorem is simply proved!
ii) With X”^p + Y”^p ?= Z”^p, if an X”1 could generate any solution of Fermat’s last theorem, there had to be correspondingly one Y” and one Z” that were solutions generated by X”, of Fermat’s last theorem. But let’s look at (8) and (9), we must have Y” = -Z”. This is impossible by further logical reasoning such as, for example:
We should have : X”^p + Y”^p ?= Z”^p , then X”^p ?= 2Z”^p or (X”/Z”)^p ?= 2. The equal sign, in (X”/Z”)^p ?= 2, is impossible.
Fermat’s last theorem is simply again proved, with the connection to the concept of
(X”/Z”)^p ?= 2. Is it interesting?
Sincerely yours,
Pham Duc Sinh.
Is it interesting?
No.
Thank you for letting me post the proof.
I suddenly thought of Fermat's last theorem at a night without sleep, while studying Quantum Physics and Maxwell equations years ago. And I believe and am rather sure that Fermat's last theorem was not just a hard-to-solve problem like many of us had thought of.
In fact, Fermat's last theorem is a base for setting up calculation methods for Quantum Physics and Wave equations if we merge the concept of continuity limit of real irrational numbers into rational numbers.
Please just correlate the notion of integers, rational numbers (the quantum is one, or some rational number!)to the basic concept of Quantum Physics (quantum-quanta).
And the concept of wave tranmission to the concept of continuity of real-irrational numbers!
If we apply the reasonings in the above proof of Fermat's last theorem to real-irrational numbers, we of course shall find out that the current concept of real-irrational numbers is still to be re-examined!
That is interesting, I think, because the concept of real-irrational numbers is basic for number theory, geometry, differential equations,trigonometry,...
Sinh
Sinh,
It appears to me as if your "proof" never uses the fact that X, Y, and Z are integers! So there is clearly a problem with it.
Posting it in ASCII makes it difficult to decipher. You also are missing some parenthesis, you write (X/Z-X) when you mean (X/(Z-X)). I am sorry to have wasted my time on it.
Hi George,
1.I do not know why you pose the question that I did not start with integers!
You can see that there are definition that X,Y,Z are integers, on the start of the proof! Then we have equivalent,to-be-solved equations with rational numbers.
Then is it OK to reason the proof in rational numbers, whatever they are different to integers!
We can also have reasonings for integers. But it is a little bit long to be explained. Then I avoid making it.
2. Thank you for pointing out that I did mistakes in writing
(Z/Z-X).It has to be (Z/(Z-X)). I am sorry.
I have bad eyesight and tend to abbreviate the formulas for convenience. I often make mistakes in typing!
But it does not change the thought line, does it?
Do you never make mistakes in writing mathematic formulas? I believe that people usually make.
Hi George,
Here is the typing correction.
One of some simple proofs of Fermat's last theorem as follows:
1.
X^p + Y^p ?= Z^p (X,Y,Z are integers, p: any prime >2) (1)
2. Let‘s divide (1) by (Z-X)^p, we shall get:
(X/(Z-X))^p +(Y/(Z-X))^p ?= (Z/(Z-X))^p (2)
3. That means we shall have:
X’^p + Y’^p ?= Z’^p and Z’ = X’+1 ,with X’ =X/(Z-X), Y’ =Y/(Z-X), Z’ = Z/(Z-X) (3)
4. From (3), we shall have these equivalent forms (4) and (5): Y’^p ?= pX’^(p-1) + …+pX’ +1 (4)
Y’^p ?= p(-Z’)^(p-1) + …+p(-Z’) +1 (5)
5. Similarly, let’s divide (1) by (Z-Y)^p, we shall get:
(X/(Z-Y))^p +(Y/(Z-Y))^p ?= (Z/(Z-Y))^p (6)
That means we shall have these equivalent forms (7), (8) and (9):
X”^p + Y”^p ?= Z”^p and Z” = Y”+1 ,with X” =X/(Z-Y), Y” =Y/(Z-Y), Z” = Z/(Z-Y) (7)
From (7), we shall have:
X”^p ?= pY’’^(p-1) + …+pY’’ +1 (8)
X”^p ?= p(-Z”)^(p-1) + …+p(-Z”) +1 (9)
Since p is a prime that is greater than 2, p is an odd number. Then, in (4), for any X’ we should have only one Y’ (that corresponds with X’) as a solution of (1), (3), (4), (5), if X’ could generate any solution of Fermat’s last theorem in (4).
By the equivalence between X’^p + Y’^p ?= Z’^p (3) and X”^p + Y”^p ?= Z”^p (7), we can deduce a result, that for any X” in (8), we should have only one Y” (that corresponds with X’’ ) as a solution of (1),(7),(8),(9), if X” could generate any solution of Fermat’s last theorem.
X” cannot generate any solution of Fermat’s last theorem, because we have illogical mathematical deductions, for examples, as follows:
i) In (8), (9), if an X”1 could generate any solution of Fermat’s last theorem, there had to be at least two values Y”1 and Y”2 or at most (p-1) values Y”1, Y”2,…, Y”(p-1),
that were solutions generated by X”, of Fermat’s last theorem. (Please note the even number (p-1) of pY”^(p-1) in (8)). But we already have a condition stated above, that for any X” we should have only one Y” (that corresponds with X”) as a solution of (1),(7),(8),(9), if X” could generate any solution of Fermat’s last theorem.
Fermat’s last theorem is simply proved!
ii) With X”^p + Y”^p ?= Z”^p, if an X”1 could generate any solution of Fermat’s last theorem, there had to be correspondingly one Y” and one Z” that were solutions generated by X”, of Fermat’s last theorem. But let’s look at (8) and (9), we must have Y” = -Z”. This is impossible by further logical reasoning such as, for example:
We should have : X”^p + Y”^p ?= Z”^p , then X”^p ?= 2Z”^p or (X”/Z”)^p ?= 2. The equal sign, in (X”/Z”)^p ?= 2, is impossible.
Fermat’s last theorem is simply again proved, with the connection to the concept of
(X”/Z”)^p ?= 2.
Sinh,
In your "proof", nowhere do you need to use the fact that X,Y and Z are integers. Therefore, your proof also shows that X^p + Y^p = Z^p has no solutions when p>2 is prime and X, Y and Z are arbitrary real numbers! This is certainly false, therefore, your "proof" must be incorrect.
George,
It is very interesting to discuss this. I wrote to Jeffrey Shallit that we can also apply this proof to real-irrational numbers, and find out abnormal new results for real-irrational number.
When the proof can be applied to whatever type of number, it is correct with that type of number! The main things to be checked are in the logical reasonings and the variables, too.
Please read:
ii) With X”^p + Y”^p ?= Z”^p, if an X”1 could generate any solution of Fermat’s last theorem, there had to be correspondingly one Y” and one Z” that were solutions generated by X”, of Fermat’s last theorem. But let’s look at (8) and (9), we must have Y” = -Z”. This is impossible by further logical reasoning such as, for example:
We should have : X”^p + Y”^p ?= Z”^p , then X”^p ?= 2Z”^p or (X”/Z”)^p ?= 2. The equal sign, in (X”/Z”)^p ?= 2, is impossible.
With your question, we can rewrite that:
With X" and Z" are rational numbers (they come from the definition that X,Y,Z are integers), the equal sign in (X”/Z”)^p ?= 2 is impossible.
(The problem (X”/Z”)^p ?= 2 has been one of the basic excercises to be proved in number theory and to be examined for the concept of real-irrational numbers).
I am happy and delighted that I can disscuss Fermat's last theorem with you.
WE WOULD HAVE A LOT TO DISCUSS IN NUMBER THEORY, GEOMETRY, DIFFERENTIAL AND INTEGRAL CALCULUS,..., DIFFERENTIAL EQUATIONS,..., ALGEBRA,...,TRIGONOMETRY,..., QUANTUM PHYSICS,...ETC
I CAN ASSURE YOU THAT!
(And I still avoid working with complex-imaginery numbers. I have different thoughts in definitions of complex-imaginery numbers!)
And this is another version of the proof:
Fermat's last theorem has an equivalent theorem (we accept the case X^4 +Y^4 ?= Z^4. It has been proved):
X^p+Y^p ?= Z^p (X,Y,Z are rational numbers, p : any prime>2). (1)
From (1)we can use the reasonings
of the above to prove that (1) is equivalent to the following theorem:
X’^p + Y’^p ?= Z’^p and Z’ = X’+1
X',Y',Z' are rational numbers.
