tag:blogger.com,1999:blog-4987609114415205593.post3765227796453581760..comments2020-01-21T11:51:52.295+00:00Comments on M-Phi: A simpler proof of Fermat's Last Theorem?Jeffrey Ketlandhttp://www.blogger.com/profile/01753975411670884721noreply@blogger.comBlogger27125tag:blogger.com,1999:blog-4987609114415205593.post-33527052872517049962019-06-12T23:28:26.951+01:002019-06-12T23:28:26.951+01:00I have written out an informal proof proving FLT o...I have written out an informal proof proving FLT on the unit circle. It was motivated by trying to find the simplest geometric solution I could find to an even more fundamental proof developed in "Computable Universe".<br /><br />It is hard enough to get people to read a proof of P vs NP based on a new non-standard model of set theory much less use a solution of FLT as a example application.<br /><br />https://www.researchgate.net/publication/331544505_ComputableUniversePneNP_Preprint2019Mar5<br /><br /><br />https://www.dropbox.com/s/chcog93jp2h25s8/FermatsLastTheroem_RevH.pdf?dl=0Unknownhttps://www.blogger.com/profile/08599180551594903694noreply@blogger.comtag:blogger.com,1999:blog-4987609114415205593.post-65322668502687861292018-05-20T22:44:49.402+01:002018-05-20T22:44:49.402+01:00Go to link
http://fermatslibrary.com/p/56b57e1c
...Go to link <br /><br />http://fermatslibrary.com/p/56b57e1c<br /><br />to see (new) method of finding all Pythagorean Triples. This is the first step of how Fermat proved his last theoremolafhttps://www.blogger.com/profile/05290481527334392019noreply@blogger.comtag:blogger.com,1999:blog-4987609114415205593.post-57264915723899135522018-04-30T14:52:48.319+01:002018-04-30T14:52:48.319+01:00Hi all,
I have a question for the statistical test...Hi all,<br />I have a question for the statistical test:<br />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":<br />0.00001, 0.0001, 0.001, 0.01, 0.1, 0.2, 0.3, 0.4, ..., 0.90, 0.99, 1.0<br />Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-4987609114415205593.post-71372454754378534862018-01-22T21:34:18.834+00:002018-01-22T21:34:18.834+00:00Is there any activity on this blog now?
I plan to ...Is there any activity on this blog now?<br />I plan to present "How Fermat did it" in a serie of posts on the site WWW.HUBBIT.NO.<br />But I need some vistors first.<br />If you are interested, post a message under "Fermats original solution from 1637". Once I see the visitors are many enough, I will start to post.<br />Why there? Why not in mathematical publishing magazine? Because I am not a professor,and those magazines will not under any circumstances treat the proof seriously. I already experienced that with Springer. But remember, Fermat himself was an amateur, and not very liked among contemporary so called professionals. So if YOU are interested, ..........olaf vethehttp://www.hubbit.nonoreply@blogger.comtag:blogger.com,1999:blog-4987609114415205593.post-11869054196931867132017-10-15T13:17:15.122+01:002017-10-15T13:17:15.122+01:00http://www.imedpub.com/articles/simple-algebraic-p...http://www.imedpub.com/articles/simple-algebraic-proofs-of-fermats-last-theorem.pdf<br />I hope the challenge you gave is met in the article above.Samuelhttps://www.blogger.com/profile/17567021604595541064noreply@blogger.comtag:blogger.com,1999:blog-4987609114415205593.post-49766307887966126312016-06-07T20:11:43.678+01:002016-06-07T20:11:43.678+01:00Pythagorean quadruples D²=C²+B²+A² can be generate...Pythagorean 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.<br />The 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. <br />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.<br />Alastair Batemanhttps://www.blogger.com/profile/14395447443056931757noreply@blogger.comtag:blogger.com,1999:blog-4987609114415205593.post-15566208467096172652016-02-13T23:19:14.862+00:002016-02-13T23:19:14.862+00:00The simplest proof of FLT follows from the Little ...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.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-4987609114415205593.post-63958326299122404732015-11-06T17:05:43.326+00:002015-11-06T17:05:43.326+00:00I have to agree with Alastair on the Pythagorean i...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.