tag:blogger.com,1999:blog-12535639.post111686936721419382..comments2020-05-15T21:04:01.369-07:00Comments on Fermat's Last Theorem: Fermat's Last Theorem: Proof for n=3Larry Freemanhttp://www.blogger.com/profile/06906614246430481533noreply@blogger.comBlogger33125tag:blogger.com,1999:blog-12535639.post-70677578015916605372017-07-23T19:40:15.556-07:002017-07-23T19:40:15.556-07:00Hi there-The PROOF OF PROOFS for n=3 - FERMAT-MURG...Hi there-<b>The PROOF OF PROOFS for n=3 - FERMAT-MURGU QUADRUPLETS</b><br />ARE A MATH BEAUTY IN CONCURRENCE NOW WITH PYTHAGOREAN TRIPLETS.<br />IT A REAL HUNTING FOR AND CONTAIN A COMPLETE MODULAR METHOD BY DEFINITION.<br />BUT ANYWAY WE SOLVED FERMAT'S LAST THEOREM WITH ABSOLUTE ACCURACY AND FOR<br />ALL n VIA <b>Fermat-Murgu Impossible Equations </b> <br />1. SENT all Fermat Equations Solutions in Irrational Field , without any doubts,<br />even for Z=Integers X,Y must to be Irrational.<br />2. Fermat-Murgu n Media Impassable for Fermat Triplets.<br />We Get it by Analyzing with Ion Murgu Math Millennium Equations all n neighbors<br />around of a supposed by absurd solutions (X,Y,Z). Anonymoushttps://www.blogger.com/profile/04741133106970399958noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-66992082850561948082017-03-07T10:52:50.329-08:002017-03-07T10:52:50.329-08:00x^n+y^n=z^n ..?
have solution !
if n=1,2,3,....
h...x^n+y^n=z^n ..?<br />have solution !<br />if n=1,2,3,....<br /><br />how much solution ?Anonymoushttps://www.blogger.com/profile/06392870113810463941noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-52791196006319544122017-01-03T01:02:06.497-08:002017-01-03T01:02:06.497-08:00Hallo evrybody,
Here is from where a simple proof...Hallo evrybody,<br /><br />Here is from where a simple proof will follow:<br /><br />(1) $A^n= \sum_{1}^{A} (X^n-(X-1)^n)$<br /><br />This by Telescoping Sum property, and it's a way to square any curve (the first, but also the follwing derivate) of the type $y'=nx^n$.<br /><br />It can be aslo be seen in set Theory as an Ordinal Number.<br /><br />This means that FLT can be rewritten as:<br /><br />$C^n= \sum_{1}^{A} (X^n-(X-1)^n) + \sum_{1}^{B} (X^n-(X-1)^n$<br /><br />but grouping the same terms of $A^n$ or $B^n$ will gives the Symmetric condition:<br /><br />$C^n= 2* \sum_{1}^{A} (X^n-(X-1)^n) + \sum_{A+1}^{B} (X^n-(X-1)^n$<br /><br />and/or<br /><br />$C^n= 2* \sum_{1}^{B} (X^n-(X-1)^n) - \sum_{A+1}^{B} (X^n-(X-1)^n$<br /><br />This condition imply also any $C^n/K$ is in bijection with other two rationals.<br /><br />But since for the (1) follows that for $n>2$ the derivate is a curve, so the "balancing point" where the missin area bellow the curve it's equal to the exceding one, BOTH in X than in Y the balancing point IS NOT in the MIDDLE, follows Fermat is right.<br /><br />To be more clear for the Tricotomic law, since we prove the distance of the balancing points always differs from the middle point for $n>2$, there cannot exist another Ordinal that state that such distance are equals.<br /><br />Stefano Maruelli<br /><br /><br /><br /> <br /><br /><br />Ganpa4061https://www.blogger.com/profile/17794361305070219426noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-80123377105221191442016-10-28T07:55:35.612-07:002016-10-28T07:55:35.612-07:00It really seems an elementary proof of Fermat'...It really seems an elementary proof of Fermat's Last Theorem has finally been found !: https://www.quora.com/Can-you-verify-this-proposed-elementary-proof-of-Fermats-Last-Theorem-in-the-commentary-section-below-this-question?srid=1zDkTK.https://www.blogger.com/profile/13806968031405513794noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-69602267703762280052015-12-15T20:29:09.915-08:002015-12-15T20:29:09.915-08:00Hi Larry
I need help with a problem. Knowing the ...Hi Larry <br />I need help with a problem. Knowing the result for FLT for n=3 I need to prove that if a positive integer n is divisible by 3 , then there are no x.y, z such that x^n+y^n=z^n.<br />What is did so far was say that if 3/n then 3=k.n, for some k positive integer. Then if (x^3, y^3,z^3) is a solution for the exponent k <br />(x^3)^k+(y^3)^k=(z^3)^k<br />and because 3.k=k.3 in integers <br />(x^k)^3+(y^k)^3= (z^k)^3<br /><br />where x^k=a ,y^k,=b z^k=c are positive integers. Then there are no , a,b,c positive integers such that a^3+b^3=Z^3 .<br />Could that be remotely correct?MaNAtaliahttps://www.blogger.com/profile/04622006124196010069noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-56699676470348627052015-07-09T01:28:02.719-07:002015-07-09T01:28:02.719-07:00Hi, Larry, I want to ask: what is the conclusion o...Hi, Larry, I want to ask: what is the conclusion of the infinite descent on n=3, is it that since we are talking about natural numbers, then infinite descent would not be applicable and thus we can conclude that there does not exist a solution to the FLT when n=3? Also in the case n=4, is the smaller solution n=2? If yes n=2 has a solution, so how does it work?