That means:
Y’^p ?= pX’^(p-1) + …+pX’ +1 (3).
We need checking (2) and (3) only with Y'(integers) and X' (real rational numbers that cannot be reduced to integers), then we can see that the equal sign in (3) is impossible.
George,
If you are still in doubt that the current concepts of real irrational numbers are still to be re-examined, let me post an exercise of real numbers that I have read in a number textbook as follows:
Let a,b,c,d are rational numbers, e is real irrational number, prove that if a+ce= b+de, then we must have a=b and c=d.
Proof:
(a-b) ?= (d-c)e. Then we can easily see that a=b and c=d, otherwise e ?= (a-b)/(d-c). An irrational number cannot be equal to a rational number by definition.
We can see that, even with basic calculations such as addition and multiplication, irrational numbers behave not quite the same as rational numbers . This exercise reminds us the independence concept of vectors in linear algebra.
Then, the posted proof of Fermat's last theorem shows us that there is actually a new hidden difference between rational and irrational numbers.
If we want to deny a proof, we must point out the false and illogical reasoning or false, illogical variables used in the proof.
We cannot arbitrarily apply a proof or function used for rational numbers to irrational numbers, right?
I wonder if there is proof that FLT can't be proven using goniometry. Otherwise that seems to me a much simpler way of approaching this problem.
Given any value of a, b and n>2, the angle between side a and b is always somewhere between 60 and 90 degrees, so at least this gives you an illustration of the problem, while you can also do some math with this using the cosine law.
Very, very doubtful, since (a) such a technique is too puny to accomplish much and (b) doesn't distinguish adequately between the properties of integers and reals.
...Which means that if some basic goniometry could be helpful in solving this problem, it would be 'mirabilem'.
Anyway, such a technique isn't too puny to demonstrate that FLT is at least plausible, which starts at having a look at the situation for n=1 and then I guess you can figure out the rest for yourself.
Why may I ask are mathematicians so adamant that there is no relatively simple proof of FLT just because it does not yield easily to their abstract mathematics when the maths of Euclid is all that is needed.
We read that Pierre de Fermat had a 'EUREKA' moment and generated the equation z^n = y^n + x^n which now bears his name. NOT TRUE!! This was a creation by subsequent mathematicians as confirmed by the following statement of Andre Weil,: “Only on one ill-fated occasion did Fermat ever mention a curve of higher genus x^n + y^n = z^n ....... ".
Fermat's revelation was that the algorithm for splitting a power x^n was identical with that for splitting the square so that he ended up with the generic equation x^n = z^2 - y^2 which is valid for EVERY INTEGER FROM 1 TO INFINITY. It is as plain as the nose on ones face for any idiot to see that the only exponent that will make all 3 exponents the same is 2 so his statement and hence his theorem is proved. The above algorithm is none other than Euclid's Elements, book2, props 5&6, available to Fermat and the basis for my simpleton reconstruction of Fermat's proof which is posted on U-tube.
Why I ask has prop 8 of the same book been demoted to a rule, 'the quarters square rule' by mathematicians and then surreptitiously resurrected as Euclid's numbers when it suits them to explain Pythagorean triples. In case the mathematicians don't know it prop 8 is just 5&6 with a twist and actually more useful when it comes to explaining the powers.
What do the binomial polynomials centred around Pascal's triangle tell the mathematicians? Nothing? What do the factored polynomials alias the 'Binomial Numbers' tell the mathematicians? Again apparently nothing.
What they do tell one is that the basic property of the integers is 'unity' which is 'oneness' and of the powers it is 'squareness ' which is 'fourness' and that the progression from one level of exponentiation to the next is smooth and precise. And NO, it didn't require the modern 1993 proof to lay to rest the myth that somewhere out there in number space was a super massive 'Fermat Equation' waiting to jump out at us and cause a 'shockwave' of a tsunami' in the mathematical community as no mathematician worth his salt ever believed there was.
You're absolutely right! How silly to think that mathematicians are the ones to consult when considering an equation! We should consult dentists and insurance salesman and crackpots instead!
Consulting the Mathsiah would be my first choice. It is said that He has all the answers.
May I take this opportunity to thank you for posting my previous comment.
Now let me think. The smallest pythagorean triple is 3^2 + 4^2 = 5^2 but hang on 4 is 2^2 so 4^2 is just (2^2)^2 which is umm.. 2^4 I think. Yes that's right. Now 2^4 is 2(2^3) which I believe is a rectangle of size 2 units by 8 units = 16 units which must be right since 4^2 equals 16. Now if I take 4 rectangles and form a square from them the square will be.. let's see... 8 +2 =10 ... that's right it will be 10^2 = 100 square units. Now the resulting internal square will be .... ehh .. 8 - 2 = 6 .... will be 6^2 = 36 square units. So 10^2 - 6^2 = 100 - 36 = 64. Now press the square root key on my Casio calculator and ... hey the answer's 8 which is a perfect square so 10^2 - 6^2 = 8^2.
Now doesn't the 'Difference of Squares' formula a^2 - b^2 =(a+b)(a-b) [EE,book2,props5&6] so that 10^2 - 6^2 = (10+6)(10-6) =(16)(4) = 64 =8^2 where the quantities in brackets look remarkably like the 4 rectangles I started with!
Now isn't forming 4 rectangles into a square just the 'Quarter Squares Rule' ... sorry I mean EE,book2,prop8 so that (8+2)^2 - (8-2)^2 = (4)(8)(2) =64 which of course is again just 10^2 - 6^2 = 8^2 which is just (4)5^2 - (4)3^2 = (4)4^2. The 'Quarter Squares Rule' should properly be called the 'Four Squares Rule' don't you think?
If however we evaluate the 'QSR' as ((8+2)/2)^2 - ((8-2)/2)^2 we get 5^2 - 3^2 which as we all know equals 4^2 so I am led to believe that EE,book2,props5&6 and EE,book2,prop8 are just two sides of the same coin. Nay, not believe, THEY ARE one and the same thing.
So 4^2 = 5^2 - 3^2 which is just the smallest pythagorean triple rearranged is identical with 2^4 = 5^2 - 3^2 which if my algebra is up to the task is just ... x^n = z^2 - y^2 ... and if my understanding of algebra is correct must apply to every integer from 1 to infinity. Hey, if I write x^n as (x^(n/2))^2 then I get ... (x^(n/2)^2 = z^2 - y^2 ... but isn't this just PYTHAGORUS'S EQUATION.
So how can I elevate ... x^n = z^2 - y^2 ... into ... x^n = z^n - y^n ... I know I'll ask an expert, a mathmagician ... " You can't Jimmy" ... "Why?" ... " Because the 'Modern Proof of 1993' say's so" ... "Funny I thought it was Fermat circa 1630 that said so" .. " Oh no they're adamant that he didn't have the least idea what he was talking about".
Well since the mathmagicians don't want to acknowledge the validity of ... x^n = z^2 - y^2 ... then I herewith claim priority for the algebraic form and hereby name it 'Bateman's Generic Algebraic Equation for the Triples of the Powers'. NOT BAD FOR A CRANK WOULDN'T YOU SAY!!
Q.E.D. Proof by demonstration.
Sorry, but the equation x^n = z^2 - y^2 has nothing to do with Fermat's last theorem, and is of no interest (since every odd number or multiple of 4 is the difference of two squares).
We beg to differ but as far as I am concerned Fermat said ”It is impossible to separate a cube into two cubes ......" which for me means that ... x^n = z^n - y^n ... because he found out that ... x^n = z^2 - y^2 ... prohibited it!! ' C'est la vie' and thank you for the comments.
Sorry for the typo but my last comment should have said ....
We beg to differ but as far as I am concerned Fermat said ”It is impossible to separate a cube into two cubes ......" which for me means that ... x^n ≠ z^n - y^n ... because he found out that ... x^n = z^2 - y^2 ... prohibited it!! ' C'est la vie' and thank you for the comments.
Alleged simple proofs of FLT are like death defying bugs that come crawling out of the cracks and think they've seen the light: no matter how many you kill, they just keep on coming.
It can be proved that a=nuwv+nw^n; b=nuwv+v^n; c=nuwv+w^n+v^n to satisfy equation a^n+b^n=c^n. When n=2 it becomes an identity (2wv+2w^2)^2+(2wv+v^2)^2=(2wv=2w^2+v^2)^2 (with u=1). For powers >2 it becomes a polynomial equation that maybe can proved having no integral roots by experienced mathematician. I proved it maybe not the best way.