<br /><br />https://www.scribd.com/doc/288757467/Fermat-s-Last-Theorem-as-a-Specific-Case-of-Right-Triangles<br /><br />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...OVVOhttps://www.blogger.com/profile/09987858293295666426noreply@blogger.comtag:blogger.com,1999:blog-4987609114415205593.post-14585704344166454182014-05-29T23:06:49.800+01:002014-05-29T23:06:49.800+01:00Hey Bert I'm back!! Incidentally Jeffrey's...Hey Bert I'm back!! Incidentally Jeffrey's kicked me off his blog cause my simple math's is giving him a headache!! <br />What 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. <br />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.<br />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.<br />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!!Alastair Batemanhttps://www.blogger.com/profile/14395447443056931757noreply@blogger.comtag:blogger.com,1999:blog-4987609114415205593.post-65575786936426001662014-05-05T13:45:22.837+01:002014-05-05T13:45:22.837+01:00PS. Sorry, the correct spelling is of course '...PS. Sorry, the correct spelling is of course 'reframing' and 'rephrasing'.<br />BTW. 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.Bert Brouwerhttps://www.blogger.com/profile/11132121296728341268noreply@blogger.comtag:blogger.com,1999:blog-4987609114415205593.post-49373599963212805032014-05-05T10:53:04.116+01:002014-05-05T10:53:04.116+01:00If you want to no where things are heading for n&g...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):<br /><br />If 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.<br />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.<br /><br />From this the direction in which way things are heading becomes clear: <br />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)<br /><br />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.<br /><br />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.<br /> <br /> <br />Bert Brouwerhttps://www.blogger.com/profile/11132121296728341268noreply@blogger.comtag:blogger.com,1999:blog-4987609114415205593.post-69395102087450398132014-04-11T09:21:23.276+01:002014-04-11T09:21:23.276+01:00There is another explanation of a simple proof of ...There is another explanation of a simple proof of Fermat’s last theorem as follows:<br />X^p + Y^p ?= Z^p (X,Y,Z are integers, p: any prime >2) (1)<br /> <br />1. Let‘s divide (1) by (Z-X)^p, we shall get:<br /> (X/(Z-X))^p +( Y/(Z-X))^p ?= (Z/(Z-X))^p (2)<br /><br />2. That means we shall have:<br /> X’^p + Y’^p ?= Z’^p and Z’ = X’+1 , with X’ =(X/(Z-X)), Y’ =(Y/(Z-X)), Z’ =(Z/(Z-X)) (3)<br /><br />3. From (3), we shall have these equivalent forms (4) and (5):<br />Y’^p ?= pX’^(p-1) + …+pX’ +1 (4)<br />Y’^p ?= p(-Z’) ^(p-1) + …+p(-Z’) +1 (5)<br /><br />4. Similarly, let’s divide (1) by (Z-Y) ^p , we shall get:<br />(X/(Z-Y)) ^p +( Y/(Z-Y)) ^p ?= (Z/(Z-Y)) ^p (6)<br /><br />That means we shall have these equivalent forms (7), (8) and (9):<br />X” ^p + Y” ^p ?= Z” ^p and Z” = Y”+1 , with X” =(X/(Z-Y)), Y” =(Y/(Z-Y)), Z” =(Z/(Z-Y)) (7)<br /><br />From (7), we shall have:<br />X” ^p ?= pY”^(p-1) + …+pY” +1 (8)<br />X”^p ?= p(-Z”)^(p-1) + …+p(-Z”) +1 (9)<br /><br />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).<br /><br />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.<br />X” cannot generate any solution of Fermat’s last theorem, because we have illogical mathematical deductions, for examples, as follows:<br /><br />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.<br />Fermat’s last theorem is simply proved!<br /><br /> <br />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:<br />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.<br />Fermat’s last theorem is simply again proved, with the connection to the concept of (X”/Z”) ^p ?= 2. Is it interesting? <br /><br />Email: thaotrangtvt3@gmail.com<br />Thảo Trầnhttps://www.blogger.com/profile/12613121951684904144noreply@blogger.comtag:blogger.com,1999:blog-4987609114415205593.post-9029221366379148842014-02-15T20:13:30.482+00:002014-02-15T20:13:30.482+00:00http://iosrjournals.org/iosr-jm/papers/Vol7-issue4...http://iosrjournals.org/iosr-jm/papers/Vol7-issue4/A0740103.pdf<br />To Anonymous who has pointed out the blog.