<br />Thanks for your time.olaleye oluwatosinhttps://www.blogger.com/profile/03813751075116226878noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-75879646385874259292013-10-17T08:16:38.636-07:002013-10-17T08:16:38.636-07:00I have proof of fermet last theorem as...I have proof of fermet last theorem as ferma proved shortly.GJhttps://www.blogger.com/profile/17356833579296894108noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-34514282084497780752013-10-17T07:36:03.796-07:002013-10-17T07:36:03.796-07:00I have proved fermat last theorem...I have proved fermat last theorem as ferma proved which is very easy method. Where should I public this ferma proof.GJhttps://www.blogger.com/profile/17356833579296894108noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-85343858826975449772013-10-09T20:26:33.993-07:002013-10-09T20:26:33.993-07:00I admit that there is a basic premise of which I a...I admit that there is a basic premise of which I am unaware. Perhaps one being that the unknown must be a positive integer? X=1, Y= -1, makes Z = 0 and also solves for XYZ = 0. But I don't know why that is relevant. (2,-2), (3,-3), etc solve the equation.Anonymoushttps://www.blogger.com/profile/16948104108235008848noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-67566896540704728092013-10-09T20:25:26.296-07:002013-10-09T20:25:26.296-07:00I admit that there is a basic premise of which I a...I admit that there is a basic premise of which I am unaware. Perhaps one being that the unknown must be a positive integer? X=1, Y= -1, makes Z = 0 and also solves for XYZ = 0. But I don't know why that is relevant. (2,-2), (3,-3), etc solve the equation.Anonymoushttps://www.blogger.com/profile/16948104108235008848noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-57846906373779622722013-07-09T17:55:44.535-07:002013-07-09T17:55:44.535-07:00An interesting feature of x^n + y^n = z^n is when ...An interesting feature of x^n + y^n = z^n is when n=1, 0 and -1 given the right formula.<br /><br />It will give values as if looking correspondingly along x, y and z and relative values of the other lines.<br /><br />i.e. with x=3 and y=4 (giving z=5 then:<br />n=1 gives x=1.8 y=3.2 z=5<br />n=0 gives x=0.36 y=0.64 z=1<br />n=-1 gives x=0.072 y=0.128 z=0.2Anonymoushttps://www.blogger.com/profile/18236135029986371097noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-66353970326180731092013-07-09T16:16:13.413-07:002013-07-09T16:16:13.413-07:00Hi Larry,
If I had a proof for FLT where would I s...Hi Larry,<br />If I had a proof for FLT where would I submit it?Anonymoushttps://www.blogger.com/profile/18236135029986371097noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-77489726485099709142013-06-03T19:52:41.080-07:002013-06-03T19:52:41.080-07:00Hi Andy,
Click the link at that step to find out ...Hi Andy,<br /><br />Click the link at that step to find out about p,q. It's explained in the lemma on the page of the link.Larry Freemanhttps://www.blogger.com/profile/06906614246430481533noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-86871213128564676632013-06-03T18:43:10.529-07:002013-06-03T18:43:10.529-07:00Hi i am in a math research class and my topic is d...Hi i am in a math research class and my topic is doing this theorem and i am reading this and i don't know where variables pq and gcd come from into the xyzAnonymoushttps://www.blogger.com/profile/11771475170532018819noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-86889396278316829262013-04-20T02:50:46.016-07:002013-04-20T02:50:46.016-07:00nice blog :)nice blog :)adminhttps://www.blogger.com/profile/07335896163735766918noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-51511173029319137682012-11-19T04:33:10.369-08:002012-11-19T04:33:10.369-08:00Sir i have a proof of Fermat's theorem for n=3...Sir i have a proof of Fermat's theorem for n=3 in new way and with the help of odd number techniques i prepare a table with the help of which we calculate the value of square of hypotenuse.Anonymoushttps://www.blogger.com/profile/09716584296048447087noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-77626048996820927022012-03-04T19:17:32.960-08:002012-03-04T19:17:32.960-08:00Hi Larry,
Another hint, for x, y, z are all integ...Hi Larry,<br /><br />Another hint, for x, y, z are all integers and power integer nth is greater than 2, then the equation <br /><br />x^n + y^n = z^n<br /><br />is as false as 1 = 2.<br /><br />Cheers,<br /><br />Joe NilaadJoehttps://www.blogger.com/profile/04475008322646082710noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-78767230812145664692012-03-03T17:55:57.583-08:002012-03-03T17:55:57.583-08:00Hi Larry,