Welcome to 'ASK A MATHEMATICIAN', not a butcher, a baker or candlestick maker but a mathematician.
Assume an all integer solution of a^n + b^ n = c^n exists and a,b,c are coprime and a > b.
Now (a^n)^2 - (b^n)^2 = (a^n + b^n)( a^n - b^n) : FACT.
Hence (a^n)^2 - (b^n)^2 = (c^n)( a^n - b^n).
By the distributive law of multiplication (a^n) (a^n) - (b^n) (b^n) = (c^n)( a^n) - (c^n)( b^n).
BUT c^n cannot equal both a^n & b^n at the same time hence 'REDUCTIO AD ABSURDUM'.
QUESTION: Where's the 'school boy error' or isn't there one?
PS: Have you named your website 'RECURSIVITY' for no particular reason or some deep mathematical reason?
It can be shown that to be integrals in a^n+b^n=c^n it's required a=nuwv+nw^n; b=nuwv+v^n; c=nuwv+nw^n+v^n. When n=2 we obtain identity (2wv+2w^2)^2+(2wv+v^2)^2=
=(2wv+2w^2+v^2)^2 (with u=1). When n>2 it becomes a polynomial equation that I believe can be proved not having integral roots by proficient mathematician . I maybe proved it not in the most elegant way
Dear George, Jeffrey,
I wonder why when you are just in doubts of some simple proofs, you can say that such proofs are or will be proven wrong. We, of course, are often in doubts of many things, not only mathematics, and such doubts are necessary!
Moreover, if a technique is puny, it is right or wrong, not by being puny or not being puny. My attempted simple proof of Fermat's last theorem has been presented to be analyzed, and I should be very glad if you give me more analyses to the proof by pointing the wrong reasonings in the theorem.
Otherwise, we will always make a haste remark to any proof of any theorem that such proof is proven wrong just by saying that the proof is a)puny and b) doubtful.
Such haste remarks and conclusions are, of course, sometimes right and sometimes wrong, but they are not the way mathematicians apply in their work!
is it essential in mathematics?
Dear George, Jeffrey,
I wonder, why when you are in doubt of some proof, you think that such proof may be wrongly proven by just saying that it is a) puny and b) doubtful.
We, of course, are often in doubts of many things in life and it is necessary.
My attempted simple proof have been presented to be analyzed and I should be very glad if you would give it more analyses.
Please give me more review to the reasonings of the proof.
1+2 ?=3 is puny, but is it a essential part of mathematics?
Thank you very much.
Please give me more review to the reasonings of the proof.
No, thanks. I have a limited amount of time to spend on things for which the prospect of success is so tiny. There are many, many amateurs who think they have proved Fermat's last theorem. So far, every amateur has failed. With such puny methods, the chances you have succeeded are virtually nil.
I would have loved a reply to my PS simply because there is a simple numerical recursive algorithm for the differences of adjacent numbers raised to the same power starting at a^0 which if fed back into the algorithm gives the differences between adjacent numbers raised to the next higher power and so I was wondering whether this would have any implications for FLT for the mathematicians.
So the challenge for anyone wishing to take it on is to find the algorithm starting with nothing more than a column of zero's. By applying some simple algebra to the results it is possible to derive the polynomials for the 'Binomial Numbers'... see 'Wolfram MathWorld'... which could fit the bill for the polynomials that result from the identities given by McPgr.
I will if no one succeeds supply the algorithm.
I guess there's only one way to silence the debate about simple proofs of FLT ones and for all: that is proving that for some principle reason it can't be done the easy way, let's say by proving that you can't make something like a distuingishing "cosine law for triplets" that could then be compared with FLT.
Dear George, Jeffrey,
Perhaps we should be reminded that in the time that Fermat was living, the notion and concept of irrational numbers did not exist or were very primitive.
Mathematicians then built up irrational numbers and still now disregard the eery behaviour of irrational in the problem such as a+be ?= c+de (a,b,c,d: rational,e: irrational) and then treat calculus of irrational numbers in addition, subtraction, multiplication and division as the same as to calculus of rational numbers.
I know that almost all mathematicians (and very much possibly including me) have failed in the attempts of proving Fermat's last theorem. But I should be very glad when someone would point out the wrong reasonings in the proof. Otherwise, we would just again reject and deny any proof by just thinking that it is puny and doubtful.
Wiles proof is related with elliptic curves. We can see that the equation in my proof has a form:
Y'^p ?= pX'(p-1) +...+1. Is it related with elliptic curves somehow?
Jeffrey Shallit said....
"There are many, many amateurs who think they have proved Fermat's last theorem. So far, every amateur has failed. With such puny methods, the chances you have succeeded are virtually nil. ....."
That is all except mine of course since my proof is none other than PYTHAGORUS'S EQUATION and at the end of the day all the modern proof has to say is that Pythagorus's theorem reigns supreme but we mere mortals knew that all the time!!!
In response to Bert Brouwer's ongoing negative and senseless remarks I would suggest that he gets his behind out of his chair and prove his own banal offering but he should understand that his resulting drivel won't baffle, confuse or put anyone off their self established truths!!
Alastair, have you noticed that nowhere in your "proof" do you use the hypothesis, n > 2? That means you have succeeded in proving that there are no relatively prime integers such that a + b = c, a truly remarkable result.
Personally I feel perfectly happy to sit back and relax and watch this wild goose chase for many decades to come. But if Mr. Shallit himself wants to prove his own 'theorem' that some puny technique won't do the trick, because it doesn't distinguish adeqautely between the properties of integers and reals, then I would of course be glad to know.
Motherf***** of god... I'm sorry, am I not allowed to curse on your blog Dr Shallit. I blew a gasket reading whatshisname's elementary "proof" of FLT.
This idiot reminds me of an elderly Indian gentleman I know down the block who pretends he is a great wise physics-poet sage. The ironic thing is, he actually had earned some degree in physics, long ago, 50 years ago, enough to get a lifelong job in applying physics to the medical field: developing CAT scans & PET scans.
But, I can tell, he never had beyond calculus 50 years ago, of which he remembers nothing.
And his poetry sucks, too.
Anyway, I've gotten so jaded & now fearful of ALL peer-reviewed journals, fearing NONE of them seriously peer review works, that I would not know to which ones to submit a piece of math (which I don't have at present anyway).
In which journal did Wiles publish his proof of the Taniyama-Shimura Theorem? Surely THAT journal has to be reputable, no?
My question to amateurs who think they have proved all the big name theorems (by elementary means) or outstanding big name conjectures that are constantly mentioned throughout pop media, without FIRST going through the rigors of proving much LESS glamorous much LESS ambitious theorems along the way?
There are HUNDREDS of unsolved "lesser" conjectures in math, needing & deserving to be solved.
What makes you think that math does NOT require the long years of practice & trial & error & building up & learning from what others have done, doing the tedious boring exercises?
Why would you think math is the exception? Why would you not think that about playing sports or mastering a musical instrument or becoming a five-star general in the army after your first day in basic training?
Sigh.
To Bert Brouwer (I assume & hope no relationship to the famous Luitzen Brouwer who proved Brouwer's Fixed Point Theorem):
guess who will be first to find an elementary proof of FLT?
Andrew Wiles: the one who first proved FLT the long slow hard highly advanced technical way. If ANYONE would know how to find an elementary proof of FLT, it will be Andrew Wiles & Gerhard Frey & Richard Taylor.
IThinkWithMy Liver can also sit back and relax, because I'm indeed not related to Luitzen Brouwer.
I notice that he has already answered his own question he asked me:
Guess who will be first to find an elementary proof of FLT?
I will take his answer on faith, so I guess the only thing left for me to do is to say "Amen" to that... although (since it is a believe without proof) one can never know for sure of course. Who knows, maybe one day a martial arts instructor with a good spatial visualization ability and some elementary knowlegde of math is up to the task as well.
Thank you for your comment Gerry and for the slap on the back of the head which I needed since 'Reductio Ad Absurdum' is actually 'Proof Ad Absurdum' because the difference equations either side of the equality do not require the numbers to be the same i.e. 900 - 800 = 600 - 500 is valid. So my insisting that the difference of two squares should be the difference of the same two squares is rubbish and it is valid for it to be equal to the difference of two rectangles which have a common side.
However the foregoing has no bearing on the validity of the equation x^n = z^2 - y^2 ... which is a proof of FLT as far as I am concerned.