<br />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,PiyadasaR.A.D.Piyadasanoreply@blogger.comtag:blogger.com,1999:blog-4987609114415205593.post-69308807331655190632013-12-26T23:07:50.650+00:002013-12-26T23:07:50.650+00:00Samuel Bonaya Buya are you kidding.Samuel Bonaya Buya are you kidding.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-4987609114415205593.post-54248490228513742192013-12-16T20:10:09.426+00:002013-12-16T20:10:09.426+00:00There is indeed a simple proof of Fermats last the...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+ARISINGSamuel Bonaya Buyahttps://www.blogger.com/profile/13888621400263034065noreply@blogger.comtag:blogger.com,1999:blog-4987609114415205593.post-36355734928994390382013-12-02T06:01:40.599+00:002013-12-02T06:01:40.599+00:00See :
http://www.iosrjournals.org/iosr-jm/pages/v7...See :<br />http://www.iosrjournals.org/iosr-jm/pages/v7i4.html<br /><br />or<br /><br />http://iosrjournals.org/iosr-jm/papers/Vol7-issue4/A0740103.pdf<br /><br />How to prove the impossibility of such condition?<br />From where we get the wrong proof proposed by Lame in 1847?<br />Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-4987609114415205593.post-6497633537418729322013-08-24T11:01:48.253+01:002013-08-24T11:01:48.253+01:00In http://www.mymathforum.com/viewtopic.php?f=40&a...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.<br /> In this thread,check the August 3th post that contains the latest version Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-4987609114415205593.post-24864878255299630472013-08-23T12:42:48.868+01:002013-08-23T12:42:48.868+01:00I am not surprised by your comment,Jeffery.However...I am not surprised by your comment,Jeffery.However, you have no right to tell anything wrong regarding my proof.<br />R.A.D.R.A.D.Piyadasanoreply@blogger.comtag:blogger.com,1999:blog-4987609114415205593.post-5754932306986841692013-08-09T00:34:36.015+01:002013-08-09T00:34:36.015+01:00O.K. I read the blog
http://recursed.blogspot.co...O.K. I read the blog <br /><br />http://recursed.blogspot.co.uk/2012/03/proof-of-fermats-last-theorem.html<br />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.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-4987609114415205593.post-24256373660320346772013-08-04T07:33:26.231+01:002013-08-04T07:33:26.231+01:00I am reading
http://recursed.blogspot.co.uk/2012...I am reading <br /><br />http://recursed.blogspot.co.uk/2012/03/proof-of-fermats-last-theorem.html<br />to see what says about the proof of R.A.D.<br />Is this Blog is controlled by Jeffery Ketland./ Pl.keep this for a few days evenAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-4987609114415205593.post-16390181031729597102013-07-22T00:29:03.454+01:002013-07-22T00:29:03.454+01:00Thanks for your comment, R.A.D.
Your alleged &quo...Thanks for your comment, R.A.D.<br /><br />Your alleged "proof" is not a proof.<br /><br />http://recursed.blogspot.co.uk/2012/03/proof-of-fermats-last-theorem.html<br /><br />Cheers,<br /><br />JeffJeffrey Ketlandhttps://www.blogger.com/profile/01753975411670884721noreply@blogger.comtag:blogger.com,1999:blog-4987609114415205593.post-88209391866738266252013-07-21T23:53:31.524+01:002013-07-21T23:53:31.524+01:00I think Hervey Friedman grand conjecture and Colin...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.<br />The 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.R.A.D.Piyadasanoreply@blogger.comtag:blogger.com,1999:blog-4987609114415205593.post-78763638617052200182013-06-28T04:10:47.115+01:002013-06-28T04:10:47.115+01:00http://www.mymathforum.com/viewtopic.php?f=40&...http://www.mymathforum.com/viewtopic.php?f=40&t=40857<br />By this do you mean proof of FLT or what? I could find nothing.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-4987609114415205593.post-52707118139788429322013-06-21T18:32:32.840+01:002013-06-21T18:32:32.840+01:00Check this:
http://www.mymathforum.com/viewtopic.p...Check this:<br />http://www.mymathforum.com/viewtopic.php?f=40&t=40857Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-4987609114415205593.post-69270639155743262562013-05-22T13:42:25.824+01:002013-05-22T13:42:25.824+01:00A simple and short analytical proof of Fermat'...A simple and short analytical proof of Fermat's last theorem on the internet supports Mclarty findings.R.A.D.Piyadasanoreply@blogger.com