McLarty calls Grothendieck's work "a toolkit," and showed, at the Joint Mathematics Meetings in San Diego in January, that only a small portion is needed to prove Fermat's Last Theorem. [...] "Where Grothendieck used strong set theory I've shown he could do with only a fraction of it," McLarty said. "I use finite-order arithmetic, where all sets are built from numbers in just a few steps. You don't need sets of sets of numbers, which Grothendieck used in his toolkit and Andrew Wiles used to prove the theorem in the 90s."According to the article, Ohio State emeritus Harvey Friedman has described McLarty's result as a "clarifying first step". McLarty retorts:

"Fermat's Last Theorem is just about numbers, so it seems like we ought to be able to prove it by just talking about numbers," McLarty said. "I believe that can be done, but it will require many new insights into numbers. It will be very hard. Harvey sees my work as a preliminary step to that, and I agree it is."This sounds like a very exciting development, and I for one look forward to seeing where it will go. Any M-Phi'ers around with insider's knowledge on the whole thing? I could not find further info over at McLarty's website.

UPDATE: Here is a relevant article by McLarty, 'What does it take to prove Fermat's Last Theorem?', published in the Bulletin of Symbolic Logic in 2010 (haven't read it yet).

There has been quite a bit of discussion of McLarty's work on the Foundations of Mathematics mailing list, and one could get a really good sense for McLarty's approach and how it developed by looking through the archives.

ReplyDeleteThanks for the pointer! I used to keep track of the FOM list, but as of recently I've had little time for it, unfortunately... I'll take a look at the archives.

DeleteA simple and short analytical proof of Fermat's last theorem on the internet supports Mclarty findings.

ReplyDeleteCheck this:

Deletehttp://www.mymathforum.com/viewtopic.php?f=40&t=40857

http://www.mymathforum.com/viewtopic.php?f=40&t=40857

ReplyDeleteBy this do you mean proof of FLT or what? I could find nothing.

In http://www.mymathforum.com/viewtopic.php?f=40&t=40857 there's an interesting post under the name "a new approach to fermat's last theorem" sent on May 30th, 2013.

DeleteIn this thread,check the August 3th post that contains the latest version

I think Hervey Friedman grand conjecture and Colin McLarty's prediction simply imply that Fermat's last theorem can be proved using elementary mathematics.I have already developed a simple proof;" A simple proof and short analytical proof of Fermat's last theorem',Canadian journal on computing in Mathematics,Natural Sciences,Engineering and Medicine, Vol.2,No.3,March 2011, p-57-63.

ReplyDeleteThe second proof I have developed is based on the fundamental theorem on Arithmetic, Binomial theorem and the Remainder theorem to be published. In this case no need of Fermat's little theorem even.

Thanks for your comment, R.A.D.

ReplyDeleteYour alleged "proof" is not a proof.

http://recursed.blogspot.co.uk/2012/03/proof-of-fermats-last-theorem.html

Cheers,

Jeff

I am reading

ReplyDeletehttp://recursed.blogspot.co.uk/2012/03/proof-of-fermats-last-theorem.html

to see what says about the proof of R.A.D.

Is this Blog is controlled by Jeffery Ketland./ Pl.keep this for a few days even

O.K. I read the blog

ReplyDeletehttp://recursed.blogspot.co.uk/2012/03/proof-of-fermats-last-theorem.html

In this blog a proof of Fermat's last theorem has been hackneyed. I must emphasize it is not the general proof of R.A.D. It is said that(in the Blog) R.A.D has proved (A Simple and Short Analytical Proof) Fermat's Last Theorem for all exponents. It is seemed that the Prof. who is running the blog did not believe in the beginning that R.A.D had proved the theorem in general.Later he has understood he had done it.'A ample and short analytical proof of Fermat's last theorem has been published' "I rest my case. " R.A.D is an amateur.He did not want to publish his results even.He is saying that second simple proof is O.K.Hello ,Jeffery Ketland first read the blog and R.A.D.s proof carefully.Then comment.Thanks.

I am not surprised by your comment,Jeffery.However, you have no right to tell anything wrong regarding my proof.