Here as promised is the simple numerical recursive algorithm for the differences of the powers starting with nothing but a column of zeros which result from (z^0 -y^0) =(1-1) =0 which we then feed into the following algorithm. The resulting answers are feed back into the algorithm after the exponent of the first term has been increased by 1.
For (z^1 -y^1/(z-y) : 0^0 +1x0 =1: 1^0 +2x0 =1: 2^0 +3x0 =1: 3^0 +4x0 =1: 4^0 +5x0 =1:
For (z^2 -y^2)/(z-y) : 0^1 +1x1 =1: 1^1 +2x1 =3: 2^1 +3x1 =5: 3^1 +4x1 =7: 4^1 +5x1 =9:
For (z^3 -y^3)/(z-y) : 0^2 +1x1 =1: 1^2 +2x3 =7: 2^2 +3x5 =19: 3^2 +4x7 =37: 4^2 +5x9 =61:
For (z^4 -y^4)/(z-y) : 0^3 +1x1 =1: 1^3 +2x7 =15: 2^3 +3x19 =65: 3^3 +4x37 =175: 4^3 +5x61 =369:
So what we have produced are the first differences of the finite difference tables and it is plain for ANYONE to see that these can be simply extended to apply to the infinity of powers applied to the infinity of integers. From the cubic onwards the differences are divergent which after feedback become more and more and more divergent. Now (z^n -y^n) = d1+d2+d3+... dm; that is the sum of the differences between z^n & y^n so why would anybody expect d1+d2+d3+... dm, the partial sum of a divergent series to equal x^n. Apparently the 'mathmagicians' did when they did their computer searches upto an exponent size of 4 million.
Incidentally, the factored polynomial equations which are the binomial numbers can be derived directly from the arithmetic which exhibits the duality to be expected from interchange of the x's & y's i.e. 3^3 +4x37 =175 = 4^3 +3x37. I'll leave the algebra to you.
Also from the 'DoS' equation we have the two identities as solutions to 'Fermat's' equation [1] {(a^2)^n -(b^2)^n}/(a^n-b^n) =(a^n+b^n) & [2] {(a^2)^n -(b^2)^n}/(a^n+b^n) =(a^n-b^n) neither of which look to be likely candidates for c^n do they?
True , the foregoing is only factual observation and not analytic proof but it does give a lay person like myself an understandable insight into what is going on, which the modern proof doesn't and never could, plus the knowledge that contrary to the denial and attempted ridicule by the disciples of 'modern abstract mathematics', the tools and knowledge did exist for a person of Fermat's ability to prove his now famous statement.
Also I ask, "Why did it take the mathematicians so long to prove such a simple and trivial statement in light of the simple foregoing observations and loads, loads more besides?"
Alastair:
I have no interest in your incoherent ravings. Post them elsewhere.
Who is the idiot has been meant by Thinkwithmyliver?Wonderful rascal having all knowledge knowing evetyhing.
Hello,Ithinkwithmyliver,I want to know who is the idiot you mean.I wonder weather you have a brain since you think with liver.
I take this opportunity to thank you for your comment on my general proof of Fermat's theorem.I got read the main points in the CMNSEM paper by an expert{prof. In Number Theory).Proof is OK and elementary.
Thank you again and may I kindly request you not to response to rascals.
Although this is again puny, but I have not found the fault in this proof. Please help me find such.
X^p +Y^p ?= Z^p (1)
(X+Y-Y)^p+ Y^p ?= Z^p (2)
X^p +(Y+X-X)^p ?= Z^p (3)
or we shall have:
(X+Y)^p -Z^p-p(X+Y)^(p-1)Y+...+
P(X+Y)Y^(p-1) ?=0 (4)
(X+Y)^p -Z^p-p(X+Y)^(p-1)X+...+
P(X+Y)X^(p-1) ?=0 (5)
Let's put (X+Y)=K, we shall have:
K^p -Z^p-pK^(p-1).Y+...+
pKY^(p-1) ?=0 (6)
K^p -Z^p-pK^(p-1)X+...+
pKX^(p-1) ?=0 (7)
Both (6) and (7) have the same form as:
K^p -Z^p-pK^(p-1)U+...+
pKU^(p-1) ?=0 (8).
Then if (6) had a solution Y=U,
(7) would too have a solution X=U =Y.
And (1) would have a corresponding solution: U^p+U^P ?=Z^p.
This is impossible for (1) with integers such as X=U=Y, Z.
To Thao Tran
It looks like you're manipulating with the same equation. Here's example for n=3 (presented in standard form)
(X+Y)^3-3(X+Y)^2*X+3(X+Y)*X^2-Z^3=0
(X+Y)^3-3(X+Y)^2*Y+3(X+Y)*Y^2-Z^3=0
After subtraction we obtain
3(X+Y)^2(X-Y)=3(X+Y)(X^2-Y^2)
Dear McPgr,
How can you be sure that we have
(X+Y)^3-3(X+Y)^2*X+3(X+Y)*X^2-Z^3=0
(X+Y)^3-3(X+Y)^2*Y+3(X+Y)*Y^2-Z^3=0 ?
We, of course , always have
(X+Y)^3-3(X+Y)^2*X+3(X+Y)*X^2-Z^3= C
(X+Y)^3-3(X+Y)^2*Y+3(X+Y)*Y^2-Z^3=C.
To Thao Tran
Subtracting left hand parts of your equations one from another you obtain zero. They are the same. There's just manipulating with characters.
To McPgr,
X and Y are two independent variables, not just characters in the equations.
Now,I understand why ITHINKWITHLIVER did condemn the elementary proof given by me, I suppose.This is the reason
So far, every amateur has failed. With such puny methods, the chances you have succeeded are virtually nil. Jeffery Shalit.I have been also treated as an amateur.I solved a big problem in physics long ago. But I am a mathematician. Whatever, it is you(ITHINKWITHLIVER) have no right to use such words as mother f......If my proof is wrong, I don't know,it is up to anyone to tell it.I was too late in understanding the reason for ITHINKWITHLIVER'S comment.R.A.D..P
Unfortunately, it seems that the paper has been removed from the journal's website. I'm guessing it used to be the paper in volume 2, number 2, pp. 317-338. If anyone has a copy of the paper, please confirm the details. Thanks.
Not unfortunately ,fortunately all papers of mine are removed.I am responsible for this.Man like you can not understand the following. We researchers, do research for the sake of researches. Not for the foolish riff-raff or their existance.
You (joelreyesnoche)might be happy if the proof given in cmnsem vol.2 No.3 (2011)march p.57-63 (for your information) is wrong. Therefore may I ask Prof.Jefrrey Shalit. I am very much thankful to you Professor if you say weather my proof is correct or wrong.This would be great help to to me as well.
RADP(Piyadasa Ranawaka)
To all that who that like to condemn the proof of Fermat's last theorem given by me.
It is your ignorance or some other bad quality
Whenever I wanted to solve a big problem I registered for Ph.D or D.Sc at a standard university.In case of Fermat's last theorem I did the same This is actually to save the time
I will tell later you why almost all amateur has been failed. Piyadasa Ranawaka
My interest in the Fermat Conjecture, (FC,) began as an interest in the Pythagorean theorem. I wasn't looking for other integral solutions to the n greater than 2 problem. I was more interested in the fact that odd integral values of 'a' resulted in 'b' being equal to (c-1.) Then, I began looking at the FC, and his statement, in the margin, that he had a solution, but it was too big to fit in the margin. I realized that with the mathematical tools in existence at that time, it was probably algebraic or geometric in expression. The solution proposed in 1995 that exceeded 120 pages, and used mathematics far in excess of what was available to Fermat was probably much more complex that needed. I began by using simple substitution to eliminate variables. It was determined that b= [(a to the 2)-1]/2, and c=[(a to the 2)+1]/2. Using an integral 'n', the problem became, "Is there a positive integer 'n', such that the equation: [(a to the n)-1]/n is an integer?" Although (a to the n) / n would result in an integral solution, because 1 is subtracted from the numerator prior to the division, there are NO positive integers for 'a' and 'n' greater than 2 that would result in b being positively non-fractional. COMMENTS? A.Conti, Allen@TheWalnutGrove.com or OWInc@att.net
At the web site
http://unsolvedproblems.org/
I am offering a prize of US$1000 for any simple proof of FLT. Also for 19 other unsolved problems in number theory, logic, and cryptography.
