ෆර්මාගේ අවසන් ගැටළුව

luxmen

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priyankaH

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You can write and publish any thing you want on the Internet but did any of the reputable Professor or any University ever accepted as your proof is correct? because when I last checked with the university still no one is able to completely solve it yet !!!
 
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silentsahan

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    Rajagiriya

    file:///C:/Documents%20and%20Settings/mn/My%20Documents/Downloads/FERMAT_LAST_THEOREM.pdf

    මේ මොකක්ද යකෝ මේ url එක..:baffled:
    දැන් උඹ කියන දේවල් එක්ක මෙහෙම දෙයක්
    හරියට බ්ලොග් එකේ දාන හැටි වත් දන්නේ නැහැ
    කියලා නේද ලෝකයා හිතන්නේ..

    අනික උබේ ඔය ස්ටැට් ටික වැඩක් නැහැ.. එක එකා
    vpn ප්‍රොක්සි දාලා එන හිට්ස් ඔය.. ඔතන සේරම
    උන් ලංකාවේ උන් වෙන්නැති...


    උඹේ වයස කියපන් බලන්න..
    අනේ මන්ද උඹලා.. :baffled:
     

    mag123

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    -----------------------------------------------------

    You can write and publish any thing you want on the Internet but did any of the reputable Professor or any University ever accepted as your proof is correct? because when I last checked with the university still no one is able to completely solve it yet !!!

    Yes , you are correct Professor Sir Andrew John Wiles proved it and his proof is over 150 pages, and uses techniques from algebraic geometry and number theory .

    -----------------------------------------------------------------------------------
    During 21–23 June 1993 Wiles announced and presented his proof of the Taniyama–Shimura conjecture for semi-stable elliptic curves, and hence of Fermat's Last Theorem, over the course of three lectures delivered at the Isaac Newton Institute for Mathematical Sciences in Cambridge, England.[1] There was a relatively large amount of press coverage afterwards.[23]
    After the announcement, Katz was appointed as one of the referees to review Wiles's manuscript. In the course of his review, he asked Wiles a series of clarifying questions that led Wiles to recognise that the proof contained a gap. There was an error in one critical portion of the proof which gave a bound for the order of a particular group: the Euler system used to extend Flach's method was incomplete. The error would not have rendered his work worthless – each part of Wiles's work was highly significant and innovative by itself, as were the many developments and techniques he had created in the course of his work, and only one part was affected.[29] Without this part proved, however, there was no actual proof of Fermat's Last Theorem.
    Wiles and his former student Richard Taylor spent almost a year resolving this issue.[30][31] Wiles indicates that on the morning of 19 September 1994 he realised that the specific reason why the Flach approach would not work directly suggested a new approach based on his previous attempts using Iwasawa theory, which resolved the issue and resulted in a CNF that was valid for all of the required cases. On 6 October Wiles asked three colleagues (including Faltings) to review his new proof,[11] and on 24 October 1994 Wiles submitted two manuscripts, "Modular elliptic curves and Fermat's Last Theorem"[32] and "Ring theoretic properties of certain Hecke algebras",[33] the second of which Wiles had written with Taylor and proved that certain conditions were met which were needed to justify the corrected step in the main paper.
    The two papers were vetted and finally published as the entirety of the May 1995 issue of the Annals of Mathematics. The new proof was widely analysed, and became accepted as likely correct in its major components.[34][35][36][37] These papers established the modularity theorem for semistable elliptic curves, the last step in proving Fermat's Last Theorem, 358 years after it was conjectured.
    ----------------------------------------------------------------------------------------------------------------
    :yes::yes::yes:-
     
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    mag123

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  • Jan 20, 2008
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    24 MAY 2016 • 8:49AM

    When Andrew Wiles received the £500,000 Abel Prize for mathematics last week, there was a general sense of “At last!” in the mathematical community.

    After all, Professor Wiles had already won almost every other prize for his 1995 proof of Fermat’s last theorem, the most notorious problem in the history of mathematics.

    As it has been mentioned in Dr Who, Star Trek, The Simpsons and the Liz Hurley blockbuster Bedazzled, I would hope that most people would know the intricacies of Fermat’s last theorem by now, but here’s a quick recap for those who are still puzzled about why there is so much fuss over solving a maths problem.

    Given that there are infinitely many possible numbers to check it was quite a claim, but Fermat was absolutely sure that no numbers fitted the equation because he had a logical watertight argument. Sadly, he never wrote down his proof. Instead, in the margin of a book, he left a tantalizing note in Latin: “I have a truly marvellous demonstration of this proposition (demonstrationem mirabilem) which this margin is too narrow to contain.”

    After Fermat’s death, mathematicians found lots of similar notes (“I can prove this, but I have to feed the cat” or “I can prove that, but I have to wash my hair”), so they set about rediscovering Fermat’s supposed proofs. They were successful in every case, except proving that (an + bn = cn) has no solutions, which is why it became known as Fermat’s last theorem, namely the last one that could be proven.

    For three centuries, mathematicians tried and failed to find a proof, which is why Wiles’s eventual success was such a major achievement, and why he has been showered with prizes and accolades. For example, there was the King Faisal International Prize (£140,000), the Wolf Prize (£70,000), a knighthood and the Oxford maths department is now housed in the Andrew Wiles Building. It was even rumoured that Gap asked him to endorse its range of menswear.
     