ReplyDeleteR.A.D.

See :

ReplyDeletehttp://www.iosrjournals.org/iosr-jm/pages/v7i4.html

or

http://iosrjournals.org/iosr-jm/papers/Vol7-issue4/A0740103.pdf

How to prove the impossibility of such condition?

From where we get the wrong proof proposed by Lame in 1847?

There is indeed a simple proof of Fermats last theorem as explained here: http://www.network54.com/Forum/666092/thread/1386755923/last-1386755923/MY+PROOF+OF+FERMAT’S+LAST+THEOREM+AND+MATTERS+ARISING

ReplyDeleteSamuel Bonaya Buya are you kidding.

ReplyDeletehttp://iosrjournals.org/iosr-jm/papers/Vol7-issue4/A0740103.pdf

ReplyDeleteTo Anonymous who has pointed out the blog.

Any one can find numbers that satisfy the condition of FLT(A^n+B^n=C^n).They are numbers but by no means integers. I challenge that no one in the world can not find integers that satisfy the condition of FLT.R,A,D,Piyadasa

http://www.imedpub.com/articles/simple-algebraic-proofs-of-fermats-last-theorem.pdf

DeleteI hope the challenge you gave is met in the article above.

There is another explanation of a simple proof of Fermat’s last theorem as follows:

ReplyDeleteX^p + Y^p ?= Z^p (X,Y,Z are integers, p: any prime >2) (1)

1. Let‘s divide (1) by (Z-X)^p, we shall get:

(X/(Z-X))^p +( Y/(Z-X))^p ?= (Z/(Z-X))^p (2)

2. 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)

3. 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)

4. 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 an 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) and (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?

Email: thaotrangtvt3@gmail.com

If you want to no where things are heading for n>2, then n=1 is your point of reference to see where it all started... Here's a simple demonstration that FLT is at least plausible, using some basic trigonometry (or even less than that):

ReplyDeleteIf n=1 the angle between side a and b is 180 degrees; for all values of a and b, all solutions for c are also integers.

If n=2, the angle between side a and b is 90 degrees; for only some values of a and b, c has solutions in integers.

From this the direction in which way things are heading becomes clear:

If n>2, the angle between side a and b must be <90 degrees (which is true); and the logical assumption is also that for none of the values of a and b, c has solutions in integers (which is the infamous FLT)

Note that if n=infinite and a=b=1, c comes infinitely close to being the integer 1, which makes FLT almost false because there's almost one solution for c if n>2.

Saying that FLT is all about numbers is questionable, because based on this approach as shown above, you can argue that it also has something to do with angles: it is clear that for n>2 the cosine law applies. And refraiming the problem can also mean refraising the formula that could ultimately proof FLT. Maybe the end result proofs that the correlation between c and for instance a, is something like (a+1)/a. Then you run quickly out of options for c being an integer, I would say.

PS. Sorry, the correct spelling is of course 'reframing' and 'rephrasing'.

DeleteBTW. What distinguishes FLT in the cosine law from a^2+b^2=c^2 is of course -2ab cos gamma, while for n=1 it can be reformulated as a+b=c.

Hey Bert I'm back!! Incidentally Jeffrey's kicked me off his blog cause my simple math's is giving him a headache!!

ReplyDeleteWhat is the difference between 14^2-13^2=27, 13^2-12^2=25, 12^2-11^2=23? Why of course 25=5^2 so the middle one is a Primitive Pythagorean Triple. So what are the other two? Well the 'mathmagicians' can't be bothered to tell us so let's call them 'pseudo Pythagorean Triples' where only two legs are integer and one is irrational. The logic for this is that we can put 25 = (25^1/2)^2 = 5^2 and therefore also (27^1/2)^2=27 & (23^1/2)^2=23.

So 14^2=13^2+(27^1/2)^2 & 12^2=11^2+(23^1/2)^2 both of which satisfy Pythagoras's equation. But 27 =3^3 so 14^2=13^2+3^3 which is z^2=y^2+x^n gives z^2-y^2= x^n=(z+y)(z-y) and is just the difference of squares as proved by Euclid, book2,props 5&6. The 'DoS' generates every integer from 1 to infinity and so generates EVERY INTEGER RAISED TO EVERY CONCEIVABLE POWER. When the integer generated is a square we get a 'PPT' and all the rest are therefore 'pseudoPT' by my definition.