The Relations of Barlow and FLT
Hi Guys
I have found The Relations of Barlow when I read Ribenboim's Book. "Playing" with it, I found another sistem of equations, and then, I used modular congruences and I found what is summarized below (we can work just with natural numbers to analize the fermat's integer soluctions,because if one or 2 of the numbers are negatives, them we can change the place in the equation in order to have just positive numbers. If al of them are negative, we can multiply the equation by -1):
I didn't make an English version*, because I am having dificulties in the mahts expression (loses its format when I paste here), but if you follow the steps below, then you will achiev the same results :
* I published the entire demonstration in a Blog I created (in this blog is better to read the maths equations) and in other forums in Brazil and others countries :
http://filosofbeer.blogspot.com.br/2015/04/o-ultimo-teorema-de-fermat-as-relacoes.html
From The Barlow Relations (See Fermat Last Theorem for Amateurs, pg 99 - but is very easy to show this) we have, in both Case 1 (n dont divide z) and Case 2 (n divide z) of FLT :
z-x = y0^n
z-y = x0^n
From this , we conclude thar
y-x = y0^n-x0^n = (y0-x0)(y0^(n-1)+....+x0^(n-1))
gdc(y-x, y0^(n-1)+....+x0^(n-1)) = y0^(n-1)+....+x0^(n-1)=p
y0^(n-1)+....+x0^(n-1) "=" 0 mod (p)
y-x "=" 0 mod (p)
Replacing x0 = (z-y)^(1/n) and y0 = (z-x)^(1/n)
[(z-y)^(n-1)]^(1/n)+....+[(z-x)^(n-1)](1/n) "=" 0 mod(p)
Replacing y "=" x mod (p)
n[(z-x)^(n-1)](1/n) "=" 0 mod(p)
ny0^(n-1)"=" 0 mod(p)
p = 1 or p divides y0^(n-1) or p=n
As x and y are both > 0, then y0^(n-1)+....+x0^(n-1) is always > 1. So p can not be 1
If p divides y0, then it will divide x. But gdc(x,y,z)=1, so p can not divide y0 or x0
If p=n, then
y0^(n-1)+....+x0^(n-1)= n
But this sum has n terms, and this is possible just if y0=1 and x0 = 1 , because both are > 0. And if one of them is larger then 1, then the sum wil be larger than n.
If x0=1 and y0=1, then using the Relations of Barlow, we conclude that x=y, wich contradicts te conditions z>y>x and gdc (x,y)=1.
The same analysis could be used in the case 2 (even if n divedes x ou y).
So, there's no integers soluction for the equation x^n+y^n = z^n
FelipeRJ
* If p divides or have common factor with x0, p will divide or have commom factor with y0, wich contradicts the condiction mdc(x,y)=1.
An English Version :
http://filosofbeer.blogspot.com.br/2015/04/the-relations-of-barlow-and-fermat-las.html
Now,I will tell you why many many people are failed in the proof of Fermat's last theorem. Look carefully, all of them starts from the original equation x^n+y^n=Z^n and then look for a contradiction.This is utterly impossible since it has been assumed that x,y,z satisfy the equation. . I will give another example. In order to solve the elliptic curve y^2 +2=x^3, a famous equation , for integer roots one needs advanced methods such as unique factorization method if you starts from the original equation.Very good intelligence needs to do by elementary methods. One of my students has solved this equation using elementary mathematics.
To the best of my knowledge, there is no elementary proof of Fermat's last theorem for n=4 without depending on the Method of Infinite Descent of Fermat.I, with one of my students, designed an elementary proof without using any mathematical tool(only high-school Mathematics used).It was difficult and tricky.However, if we made a tiny mistake,then the proof followed at once.This is another evidence for my saying regarding the difficulty of proving Fermat's last theorem for all indices.
Thank you very much Prof. Jefferey for maintaining this Blog, and wish you Merry Christmas in advance.
A simple proof of FLT is one of the many problems eligible for a prize of US$5000 on the Unsolved Problems web site at
http://unsolvedproblems.org/
Tim
Thank you unknown. Would you offer such an award to the 17th century problems of finding all the integer integer roots of the simplified elliptic curves Y^2=x^3-2 , y^2+4=x^3 using which Fermat challenged the Europeans? I have already written the solutions using 17th century mathematics and Fermat's little theorem.
Any way,Than you again for your information.
I just need almost a margin (please share) https://www.researchgate.net/publication/301674377_Fermat%27s_Last_Theorem_Proof_fixed_edition?ev=prf_pub
To S. Carlos.
I read your proof.But it is not clear at all to me.However, I appreciate your effort.
Piyadasa R.
Before you go to a 'marginal proof' see weather you can prove Fermat's last theorem for n=4 without depending on Fermat's method of infinite descent.One of my students and myself have already done this, third one of that type.Not only this, but I have designed the proof of the FLT also.This is my third proof of Fermat's last theorem and this would be the proof of Fermat, who is actually ingenious.
I would not be worried about hatefull comments on somebody's short proof
of Fermat's last theorem. I developed a short proof of integrability of
a continuous function defined on a closed finite interval. Proofs available
in textbooks are usually 15 pages long in some cases a chapter. By the length
of a proof I mean all the results one needs to prove the theorem starting
from the definition of continuity. It is an interesting experience to see
those idiotic reports from referees and letters from editors. I claim that
if somebody writes an idiotic report, he is an idiot. The question is if
they are so stupid they do not understand the trouble of proving integrability.
The answer is clear, they do not want shorter proofs. Let me ask them directly:
"Why do you prefer long proofs to short ones?" This is a retoric question.
I believe the same kind of people does not want to accept any short proof
of FLT.
Reference: Mathematics Student, last year.
Sincerely,
Josef Bukac
Simple proofs of FLT can be posted on the Unsolved Problems web site at
http://unsolvedproblems.org/
Any correct proof will be awarded a significant monetary prize, as detailed at the site.
Fermat's last theorem have been made difficult by mathematicians.See how simple is the proof.
Proof of the theorem
Let us prove the theorem for n=4 first.
proof of Fermat’s last theorem for n=4
Fermat’s last theorem for n=4 can be stated thus;
There are no non-trivial x,y,z integers satisfying the equation
z^4=y^4+x^4,(x,y)=1 (1)
Without loss of generality, we can assume that x,y,z>0 ,and also we have labeled the integer triple such that z>y>x>0 .
Consider the equation z^4/x^4 =y^4/x^4 +1 which can be written as
g^4=h^4+1, h= y/x>1 (2)
where g,h are rational numbers .
From (2),
(g^2-h^2 )(g^2+h^2 )=1 (3)
and let (g^2-h^2 )=d. Then g^2+h^2 =d+〖2h〗^2=1/d>0 . In other words,d>0 and
d^2+2h^2 d-1=0 (4)
We must have from (4) that
0<2h^2 d<1 (5)
since〖 d〗^2>0,and d^2+2h^2 d=1
〖00 This means that the quadratic in h^2 is positive and hence the discriminant ∆=〖4g〗^4-8<0 ,g^4<2 ,and therefore from (2), h^4<1.This is impossible since h= y/x>1. Hence, (1) has no non-trivial integer triple solution. After proving FLT for n=4 , one has to prove for FLT for any odd prime p in order to prove FLT in general[2]. Now, consider the equation
z^p=y^p+x^p,(x,y)=1 (6)
where p is any odd prime. By rearranging and relabeling of the components in (6), we can assume without loss generality that y>x>O. From(6), we obtain
z^p/x^p =y^p/x^p +1 and this can be written as
g^p=h^p+1, where g,h are positive and h>1 If y≠0. Definitions of g,h are obvious. Therefore g^p-h^p=1 and this gives
(g-h)(g^(p-1)+〖hg〗^(p-2)+⋯…..+h^k g^(p-k-1)+⋯..+h^(p-2) g+h^(p-1) )=1 (7)
Let g-h=d which is positive since the longer expression within bracket in (5) is positive, that is 1/d is positive. Then from (7),
〖(h+d)〗^(p-1)+h〖(h+d)〗^(p-2)+……….+h^(p-2) (h+d)+h^(p-1)=1/d and this gives
d^p+hd^(p-1)+⋯………+h^(p-2) d^2+ph^(p-1) d-1=0 (8)
Note that all terms in (8) are positive except the constant term (-1). Now, we have h^(p-2) d^2+ph^(p-1) d-1<0 . In other words,
〖0<(-h〗^(p-2))d^2-ph^(p-1) d+1 (9)
and (9) is a quadratic in d if h≠0 and its discriminant should be negative. In other words∆<0,
〖∆=p〗^2 h^(2(p-1))+4h^(p-2)<0 (10)
This is impossible since h is positive, and therefore we conclude that there are no non-trivial x,y,z>0 integers satisfying the equation (6). Even if we consider the quadratic expression (9)as a function of g we will arrive at the same conclusion.
