Fermat's Last theorem short proof
in [Math]
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From: karzeddin on 15 Mar 2007 06:52 Gerry Myerson ÃÑÓáÊ: > In article > <11558709.1173924022873.JavaMail.jakarta(a)nitrogen.mathforum.org>, > Joseph Parranto <jparranto(a)yahoo.com> wrote: > > > Let me be very clear that I cannot comment on your formulas...I do not have > > enough skill there. > > If you search out the proofs of others, and actually read them, you will be > > able to tell if they are copies of yours. But only you can do that - no one > > else. WHY ? I'm I only the mathematician on this sphere? OR COULD IT BE THE MATHEMATICS IS ONLY ME! > > > > Wiles Proof can be downloaded from the publisher, and there are many > > commentaries on it and explanations from many sources. You have access to the > > whole world from your computer - time to use it to search. > > > > Case 1 was proved in 1910...and in 1930 the basis of Wiles proof was > > established but never finished. Your work is like that - unfinished = Not > > complete. > Are you sure? > Say, what? > > If by Case 1 you mean what used to be called The First Case of Fermat's > Last Theorem, that is, the case where the prime exponent does not divide > any of the bases, then I think you're wrong about it having been proved > in 1910. What's your source? > > As for your comment about 1930, I have no idea what you are talking > about. > I also wonder as Gerry Myerson where is your source please > -- > Gerry Myerson (gerry(a)maths.mq.edi.ai) (i -> u for email) today something happened to my computer, where I couldn't get into sci.math as usually, nor coud I google the words "fermat's last theorem short proof" try it your self if you don't belive. What is going on ?, can any body sincerely explain ?, My Regards B.Karzeddin
From: karzeddin on 15 Mar 2007 07:13 Roman B. Binder ÃÑÓáÊ: > > 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... > Actually your method resemble mine to a large degree, though you are adding many symbols, but the issue is the same, that is presenting FLT in another form to make it easy to catch by a cotradictory illogical logic, even to non-mathematicians, where many might have secretly got it. You can't also get a counter example. Best Regards B.Karzeddin > 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: Roman B. Binder on 15 Mar 2007 12:33 King Bassam Karzeddin : > Roman B. Binder أرسلت: > > > 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... > > > > Actually your method resemble mine to a large degree, > though you are > adding many symbols, but the issue is the same, that > is presenting FLT > in another form to make it easy to catch by a > cotradictory illogical > logic, even to non-mathematicians, where many might > have secretly got > it. > You can't also get a counter example. > > Best Regards > > B.Karzeddin > > 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 Hi, I used to be busy when You've comment my notes: When counterexamples so for n=3 they are written with my formula above. For bigger values of n there is problem for to find ideal number once Ext(n) = p^n or n^(nu-1) p^n but it could be possible... On the other hand You've not comment my P.S. Actually I have clear and almost elementary way for to proof for n=3 that: t^3 -2.3^u.abt -3^(3u-1).a^3 -b^3 = 0...........(*) can not be completed for proper natural values of a;b;t but I am just climbing up to bigger values and once it is too much fuss with FLT now I'll wait for to publish it in more clear circumstances. At the moment I'll show You why "u" should be bigger than 1: Into eq.(*) lets input: t = B + R we'll achieve: R^3+3bR^2-3b[2.3^(u-1) -b]R-2.3^u.ab^2-3^(3u-1).a^3 = 0 Now according to Eisenstein criterion such equation is not in the field of rational numbers for u=1 once the absolute therm Abs: Abs =-[2.3^u.ab^2 +3^(3u-1).a^3]= =-3^u.a[2.b^2 +3^(2u-1).a^2] and for u=1 will be divided only by 3 ... Best Regards Ro-Bin
From: Joseph Parranto on 15 Mar 2007 13:39 No
From: Joseph Parranto on 15 Mar 2007 13:59
source is http://www.mathpages.com/home/kmath367.htm It is a mis-read, however, as the comment does not say E. Dubois proposed a proof. 1930 was the start of ring theory.... "http://www-groups.dcs.st-and.ac.uk/~history/HistTopics/Ring_theory.html" If you have no idea...you have no idea. But if you want an idea, critique is not the route. Correction works... There are about 26,100 references from google on "Fermat Last Theorem 1910." There are about 47,100 references from Google on "Fermat Last Theorem 1930." They are readily available for all to see... |