As Fermat said 'It is not possible to SEPARATE ... any power greater than the second into two powers with the same exponent ..'. Why did he say that? Because he found that he only ever got the 'DoS' solution z^2-y^2= x^n and he knew and we know that it is impossible to elevate the equality to an all integer solution of exponent n because ... no, not the modern proof ... but the 'Distributive Law of Multiplication' prohibits it.

As I've commented elsewhere before 'you can't push a square peg into a round hole' and likewise for the integers, you can't push a cubic or higher power solution into what is after all a Pythagorean equation. So Pythagoras reigns supreme. 350 years to solve such a trivial problem, I can't get my breath!!

I have to agree with Alastair on the Pythagorean influence. In fact, when you generalize the law of cosines you find Fermat as a special case of right triangles. You can show how the isosceles triangle is a trivial support to FLT due to Pythagoras' constant and how the generalized law of cosines holds for equilateral triangles.

ReplyDeletehttps://www.scribd.com/doc/288757467/Fermat-s-Last-Theorem-as-a-Specific-Case-of-Right-Triangles

What would really be of interest is if you could demonstrate the relationship between right triangles and the elliptical curves Wiles used to simply the proof...

The simplest proof of FLT follows from the Little Theorem. The proof however takes a backseat to an amazing pattern that forms the Yin Yang of mathematics. The simplicity is stunning.

ReplyDeletePythagorean quadruples D²=C²+B²+A² can be generated from (a+b)²= a²+2ab+b² where D=(a+b)² : C= a² : B= 2ab : A= b². But what if the inputs into (a+b)² are themselves squares or more precisely pythagorean triples a²+b²=c². Then you will find we generate 3 related pythagorean quadruples, 2 with all 4 terms positive and the other with 3 positive and 1 negative term. (This later quadruple is WRONGLY NOT classified as a pythagorean quadruple by mathematicians). So for 3²+4²=5² we get: (i) 25² =20²+12²+9² : (ii) 25² =16²+15²+12² (iii) 20² =16²+15²-9² and this applies to every single pythagorean triple.

ReplyDeleteThe Fermat triple is easily shown to algebraically generate the 3 self same quadruples but with the exponent 2 replaced by n but it should be observed that 2 terms in each quadruple are perfect squares since (aⁿ)(aⁿ)=a²ⁿ=(aⁿ)² and therefore all 3 quadruples can be stated as the difference of 2 squares. I have named named them Fermat - Bateman quadruples. Furthermore every square integer and therefore every (aⁿ)² can be expressed as a NEGATIVE pythagorean quadruple whose terms are simple to derive so for example: 20² =16²+13²-5². Taking D² =C²+B²-A² then D-C=4 : B-A=8 : B+A= (D+C)/2. For D odd then B & A are rational with a decimal fraction of 0.5.

Also every integer of the form (4n)² is the sum of 4 adjacent odd integers squared minus 20=4²+2² hence 20² =13²+11²+9²+7²-20 and which includes every (4nⁿ)². Note that the sum of the 2 largest terms which for the example is 13+11=24 and is always 4 more than the total sum. So every term of a Fermat triple if one existed would have a simple solution in terms of squares after squaring once or infinitely many times so as Pierre de Fermat said 350 years ago triples above the second power cannot exist. So for it to be repeatedly stated and dogmatically defended that the tools for him to have proved his theorem did not exist in his day is at the least erroneous and at worst a lie. It is for the reasons given that no quadruples above the 4th power are known simply because they do not and cannot exist. It is worth stating that cubic quadruples also contain infinitely many related trios but none are algebraically of the construction of Fermat-Bateman quadruples. For example: 12³=10³+9³-1³ : 9³=8³+6³+1³ : 12³-10³=9³-1³=8³+6³ therefore 12³=10³+8³+6³ divide by 2³ gives 6³=5³+4³+3³ the smallest all integer cubic quadruple.