May I ask the world not to make that Fermat's last theorem difficult by saying this and that.
Hi Piyadasa, if your proof is correct and you post it at the Unsolved Problems web site at http://unsolvedproblems.org/, you will win US$2,500 and a bottle of champagne.
Hello Unknown,I do not want to sell my research work for money whatever the amount may be.I know the proof is correct and i am addressing only the researchers of high potential to ask not to think that Fermat's last theorem is difficult and not to waste time by following the conventional approaches.
Your proof is not correct and no one has any interest in it. Sorry.
Thank you Prof.Jeffery.In case of n=4,I assumed there are non trivial integer triples and thereby new proof has been proposed.It is true that I have made a mistake in case of the index p.The equation last mentioned should be corrected as
〖d^p+⋯….+(p(p-1)/2)h〗^(p-2) d^2+p〖dh〗^(p-1))=1
However, arguments are the same.
This is only a proof designed to show that proof of FLT is simple.
Hello Piyadasa, if you truly believe your proof to be correct, you should send it to a peer-reviewed journal. There are at least 20 to choose from in the area of number theory. Or if you just want feedback, for free, you can post it to the Yahoo! group associated with the Unsolved Problems site.
This proof is designed to make the researches know that proof of FLT should be done within a few steps only.I have another rigorous equally short proof and it will be published.
Dear Prof.Jeffery pl. let me add the corrected proof of FLT.
Proof of the theorem
proof of Fermat’s last theorem for n=4
Fermat’s last theorem for n=4 can be stated thus;
There are no non-trivial x,y,z integers satisfying the equation
z^4=y^4+x^4,(x,y)=1 (1)
Let us assume, on the contrary ,that there are x,y,z integers satisfying the above equation.
Without loss of generality , we can assume that z>y>x>0.
Now,consider the equation z^4/x^4 =y^4/x^4 +1,which can be written as
g^4=h^4+1, h=y/x>1 (2)
where g,h are rational numbers .
From (2),
(g^2-h^2 )(g^2+h^2 )=1 (3)
and let (g^2-h^2 )=d. Then g^2+h^2=d+〖2h〗^2=1/d>0 . In other words,d>0 and
d^2+2h^2 d-1=0 (4)
We must have from(4) that
0<2h^2 d<1 (5)
since〖 d〗^2>0,andd^2+2h^2 d=1.
〖00 This means that the quadratic in h^2 is positive and hence the discriminant ∆=〖4g〗^4-8<0,g^4<2,and therefore from (2),h^4<1.This is impossible since h=y/x>1. Hence, (1) has no non-trivial integer triple solution.Now, consider the equation
z^p=y^p+x^p,(x,y)=1 (6)
where p is any odd prime. As in the previous case, let us assume that there are x,y,z integers satisfying (6).Rearranging relabeling(6), we can make z>y>x>0.From (6), we obtain
z^p/x^p =y^p/x^p +1 and this can be written as
g^p=h^p+1, where g,h are positive and h>1.g^p-h^p=1 which gives
(g-h)(g^(p-1)+〖hg〗^(p-2)+⋯…..+h^k g^(p-k-1)+⋯..+h^(p-2) g+h^(p-1) )=1 (7)
Let g-h=d which is positive since the longerexpression within bracket in (7) is positive, that is,1/dis positive. Then from (7),
〖(h+d)〗^(p-1)+h〖(h+d)〗^(p-2)+……….+h^(p-2) (h+d)+h^(p-1)=1/d and this gives
d^p+hd^(p-1)+⋯………+h^(p-2) (p(p-1)/2)d^2+ph^(p-1) d-1=0 (8)
Note that all terms in (8) except the constant term (-1) are positive. Since p is odd,(8) has one real root. Now, we have h^(p-2) 〖p(p-1)/2)d〗^2+ph^(p-1) d-1<0 .In other words,
〖0<(-h〗^(p-2))(p(p-1)/2)d^2-ph^(p-1) d+1 which is a quadratic in d if h≠0 a its discriminant ∆
〖∆=p〗^2 h^(2(p-1))+4.(p(p-1)/2)h^(p-2)<0 (9)
This is impossible since h is positive, and therefore we conclude that there are no non-trivial x,y,z>0integers satisfying the equation (6). Even if we consider the quadratic expression is a function ofg we will arrive at the same conclusion.
Thank you very much for your comment again.
Your reasoning for an odd prime is correct up until and including equation (8). Unfortunately, it then goes off the rails. The discriminant for a quadratic is b^2-4ac, not b^2+4ac. Both a and c are negative in this case. So your equation (9) is incorrect.
A simple proof for FLT remains unknown.
Unknown, don't be hurry.
Note that all terms in (8) are positive except the constant term (-1). Now, we have h^(p-2) p(p-1)/2d^2+ph^(p-1) d-1<0 . In other words,
〖0<(-h〗^(p-2))(p(p-1)/2)d^2-ph^(p-1) d+1 (9)
and (9) is a quadratic in d if h≠0 and its discriminant should be negative. In other words∆<0,
〖∆=p〗^2 h^(2(p-1))+4h^(p-2) {p(p-1)2}<0 (10)
I have obtained result b^2-4ac<0 from(9)
Any way I am not surprised by your comment.
It is now clear that high school mathematics is sufficient to prove Fermat's last theorem.The way the proof has been done is completely different from the conventional approaches.the conventional approach: Assume that z^n=y^(n )+x^n,(x,y)=1,n>2 and try to draw a contradiction.This method was not successful for few hundred years.Fermat himself had a proof(I believe) similar to the proof we have suggested or a proof in which it is proved that one of x,y is zero after assuming the above equation.
The saying of Andrew Wiles, I do not believe that he had a proof.He fooled himself to think he had a proof. has no foundation.
Challenge to the world
Proof of Fermat's last theorem is very simple as i pointed out. Following the same argument to a certain extend and the very fundamental of algebraic number theory, I have already designed the third proof of Fermat's last theorem which is also very short.If any one doubt or challenge the proofs(at least given on the above) I am very happy. My job on Fermat's last theorem is not to gain anything but for the sake of mathematics is over.
I thank prof. Jeffery for keeping this blog.
Simplest proof of Fermat's last theorem
Proof of the theorem
Fermat’s last theorem for n(>2) can be stated thus: There are non-trivial integers x,y,z satisfying the equation
z^n=y^n+x^n,(x,y)=1,n>2 (1)
Rearranging and relabeling the integers, we can assume without loss of generality that z>y>x>0. From (1), we obtain
g^n=h^n+1 (2)
where g=z/x h= y/x. From the equation(2), we obtain
〖(g-h)[g^(n-1)+g^(n-2) h+⋯……..+gh^(n-2)+h^(n-1)]=g〗^n-h^n=1, and if g-h=d , we have g^(n-1)+g^(n-2) h+⋯……..+gh^(n-2)+h^(n-1)=1/d>0 and therefore d>0.
d([(d+h)^(n-1)+h(d+h)^(n-2)+⋯…….+(d+h) h^(n-2)+h^(n-1)])=1 Hence,
d^n+nhd^(n-1)+⋯……+(n(n-1)/2) h^(n-2) d^2+nh^(n-1) d-1=0 (3)
Since h,d>0,
(n(n-1)/2) h^(n-2) d^2+nh^(n-1) d-1<0 in other words
-(n(n-1)/2) h^(n-2) d^2-nh^(n-1) d+1>0 (4)
(4) is a quadratic expression in d and is positive and therefore the discrininant should be negative ,
i.e n^2 h^(2n-2)+2n(n-1) h^(n-2)<0 (5)
which never holds and we conclude that (1) is not satisfied by non trivial integer triples x,y,z.
Is it true that one Sri Lankan has proved Fermat's Last theorem very easily
No. No easy proof is known.
Hello Unknown, Is the simplest proof of Fermat's last theorem given above in this blog difficult for you?.If so, you have no high school mathematics knowledge.I can't help.
R.A.D.Piyadasa(D.Sc. Kyushu)
Probably, you are a troll, and realize that your "proof"is incorrect. However, I will give you the benefit of the doubt. In which case, does it not concern you that your "proof" never uses the fact that the quantities involved must be integers? So you have "proved" that there are no solutions to FLT even if the numbers involved are irrational. Which is, of course, nonsensical.