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    luxmen

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    මේක වැරදි

    From (1)---- c^n/2<a^n/2+b^n/2
    From(2) ----c^n/2<(a+b)^n/2
    So-----a^n/2+b^n/2=(a+b)^n/2 ----(3)


    eqn (3) සාදාරන අවස්ථාවක් නෙවෙයි.
    ----------------MAMA MEA LIYALA THIYENA DEYA GANA AVADANAYA YOMU KARANNA.----------INEQUALITY [2] COMES FROM INEQUALITY [1].SO LEFT SIDE IS EQUAL , RIGHT SIDE SHOULD BE EQUAL.
     

    luxmen

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    Ferma ge time eke Maths developed vela thibune naha Adha vage. Eka nisa Ferma ge proof eka godak sankirrna amaru proof ekak venna baha. Andrew Wilis vage pages 150 k long proof ekak venna baha kohethma, Andrew willis vage puka ira gena karapu proof ekak venna baha Ferma ge proof eka. MONAVA UNATH ANDREW WILLIES 1982 YEAR EKEDI LONG HARD PROOF EKAK DEELA ---PRIZE MONEY----EKA ARA GATTHA NEDHA?? MATA EMA PRIZE MONEY EKEN KOTASAK HARI DENAVANAM, MAMA ONAMA UNIVERSITY EKAKATA AVILLA ONAMA PROFESSOR KENEK ONAMA LOKKEK IDIRIYEDI MEA PROOF EKA PAHADILIVA IDIRIPATH KARANNAM, BUT THEY SHOULD GIVE ME A PART OF PRIZE MONEY.
     
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    luxmen

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    Mage blog eke proof eka balala thiyena aya samanya minissu nevene.Educated aya nea. US VAGEMA RUSSIA VALA AYATH BALALA THIYENAVA. RUSSIA VALA EKA DAVASATA 40 ,50 ,80 VIEWS THIYENA davas thibuna. LANKAVE AYA UNATH Maths university poraval mea elakiriye unath innavane.
     
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    ILSL

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    "Lowa kalambu ganitha gatalu" kiyala pothak thiyenawa machanla...mama eke oya prameya asurin haduna athuru gatalu godakata proof hadala thiyenawa...eka aran balanna kamathi aya...godak hoda pothak
     
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    mag123

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    ----------------MAMA MEA LIYALA THIYENA DEYA GANA AVADANAYA YOMU KARANNA.----------INEQUALITY [2] COMES FROM INEQUALITY [1].SO LEFT SIDE IS EQUAL , RIGHT SIDE SHOULD BE EQUAL.

    what you mean is because INEQUALITY [1] and INEQUALITY [2] left sides are equal therefore Right sides of the Inequalities' should be equal ? Is it ?
     

    luxmen

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    what you mean is because INEQUALITY [1] and INEQUALITY [2] left sides are equal therefore Right sides of the Inequalities' should be equal ? Is it ?
    inequality [ 2] comes from inequality [1]--------SEE INEQUALITY [1], WHEN n=2, ----c<a+b------so c^n/2<[a+b]^n/2----------------------------[a,b,c constant ,when n is variable]
     

    mag123

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  • Jan 20, 2008
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    what you mean is because INEQUALITY [1] and INEQUALITY [2] left sides are equal therefore Right sides of the Inequalities' should be equal ? Is it ?

    inequality [ 2] comes from inequality [1]--------SEE INEQUALITY [1], WHEN n=2, ----c<a+b------so c^n/2<[a+b]^n/2----------------------------[a,b,c constant ,when n is variable]

    but you did say before ;
    ----------------MAMA MEA LIYALA THIYENA DEYA GANA AVADANAYA YOMU KARANNA.----------INEQUALITY [2] COMES FROM INEQUALITY [1].SO LEFT SIDE IS EQUAL , RIGHT SIDE SHOULD BE EQUAL.

    Is it not ?




     
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    Sataninhell

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    ----------------MAMA MEA LIYALA THIYENA DEYA GANA AVADANAYA YOMU KARANNA.----------INEQUALITY [2] COMES FROM INEQUALITY [1].SO LEFT SIDE IS EQUAL , RIGHT SIDE SHOULD BE EQUAL.

    From (1)---- c^n/2<a^n/2+b^n/2
    From(2) ----c^n/2<(a+b)^n/2
    So-----a^n/2+b^n/2=(a+b)^n/2
    (a+b)^n/2 =a^n/2+b^n/2 ----------- (3)

    kohomada kiyane inequality form ekaka LHS samana una kiyala RHS samana wennama oni kiyala ??

    methanin ehata me sadanaya karana eke therumak naha. oya widiyata 3 equation eka liyanna bahane .

    oba n=2 awasthawata pamanak meya sadanaya karanawa nam , nawatha n sadaha wena agayan yodanna baha, mokada e awasthawa sathya wanne n=2 awasthawata pamanak wana nisa.
     
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    mag123

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    From (1)---- c^n/2<a^n/2+b^n/2
    From(2) ----c^n/2<(a+b)^n/2
    So-----a^n/2+b^n/2=(a+b)^n/2
    (a+b)^n/2 =a^n/2+b^n/2 ----------- (3)

    kohomada kiyane inequality form ekaka LHS samana una kiyala RHS samana wennama oni kiyala ??

    methanin ehata me sadanaya karana eke therumak naha. oya widiyata 3 equation eka liyanna bahane .

    oba n=2 awasthawata pamanak meya sadanaya karanawa nam , nawatha n sadaha wena agayan yodanna baha, mokada e awasthawa sathya wanne n=2 awasthawata pamanak wana nisa.

    That is true When you ask direct question from him , whether what he said was correct, he answered with something else .