Fermat's Last theorem short proof
in [Math]
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From: David Harden on 24 Mar 2007 16:37 On Mar 20, 5:58 am, bassam king karzeddin <bas...(a)ahu.edu.jo> wrote: > > On Feb 23, 2:13 pm, Dan Cass <d...(a)sjfc.edu> wrote: > > > > On Feb 21, 9:51 am, bassam king karzeddin > > > > <bas...(a)ahu.edu.jo> wrote: > > > > > Fermat's Last theorem short proof > > > > > > We have the following general equation (using > > the > > > > general binomial theorem) > > > > > > (x+y+z)^p =p* (x+y)*(x+z)*(y+z)*N(x,y,z) + > > > > x^p+y^p+z^p > > > > > > Where > > > > > N (x, y, z) is integer function in terms of (x, > > y, > > > > z) > > > > > Are you claiming this is true in general? > > > > > Counterexample: > > > > x=3, y=4, z=5, p=5. > > > > (x+y+z)^p = 248832 > > > > (x^p + y^p + z^p) = 4392 > > > > (x+y)(x+z)(y+z) = 560 > > > > Correction: (x+y)(x+z)(y+z) = 504, > > > and then (248832 - 4392)/504 is 485 = 5*97 > > > > Based on a few Maple calcs, I think the identity > > > (x+y+z)^p > > > =p*(x+y)*(x+z)*(y+z)*N(x,y,z) + x^p+y^p+z^p > > > gives an integer coefficient polynomial for > > N(x,y,z) > > > whenever p is odd and at least 3. > > > It works in general. A very easy proof uses the fact > > that > > Z[x,y,z] is a unique factorization domain: > > > Since p, x+y, x+z and y+z are distinct primes in this > > ring, showing > > D(x,y,z) = (x+y+z)^p - (x^p+y^p+z^p) is a multiple of > > their product > > just amounts to showing it is a multiple of each of > > them. > > > First of all, the binomial theorem definitely gives > > D(x,y,z) as a multiple of p in this ring. > > > Next, it suffices to show that D is a multiple of > > x+y: > > D is symmetric in x,y and z and therefore is a > > multiple of x+z and y+z > > if it is a multiple of x+y. > > > Write Z[x,y,z]= (Z[y,z])[x]. Since x+y is a monic > > polynomial in x, > > division by x+y can be applied to write D(x,y,z) = > > Q(x,y,z)*(x+y) + > > R(y,z). > > We need to show that R(y,z)=0. > > Since R(y,z) is independent of x, it suffices to show > > that R(y,z)=0 > > when x = -y: > > R(y,z) = D(-y,y,z) = (-y + y + z)^p - ((-y)^p + y^p + > > z^p) = > > -(1+(-1)^p)y^p. > > Since p is odd, this last expression equals 0. > > > ---- David > > > To send me email, move the r from the beginning to > > the end of the part > > before the @ and insert "alum." at the beginning of > > the part after the > > @. > > And finally comes > > David Harden with a very simple proof in the language mathematicians understand, and put it bravely on the net-against the rules of JOURNALS. where he is not plying a game... > > but no comments from the experts or big heads!! Um... I just filled in a gap most people agreed could be filled in easily. I did not complete a proof of Fermat's Last Theorem. You do not have one. You do not come close to having one, as far as I or anyone else can see. You seem to have proven, as Randy Poe says, that FLT is true for p=3 when x,y,z are nonmultiples of 3. PLEASE explain the rest of your logic. Don't put words in my mouth as you have; these taste really bad! > > Did you surrender? > then rise up the White Flag. I will not raise it and therefore it will not rise here. ---- David
From: Roman B. Binder on 27 Mar 2007 12:58 > > In article > > > <1174322686.872938.90260(a)l75g2000hse.googlegroups.com> > > > , > > "Randy Poe" <poespam-trap(a)yahoo.com> wrote: > > > > > On Mar 19, 6:22 am, bassam king karzeddin > > <bas...(a)ahu.edu.jo> wrote: > > > > Dear All > > > > > > > > I don't know if this question had been asked > > before ! > > > > > > > > Is Infinity is odd or even or WHAT? > > > > > > Neither odd nor even. Those are properties of > > integers, > > > and "infinity" is not an integer. > > > > Never even is a palindrome. > > oo is a palindrome. > > Therefore, oo is never even. > > > > -- > > Gerry Myerson (gerry(a)maths.mq.edi.ai) (i -> u for > > email) > > Hi Gerry Myerson > > It will take me too long to understand this paindrome > concept. > > My Regards > B.Karzeddin In contrary it is so easy as between 0 and 1 there are infinite many real numbers so what chance is for infinity to be some natural number ? Regards Ro-Bin