It is up to you to think in the way that you think.
'A simple and short analytical proof of Fermat's last theorem' also not a simpler proof according to you,Unknown.
That is not a response to the point, Piyadasa.
Please give me a little time to respond.
Hello,Unknown.
What you say is complete nonsense. In the beginning, I have assumed that there are x,y,z>0 integers satisfying the equation
X^n+y^n=z^n where n>2 and (x,y)=1 (1)
After that I get the equation
1+h^n= g^n where h>1 (2)
according to our fare assumption that 00 ,h cannot be never negative and this is a contradiction and it is sufficient to conclude that there are no positive integer solution for x^n+y^n=z^n . I have already proved that if there are positive integer solution for the Fermat equation (1) and only real solution we can have is h=0 and g=1.This means no positive integral y. Pl. understand that h=0 implies y=0.
I do not argue with you hereafter. Thanks professor Geffry I am sorry for being late in answering.
I do not know weather I would be killed by the so called University students. Long live CMNSEM?? .
I will be very happy if Andrew Wiles publish
or republish his work in such a way that a High School Student can understand the proof.
You're going to be waiting a long time.
6:40 AM, December 31, 2013
Thank you very much for the above comment.I have already proved and published in an international conference the proof which can be understood by an advanced level(AL) student in my country or high School student you have meant.Thank you so much. Long live CNMSEM? and that proof, you will find later.
I have designed a proof of Fermat's last theorem using the mathematics of Fermat's time. Read on the internet ' The Simplest proof of Fermat's last theorem,University of Kelaniya'
Thank you all.
Piyadasa Ranawaka
Fermat's last theorem can be proved using 17th century mathematics.What i have said is O.K.That is Fermat is ingenious is O.K I have been able to prove the theorem (Mean's theorem) g^p=h^p+1,where g=z/x ,h=y/x has only one rational solution only when g=1,h=0.That is corresponding Fermat equation has no non trivial integer triples. I think this is enough.
Thank you prof. regarding your comments on the open access journals. However,Fermat's Last theorem,Beal's conjecture,Riemann Hypothesis have all been proved,in open access journals.I have read four papers.Wonderful!!.
Thank you again for keeping this blog,Professor.I have designed three simple proofs of Fermat's Last theorem(FLT). Simplest proof on the internet is O.K.I got to prove Beal's Conjecture.
'Difference of two rational numbers can not be an integer',in one proof of FLT and BEAL's Conjecture in an open access journal.I challenge the world with respect to any proof of mine(mtself only) on the internet.Pl.Keep this blog,prof.Thank you again.
Hi all,
I have a question for the statistical test:
If we assume that the event "Fermat really had the proof" is a random variable, what probability measure would you be able to propose for the acceptance of this null hypothesis "Fermat really had the proof":
0.00001, 0.0001, 0.001, 0.01, 0.1, 0.2, 0.3, 0.4, ..., 0.90, 0.99, 1.0
Hi All,
Please go through the following link for finding counter example to FLT.
http://www.iosrjournals.org/iosr-jm/papers/Vol7-issue4/A0740103.pdf
http://www.iosrjournals.org/iosr-jm/papers/Vol7-issue4/A0740103.pdf
Hi,Unnkrishnan, wish you all the best first.I am reading your paper. According to your paper(Open access)FLT is wrong.
If not so pl.tell me.
Piyadasa ranawaka
Sir, It may be false by finding a chance for counter example using the criterion.
If such criterion is not possible through any other methods, then FLT is true using 3 pages of elementary proof.
Thanking you,
Unnikrishnan
I understand what you say.Thank you.If you pl send me your E-mail address to me, it is a very good help. My email.address is Piyadasa97@gmail.com.
I appreciate your paper very much.
Piyadasa Ranawaka
Thank you very much,Dr. Unnikrishnan for sending me your E-mail Address.It seems that you are not a Number theorist. However,you have done the part of the use of the Remainder theorem well. I wonder,weather you have done App. Mathematics as well?.Pl. note that this Blog. is centered over the use of the so called open access journals to publish papers on the most difficult theorem in all.
The rest of my comment will be sent you later. Wish you all the best.
With best regards.
Piyadasa Ranawaka.
Dear Prof.Jeffry,
Your view regarding the proofs of Fermat's last theorem and The so called open access journals
seem to be correct. In this regard it seems that some have already used the well developed countries names also.
Canadian,Brutish, and so on. Thank you very much.
I have already done three simple proofs.
One has already published in 2011,.....................................
Sir, Did n't received any Comments further. Have you gone through it ?
Hey Jeffry I’m back after 6 years to see how you’re going on.
Hey! What the hells this! The ONLY INTELLECTUAL COMMENTS on this blog NAMELY MINE which rules out you and your cronies, have been attributed to anonymous. ANONYMOUS!
Jeffrey, Jeffrey, Jeffrey I think that’s deceitful and IT STINKS but then I shouldn’t be surprised really because at the end of the day you’re a member of that intellectually superior brigade, ‘The Mathmagician’ whose arrogance means they will never admit they are wrong and that mere mortals could never possibly be right when it contradicts or undermines them in any way.
I find it hard to understand how so called professional mathematicians, some of whom are university professors have difficulty comprehending A+B=C. So if we multiply thro’ by A then B we end up with A^2B +AB^2 = ABC. This we will have to give a label, a homogenous triple composite say, to distinguish it from a heterogeneous composite triple where A+B is not equal to C. So far so good!
It is obvious to any normal person that this applies to exponent 1 and by numerical example to exponent 2 to give (A*2)(B^2)(C^2). Applied to the ‘Fermat equation’ we end up with (A^n)(B^n)(C^n). I would mention at this point that the ‘Fermat equation’ is true FOR ALL exponents when two of the base numbers are integer and the third is obtained as the nth root of their equality and would result in a homogeneous triple composite. Of course the only downside is that it would not be an all integer solution.
Further more the composite triple comes with 3 equalities that can be arranged to give 3 quadruples, 2 with all positive terms and one with one negative term. These are known not surprisingly as Fermat-Bateman quadruples. I would refer you to my numerous YouTube videos on the subject of cubic quadruples, They can be derived algebraically (proved) in numerous ways. The exponents of the quadruple can be given as all nth power terms or a mix of both nth power and square terms. So if a ‘Fermat triple existed it would, apart from its numerical input, be a clone of the Pythagorean triples. The F-B quadruples give rise to (A+B)^2=A^2+2AB+B^2. Recogonise it? What an absurdity in light of the binomial expansions of the powers of the integers which in themselves show how stupid the subsequent rubbishing of Fermat’s statement was by you.
I will now take issue with your reply to me in 2014, …….
….. Sorry, but the equation x^n = z^2 - y^2 has nothing to do with Fermat's last theorem, and is of no interest (since every odd number or multiple of 4 is the difference of two squares). ........
because although I didn’t say so at the time trying to be respectful, it beggars belief that a so called university professor of mathematics could utter such a mathematically hypocritical remark!
The equation you dismiss so off handily like an ignoramus is none other than the ‘Quarter Squares Rule’, more properly referred to as Euclid’s ‘Elements’, book2 proposition 8, a theorem and indisputable TRUTH and PROOF. So Z & Y are simply n raised to 2 powers, say p & q so that n=p+q.
To show how stupid I consider you and your ‘Band of Brothers’ can be despite your superior intellect, it is a FACT easily demonstrated at least two ways, one of which I have just shown, that the number of differences of two squares for any x^(2n) or (2n+1) is n so the bigger the exponent the more DoS solutions exist. Hardly the scenario for a Fermat equation to co-exist. So why was it deemed necessary to have a proof that applied to every PRIME power above the second when a proof for powers 3 & 4 was all that was necessary?
Further more it is a well established fact, but obviously not to you and your half wit cronies, that the partial sums of the cubes sum to a triangular number squared so that the difference of two adjacent triangular numbers squared is the cube of the difference between the 2 numbers. We can then multiply repeatedly through the expression by the square of the difference to give the 5,7,9,11,13,15..... power of x all the way to infinity. AND YOU SAY IT HAS NOTHING TO DO WITH THE FERMAT EQUATION ! B*LL*CKS!
YouTube has been awash with videos for the past 6 months regarding a new method for obtaining the roots of quadratics due to Po-Shen Loh and excellent videos, particularly his, they are as well. However I have had to comment that despite over 2 million combined views nobody, as far as I know, has commented to the fact that his method is none other than the application of the ‘Quarter Squares Rule’. At last recognition for the theorem ‘that is of little use’, a comment by Hall & Stevens in their 1893 edition of Euclid’s ‘Elements’.
