Fermat's Last theorem short proof
in [Math]
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From: maruf.syfullah on 8 Mar 2007 14:32 brother karzeddin please do everything as fast as possible to reveal your proof of FLT .we are eagerly waiting for a proof from u. best of lucks MARUF SYFULLAH DEPT: CSE L / T:4/2 BUET (BANGLADESH UNIVERSITY OF ENGINEERING AND TECHNOLOGY) DHAKA ,BANGLADESH
From: Josseph on 8 Mar 2007 10:22 I can see why you are so frustrated with your readers - You do not have a "Proof" but you do have an interesting "algorithm." The problem is language and use of terms. In English "better method" is what you have, but that is not a "proof = no doubt of outcome" without all the calculations written down -- completely. Keep in mind that your F(x) does not vary much from Wiles' proof. Converting "c^z" to "b" is so minor a change... And you cannot use this to "prove" Beal's Conjecture either - x is a common factor at the outset. If you start at the end, you only get back to the beginning. You cannot "prove" what you are trying to prove by using the thing you are trying to prove. That is "Circular logic" and that is what you do have - in abundance. Sorry... Joe
From: Roman B. Binder on 9 Mar 2007 04:47 > 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) > > P is odd prime number > (x, y, z) are three (none zero) co prime integers? > > Assuming a counter example (x, y, z) exists such that > (x^p+y^p+z^p=0) > > > (x+y+z)^p =p* (x+y)*(x+z)*(y+z)*N (x, y, z) > > CASE-1 > If (p=3) implies N (x, y, z) = 1, so we have > > (x+y+z)^3 =3* (x+y)*(x+z)*(y+z) > > Assuming (3) does not divide (x*y*z), then it does > not divide (x+y)*(x+z)*(y+z), > So the above equation does not have solution > (That is by dividing both sides by 3, you get 9 times > an integer equal to an integer which is not divisible > by 3, which of course is impossible > I think proof is completed for (p=3, and 3 is not a > factor of (x*y*z) > > My question to the specialist, is my proof a new one, > more over I will not feel strange if this was known > few centuries back > > Thanking you a lot > > Bassam King Karzeddin > Al-Hussein Bin Talal University > JORDAN Hi, I am not so big specialist but for this case of proof I'd like to point for to use of following specific parameters: once x,y,z are defined as integers so for the troth of x^n + y^n + z^n = 0 (where n odd prime numbers) let x=X; y=Y; z=-Z where X;Y;Z are natural numbers and of gcd = 1 now for X^n + Y^n -Z^n = 0 .......................(1) we can use following TAB input: X = T + B; Y = T + A; Z = T + A + B; Now eq.(1) will be simplified into: T^n - nAB(2T+A+B)Ext(n) = 0 ..............(2) where for n=3 Ext(3) = 1 for n=5 Ext(5) = 2T^2 + 2(A+B)T +A^2 +AB +B^2 and generally Ext(n) = F(T,A,B) once for n=3 T^3 = 3AB(2T+A+B) and 2T+A+B = X+Y so we'll find: T=3^u abt always once let X not divided by 3 so B = b^3; then for Y divided by 3 A = 3^(3u-1) a^3; so Z not divided by 3: 2T+A+B = t^3; or for Y not divided by 3 A = a^3; but then Z divided by 3: 2T+A+B = 3^(3u-1) t^3; Now Your case for n=3: (x+y+z)^3 = 3(x+y)(x+z)(y+z) with integers .........(3) once using proper signs for natural values X;Y;Z : Ls = (X+Y-Z)^3 = 3(X+Y)(Z-X)(Z-Y) = Rs then once X+Y-Z = 2T+A+B-T-A-B = T = 3^u abt Z-X = T+A+B-T-B = A Z-Y = T+A+B-T-A = B and now very easy we'll complete: Ls = 3^3u a^3 b^3 t^3 = Rs for X+Y = t^3; Z-X = 3^(3u-1) a^3; Z-Y = b^3; or for X+Y = 3^(3u-1) t^3; Z-X = a^3; Z-Y = b^3 Therefore there are infinite amount of such sets of numbers fulfilling questionable by You eq.(3) For bigger values of n I used to express: Ext(n) = p^n or Ext(n) = n^(nu-1) p^n Therefore according to eq.(2) T = n^u abtp always once Z-Y = B = b^n (let X be the number not / by n) but appropriate:Z-X = A = a^n or A = n^(nu-1) a^n X+Y = t^n or X+Y = n^(nu-1) t^n and Ext(n)= n^(nu-1) p^n or Ext(n) = p^n now in Your general equation shaped with natural values: (X+Y-Z)^n = n(X+Y)(Z-X)(Z-Y)N(n) we can recognize: X+Y-Z = T = n^u abtp; X+Y = t^ n or n^(nu-1) t^n; Z-X = a^n or n^(nu-1) a^n; Z-Y = b^n or n^(nu-1) b^n; N(n) = p^n or n^(nu-1) p^n; Such values fulfilling eq.(2) and only problem resists is to claim, that it is not possible for all of them as natural values: a;b;t;p. ( u is some natural value for sure and even it is to take u>=2 once for u=1 could be used Eisenstein criterion ) I will not present specific values of numbers as counterexamples but with the help of my formulas and little bit more time You can find so much as You wish... With the Best Regards Ro-Bin P.S. On the other hand I can imagine, that You have some method for to solve equations of the shape like ax^n + bx + c = 0 how it could be applicated to eq.: x^3 -2*3^u abx - 3^(3u-1) a^3 - b^3 = 0 and also in to case for to retrieve on of x roots as natural number for some proper and natural a;b;3 of gcd =1
From: bassam king karzeddin on 11 Mar 2007 05:55 > brother karzeddin > > please do everything as fast as possible to reveal > your proof of > FLT .we are eagerly waiting for a proof from u. > > > best of lucks > > MARUF SYFULLAH > DEPT: CSE > L / T:4/2 > BUET (BANGLADESH UNIVERSITY OF ENGINEERING AND > TECHNOLOGY) > DHAKA ,BANGLADESH > Dear Maruf http://www.mathpages.com/home/kmath367.htm I was surprised today to find the main ideas of my issue at the above link, And there was no mean to contact the math world to ask about it, therefore I will be grateful for some one who can explain to me what is going on, however it seems that the issue date was created after my posting this thread, unless I miss something, or may be Im the most ignorant person on this planet or may be I have to learn much, especially I think I have an idea that is the last nail to this issue, but how can you do that if every thing is simply displayed by others as their own work, OR should we READ every thing that was written about the subject which may require thousand years. I'm really confused. And in this case why should I post any thing I derive with lots of pain. Again, can any body help sincerely explain please? Thanking you B.Karzeddin
From: bassam king karzeddin on 11 Mar 2007 06:02
Dear All http://www.mathpages.com/home/kmath367.htm I was surprised today to find the main ideas of my issue at the above link, And there was no mean to contact the math world to ask about it, therefore I will be grateful for some one who can explain to me what is going on, however it seems that the issue date was created after my posting this thread, unless I miss something, or may be I'm the most ignorant person on this planet or may be I have to learn much, especially I think I have an idea that is the last nail to this issue, but how can you do that if every thing is simply displayed by others as their own work, OR should we READ every thing that was written about the subject which may require thousand years. I'm really confused. And in this case why should I post any thing I derive with lots of pain. Again, can any body help or sincerely explain please? Thanking you B.Karzeddin |