It's been 3 years since I made comm...Hi Larry,<br /><br />It's been 3 years since I made comments about Fermat might have it right about his comments.<br /><br />I can prove it but that didn't mean it would be the same as Fermat had thought. Everybody were trying to find the root. But if the equation is invalid when tripletts are integers to begin with, then what is the point for looking for the root.<br /><br />This year will be 375 years since he conjected. This number is nice be cause it is 3*5^3. I think it's time to reveal the riddle.<br /><br />I believe that with your ability you probably could come up with a proof to FLT. Think outside the box: FLT is a mathematical false statement with integers.<br /><br />Cheers,<br /><br />JoeJoehttps://www.blogger.com/profile/04475008322646082710noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-35117927296228031492011-04-08T16:39:50.250-07:002011-04-08T16:39:50.250-07:00A positive integer answer would have to be somethi...A positive integer answer would have to be something where a=c such that:<br /><br />a^3 + 0^3 = a^3<br /><br />-LarryLarry Freemanhttps://www.blogger.com/profile/06906614246430481533noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-58391928864183135032011-04-08T15:00:05.946-07:002011-04-08T15:00:05.946-07:00By a well known proof, we know that the only integ...By a well known proof, we know that the only integer answer is where abc=0.<br /><br />So, that we could have for example:<br /><br />a = 3, b = -3, c = 0 so that we have:<br /><br />(3)^3 + (-3)^3 = 0^3<br /><br />27 + -27 = 0<br /><br />If abc != 0, then it follows that a,b,c cannot all be integers if the equation holds true.Larry Freemanhttps://www.blogger.com/profile/06906614246430481533noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-59152473220175614312011-04-08T14:46:19.044-07:002011-04-08T14:46:19.044-07:00Hi, this is Bisher, I wanted to know if A^3+B^3=C^...Hi, this is Bisher, I wanted to know if A^3+B^3=C^3 is actually a possible equation with a positive integer answer.Unknownhttps://www.blogger.com/profile/13950486230356599848noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-71885735691782752822009-10-16T17:57:18.318-07:002009-10-16T17:57:18.318-07:00Proof of Case I of Fermat’s Last Theorem for n = 3...Proof of Case I of Fermat’s Last Theorem for n = 3<br />The following is the most elegant solution imaginable.<br /> <br />(x+y+z)^3 can be neatly rearranged as follows;<br /><br />(x+y+z)^3 = x^3 + y^3 + z^3 + <br /> 3(x+y)(y+z)(x+z)<br /><br />Therefore, assuming that x^3 + y^3 + z^3 = 0<br /><br />(x+y+z)^3 = 3(z+x)(z+y)(x+y) <br /><br />clearly, 3 only appears on the right hand side once as it cannot appear in (z+x) or (z+y) or (x+y) under case I.<br /><br />This is because all prime factors of (x+y) must be contained in -z^3 and hence z.<br /><br />Similarly, all prime factors of (x+z) must be contained in –y^3 and hence y<br /><br />Also all prime factors of (y+z) must be contained in –x^3 and hence must be contained in x<br />and under this particular iteration of Case I, none of x, y and z is divisible by 3.<br /><br />Therefore (z+x), (z+y) and (x+y) cannot be divisible by 3 if x, y and z cannot be divisible by 3<br /> <br />Therefore the right hand side can never be a cube, even if (z+x), (z+y) and (x+y) are all cubes<br /><br />Therefore (x+y+z)^3 = 3(z+x)(z+y)(x+y) is impossible for case I<br />where 3 does not divide x, y, z <br /><br />which means x^3 + y^3 + z^3 cannot equal 0<br /><br />This proves Case I of Fermat’s Last Theorem for n=3neat_mathshttps://www.blogger.com/profile/09923473294000429170noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-53064553088676229832009-09-24T09:21:55.012-07:002009-09-24T09:21:55.012-07:00Hi Gangerolf,
I posted a comment on your blog.
O...Hi Gangerolf,<br /><br />I posted a comment on your blog.<br /><br />On this blog, I will only talk about the history of FLT.<br /><br />Cheers,<br /><br />-LarryLarry Freemanhttps://www.blogger.com/profile/06906614246430481533noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-10654552062019077092009-09-23T17:42:03.145-07:002009-09-23T17:42:03.145-07:00Hi Larry,
Please comment on the blogger Gangerolf...Hi Larry,<br /><br />Please comment on the blogger Gangerolf's "proof".<br />http://gangerolf-gangerolf.blogspot.com/Gangerolfhttps://www.blogger.com/profile/03136072056807803751noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-18149775841326926642009-02-09T10:33:00.000-08:002009-02-09T10:33:00.000-08:00Hi Joe,Thanks for your question.The proof by Andre...Hi Joe,<BR/><BR/>Thanks for your question.<BR/><BR/>The proof by Andrew Wiles is over 100 pages long.<BR/><BR/>It is possible that there is a major insight that takes just 1 page but it is very unlikely.<BR/><BR/>Cheers,<BR/><BR/>-LarryLarry Freemanhttps://www.blogger.com/profile/06906614246430481533noreply@blogger.com