Whilst I’m on the topic of STUPIDITY what about ‘Fermat’s Sums of Two Squares Theorem’. EVERY PRIME of modularity 4n+1 we are able to check is the sum of two squares as are their composites which are the sum of two or more squares and likewise their powers. We can arrange the ‘Sums of Two Squares’ into orderly, logical sequences that cover every possible one of them just like the ‘Times Tables’ we learn at school. We know the quadratic equations that describe the sequences horizontally, perpendicular and diagonally and yet it seems that this is just not good enough for the mathmagicians who bless us with at least 50 supposed abstract proofs including Zagier’s ‘One Sentence Proof’; “ a proof for professional mathematicians” according to German Professor Grieg!
And now we have proof by computer alias ‘The Windmill Proof’ which is purported to be a graphical representation of the one sentence proof. For me, and I don’t doubt most other people, it is just meaningless, childish graphics.
If the outline of the windmill is taken and the 4 vanes connected to form a square, we have the diagram for the ‘Pythagoras Theorem’ as given by the algebraic identity (A-B)^2=A^2-2AB+B^2. The diagram applies to everyone of the 4n+1 integers. So what the hell does it prove?
In a nutshell Jeffrey anyone who even considers that there may be a 4n+1 prime that is not a sum of two squares is without doubt just plain stupid.
Well nice to have touched base with you again and there is no charge for the maths lesson.
Alastair Bateman alias ‘The Simpleton’.
Alastair:
Your ravings are most intriguing. Have you considered comedy as a profession?
I have published my research paper entitled An elementary proof for Fermat's Last Theorem using Ramanujan-Nagell equation in Turkish Journal of Computer and Mathematics Education Volume 12 No14 2021 pp.3631-3643.
Jeffrey, please accept my profound, profound apologies for my audacity in contaminating the pages of your 'BLOG' but I feel it can't be avoided as I need your HONEST and professional evaluation of the following SIMPLE ALGEBRA.
If we take the equation z^(2n+1) = y^(2n+1) + x^(2n+1) then z^(2n+1) = (a+b)^(2n+1) for every (a + b ) = z every one of which can be expanded by the well established BINOMIAL THEOREM.
Now y^(2n+1) + x^(2n+1) is divisible by (x+y) ALWAYS resulting in two factors the second of which is an algebraic expansion commonly referred as the BINOMIAL NUMBERS, of which I am sure you are very well aware.
Now it is blatantly obvious to any halfwit or crank like myself that the BINOMIAL THEOREM EXPANSION and the BINOMIAL NUMBERS EXPANSION can never be equal being two mathematically different beasts hence z^(2n+1) = y^(2n+1) + x^(2n+1) is FALSE therefore FERMAT'S LAST THEOREM is TRUE.
Pray what do you have to say about that?
Jeffrey, I see that my comment of the 23rd February has not survived.
[1] Is the algebra to complicated for you?
[2] Are you still reviewing it before sanctioning it's existence on your 'blog'.
[3] Is what I said TRUE and it pains you to admit it!
[4] Not what I would expect from someone who overtly attacks anything you don't agree with and justifiably so in most cases I have to admit.
The SILENCE IS DEAFENING yet it speaks a 1000 words to me!
Dear Prof. J. Shalit
try to read the first paper contained
in work https://arxiv.org/abs/1704.06335
You will see that Fermat had certainly not lied.
Sincerely.
Andrea Ossicini
Alastair: What are "mathematically different beasts" and where is your proof that two "mathematically different beasts" can never have the same value?
It's like saying "the Fibonacci numbers and integer squares are `mathematically different beasts' and therefore no Fibonacci number can be a square". Oops, we have F(12) = 144 = 12^2.
It's pure crackpottery!
Andrea: the fact that there are 72 different versions of your paper does not exactly inspire anyone with confidence in your work.
My dear Jeffrey, for a mathematics professor you couldn't have picked a worse example to produce a smoke screen to cloud the issue.
The Fibonacci numbers and the squares are in the context of our discussion discrete quantities, namely integers, and would both therefore plot as a series of dots which must either diverge and never meet or converge and at the point of intersection two dots will either coincide or not as the case may be. In the case you have given they coincide exactly.
The coincidence is not just chance, far from it, but due to the most important of ALL MATHEMATICAL THEOREMS, the Pythagoras theorem.
You will find if you take the trouble to investigate that the Fibonacci numbers have factors that result in a continuous, alternating and innate modularity of 4n+1 then 4n+3 ad infinitum. Hence as Fermat stated all numbers of modularity 4n+1 are the sum of two squares. Now every integer is the difference of two squares. FACT. So all the F.Nos that include a factor of modularity 4n+3 can only be the difference of two squares.
Now we have 89,144,233 were 89 and 233 have modularity 4n+1 so that 89=8^2+3^2 and 233=13^2+8^2. Now 144 has factor 3 of modularity 4n+3 so that 144=13^2-5^2 alias 13^2=12^2+5^2 the second primitive Pythagorean triple. Can you spot the pattern for the base numbers. So if we express every F.No. as the square root squared then every one forms a Pythagoras equation. Pythagorean triples are the templet for the Fermat equation and there is an infinity of infinite series of them, all a consequence of Fermats 4n+1 theorem and the maths for producing them being well defined, known and wholly dependent on triangular numbers no less. Where's the maths for the Fermat triples?
Now z^(2n+1) can be expanded by the binomial theorem as (a+b)^n as many times as there are binary partitions of (a+b)=z. Likewise for z=(c-d). These expansions are characteristied by precisely defined coefficients that are generated in a precise prescribed manner by the combination formula and by precisely prescibed finite difference formulas that terminate in a constant term that is a precisely prescribed factorial.
Now x^(2n+1)+y^(2n+1) is divisible by (x+y) ALWAYS giving binomial number expansions of the form x^2-xy+y^2 and x^4-x^3y+x^2y^2-xy^3+y^4 for the 3rd and 5th powers respectively. I call these expansions 'slip knot equations' because just as pulling the two ends of the knot causes the knot to disappear, so too does multiplying through the expansion by x and y where we just raise the power of the first and last terms and we get duplicate internal terms but of opposite sign which all cancel out just leaving the original expression we started with before division by (x+y).
What else can the algebra do? What else would a truly knowledgable mathematician expect?
The two are truly the mathematical antithesis of each other and if this doesn't constitute proof then I don't know what does since it is an absolute truth.
So to answer your question 'What are mathematically different beasts'. Why the binomial theorem expansion and the binomial number expansions and NO your generalisation for the F.Nos and squares do not qualify as 'mathematically different beasts'.
Perhaps the FICTIOUS Frey equation and Fermat triples qualify to be descibed as 'mathematically different beasts' but then I must step back and leave that one upto you since the Frey equation is part of your mathematical reality but certainly not part of mine.
So fine I'm a crank and what I do is crackpottery or in some one elses eyes a crock of pooh but fine cause like a pig I'm happy to wallow in the mire that is my mathematical realisations of reality and truth and not those of intellectual elitists who live in abject fear of being proved wrong and would never, ever admit to it.
You seem extremely confused. Being a square has virtually nothing to do with being the sum of two squares.
I am afraid I have no possible remedy for this level of confusion and word salad like "both therefore plot as a series of dots which must either diverge and never meet or converge and at the point of intersection two dots will either coincide or not as the case may be".
Time for you to go elsewhere.
As a 'mathematician'? I see you have a lot in common with Leonard Euler in so far as he worked for Catherine the Great, whilst you undoubtedly have worked in the Kremlin.
Unlike Leonard Euler is the fact that he was mathematically honest.
Puuuuurrrrrrrrrrrrrrrrrrr!
Did you not understand "Time for you to go elsewhere"?
the point of intersection two dots will either coincide or not as the case may be".
Someone doesn’t understand the meaning of intersection — among almost everything else it seems.
Dear Unknown:
The journal you published your paper in claims to be concerned with the following topics:
*training in-service mathematics teachers on IT through in-service courses,
*training pre-service mathematics teachers on IT through undergraduate programs,
*understanding how students learn mathematics and solve problems,
*identifying what student learn and think in mathematics,
*assessing and evaluating mathematics curriculum,
*designing and producing distance learning tools for teachers and students.
Yet your paper covers none of these. This appears to be misconduct by someone.
Post a Comment