From: bassam king karzeddin on 1 Apr 2007 09:02 Dear All This is about The Trinomial equation introduction solution I have presented in this Thread x^n +a* x^m +b = 0 Where, I shall mention the names of the persons that provided me with their replies I also went through a copy that was submitted by me to the Third World Academy of Sciences (TWAS) prize for 1994, in cooperation with the Royal Scientific Society (RSS), in JORDAN, their reference (7) 253/39/3/19177 date Oct 30, 1994, and (7) 253/39/3/19743 dated 6/11/1994, Signed by Dr. Hani Mulki, President Where my formula previously stated was proved and derived with very elementary methods, Here are also some of the reputable Journals replies to me about this issue: Journal of Algebra, Dept. of Math. Yale University, their replies dated (Jan. 16, 1986, and July 25, 1990) Signed by Dr. Walter Feit, Editor in chief Monash University, Dept. of Math. Australia, their reply dated 25 October 1990 Signed by Dr. Michael A. B. Deakin Cambridge University Press, New York, their replies to me dated (7 and 29), May 1990 Signed by Dr. Nancy A. Selzer, Editorial Assistant Bulletin of the Australian Mathematical Society, their reply dated 20th,July, 1990, paper number 0727 Signed by Dr. Alan S. Jones, Editor American Journal of Mathematics, The Johns Hopkins University, Their reply dated, June 8, 1990 Signed by Dr. Jun. Ichi. Igusa, Editor New York University, Courant Institute of Mathematical Sciences, their reply dated April 25, 1990 Signed by Dr. Will Klump, Executive Editor The University of Western Australia, Nedlands, Dept. of Math. Their reply dated 12, June 1990 Signed by Dr. Alistair Mees, Head of Department School of Mathematics, University College of North Wales, Bangor, UK, their reply dated 10/4/1990 Signed by Professor R.Brown Washington State University, Dept. of Pure and Applied Mathematics, there reply dated April 13, 1990 Signed by Professor Jack Robertson, Editor, Mathematics Notes The Australian National University, their reply dated 6, June 1990 Signed by Dr R. A. Bryce The American Mathematical Monthly, their reply dated, May 2, 1990 Signed by Dr. Paul T. Bateman Quarterly Journal of Mathematics, Oxford University Press, Mathematical Institute, their reply dated 5/4/1990 Signed by the Editors in hand writing without names Abd al Hameid shoman establishment, JORDAN, Their reply dated 28/3/95, Signed by Dr. Asaad Abd al rahman There is also interesting reply from Monash University in the year 2001, I will tell you about it later Signed by Dr. Michael A. B. Deakin But, unfortunately, it seems that most of them could NOT understand it. Any way they have expressed many thanks to me since it wasn't suitable for them ALL. And OF COURSE without mentioning any reason. My Best Regards Bassam Karzeddin Al-Hussein Bin Talal University JORDAN
From: bassam king karzeddin on 2 Apr 2007 02:47 Dear All I'm afraid that Wiles Proof of FLT is not a proof! Do you know why? I guess because (all or most) of you couldn't understand it, beside the lots of talks and attempts by the so many mathematicians to get the proof easily even after the proof was been announced, as if it was not a proof, but any way authorized proof is a proof, isn't it? I shall also design a mathematical bombs whenever I get free time, and throw them here in this thread, sooner or later they will blast, and hit some of those stony heads, and those who have already booked a nice seats in the history, not only that but also will shake and awake those who had been sleeping peacefully in the history, and you will have no choice, because mathematics is not mathematicians, in fact it is anti mathematicians, or it is a revolution for ever until you are balanced with its most beautiful operation "=", that had been neglected for so long. My Regards Bassam Karzeddin
From: Tonico on 2 Apr 2007 09:25
On 2 abr, 13:47, bassam king karzeddin <bas...(a)ahu.edu.jo> wrote: > Dear All > > I'm afraid that Wiles Proof of FLT is not a proof! > Do you know why? > I guess because (all or most) of you couldn't understand it, beside the lots of talks and attempts by the so many mathematicians to get the proof easily even after the proof was been announced, as if it was not a proof, but any way authorized proof is a proof, isn't it? > > I shall also design a mathematical bombs whenever I get free time, and throw them here in this thread, sooner or later they will blast, and hit some of those stony heads, and those who have already booked a nice seats in the history, not only that but also will shake and awake those who had been sleeping peacefully in the history, > > and you will have no choice, because mathematics is not mathematicians, in fact it is anti mathematicians, or it is a revolution for ever until you are balanced with its most beautiful operation "=", that had been neglected for so long. > > My Regards > > Bassam Karzeddin **************************************************** An arabian sibling of JHS, uh? Ok, enjoy the family...:>) Tonio |