Fermat's Last theorem short proof
in [Math]
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From: bassam king karzeddin on 4 Mar 2007 23:34 Dear All Another Hint Given an arbitrary triangle with the following integeral sides : m^3,m*(n^2-m^2),n*(n^2-2*m^2) where : 4*m^2 >= n^2 > 2*m^2 >= 2 will give you a triangle with one of its angles is trisected EXACTLY in the same triangle. My Regards B.Karzeddin Al Hussein Bin Talal University JORDAN
From: bassam king karzeddin on 5 Mar 2007 00:18 Dear All This is about The Trinomial equation x^n + x^m +1 = 0 Yes make sure it works only for all odd integers, and was proved by me many years back, around 1990 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, where my formula was proved and derived with very elementary methods Here are also some of the reputable Journals replies references about this issue: Journal of Algebra, Dept. of Math. Yale University, their replies dated (Jan. 16, 1986, and July 25, 1990) Monash University, Dept. of Math. Australia, their reply dated 25 October 1990 Cambridge University Press, New York, their replies to me dated (7 and 29), May 1990 Bulletin of the Australian Mathematical Society, their reply dated 20th,July, 1990 American Journal of Mathematics, The Johns Hopkins University, Their reply dated, June 8, 1990 New York University, Courant Institute of Mathematical Sciences, their reply dated April 25, 1990 The University of Western Australia, Nedlands, Dept. of Math. their reply dated 12, June 1990 School of Mathematics, University College of North Wales, Bangor, UK, their reply dated 10/4/1990 Washington State University, Dept. of Pure and Applied Mathematics, there reply dated April 13, 1990 The Australian National University, their reply dated 6, June 1990 The American Mathematical Monthly, their reply dated, May 2, 1990 Quarterly Journal of Mathematics, Oxford University Press, Mathematical Institute, their reply dated 5/4/1990 ????? ??? ?????? ?????- ??????, ?? ?????? 28/3/1995, ???? ??? : ? ?/8/471/95 There is also interesting reply from Monash University in the year 2001, I will tell you about it later But, unfortainatly, It seems that non of them could understand it. Any way they have expressed many thanks to me since it wasn't suitable for them ALL. and without any reason. Yes, there are -only- infinetly many solvable quintic equations forms, check if I can remember this form x^5 + (3*a^2*b - b^2 -a^4)*x + a*b*(2*b -a^2) = 0 Where, (a & b) belong to C, complex numbers My Best Regards Bassam Karzeddin Al-Hussein Bin Talal University JORDAN
From: bassam king karzeddin on 5 Mar 2007 05:39 Dear All Did you really think about Little Fermat's theorem? A^p = p*B +C, as a hint - assuming (C=0), then you get A^P= P* B, where A = x+y+z B = (x+y)*(x+z)*(y+z)*N (x, y, z) C = x^p + y^p + z^p And (x, y, z) are three (non-zero), distinct coprime integers Now, you can see it, clearly, WHY (A^p = p*B), does not have any solution So , Fermat had shown us the part (C) of his eqn. and kept hidden A & B the complete picture of his original little theorem, may be for a reason, because once others see it, then immediately all his work is vain, and may be he was waiting them to admit his talent in open and ask him , but again they won't because it is a matter of dignity and they have never learnt yet, that Mathematics is mainly by BIRTH Of course, he knew that A^P = C Mod p , and C = A Mod p Then he asked himself the following question, as an argument to complete the picture, is it true that holds in any case if (p (is or is not) a prime factor of A, then A^P = (A Mod p) Mod p must hold Which of course can not hold in both cases, and I hoop the proof is complete now, but the Question is open now, will you accept it , especially it is free of cost and not from JOURNALS Can You Dare to say the TRUTH , AND THAT IS A COURAGE, I have already tried Journals many years back, before, but not in this issue because I never heard about it (FLT), before I started posting here in 2004, , and it seems that will not work, they may be looking to the longest ever proof found in order to make more money My Regards Bassam.Karzeddin Al-Hussein Bin Talal University JORDAN
From: Joseph Parranto on 5 Mar 2007 20:41 I think you will find yourself between a rock and a hard place following a course that insists on "factors-in-common." The rock is FLT and the hard place is The Beal Conjecture: "(A^x) + (B^y) = (C^z) must have factors in common." So far it has defied proof, but those that have been found do satisfy the Conjecture. Having taken this route to a proof, you now have to dispute one to prove the other. This is a true "double-bind" that is not therapeutic. It is far more likely that you may prove Beal - and if you do, there is a pay-off - he is still alive and has posted a reward for a proof that is accepted as true. Good Luck. Joe
From: bassam king karzeddin on 5 Mar 2007 23:45
> I think you will find yourself between a rock and a > hard place following a course that insists on > "factors-in-common." The rock is FLT and the hard > place is The Beal Conjecture: "(A^x) + (B^y) = (C^z) > must have factors in common." So far it has defied > proof, but those that have been found do satisfy the > Conjecture. Having taken this route to a proof, you > now have to dispute one to prove the other. This is a > true "double-bind" that is not therapeutic. It is far > more likely that you may prove Beal - and if you do, > there is a pay-off - he is still alive and has posted > a reward for a proof that is accepted as true. Good > Luck. > > Joe My formula is the following Trinomial Equation Many years back, that was even before (1990), I have derived the following solution, one real root for any odd degree of the Trinomial Equation that was written in a book- with a copy right-from Ministry of culture at JORDAN, Solution of equations by power series and submitted three copies to the Royal Scientific Society (RSS) in JORDAN, with cooperation with the Third World Academy of Science (TWAS), in 1994. The same formula I have provided the Journals I have already mentioned in 1990, but can they tell me what did they do about it, did they really understand it?? Here is my introduction solution to the trinomial equation, I hope time will help me to show you the importance of this equation to some unsolved problems (including The Peal's Conjecture as I do realize now), and how this can be converted to a closed form solution in many cases such as fifth degree equation, so you may try, but I will not, because I don't posses a common language with JOURNALS, and I really wish to finish their RULES. So, try and good luck Let f (x) be a rational integral function of x, (n, m) are two odd positive integers such that (n > m), and (a, b) are two non-zero rational numbers such that f (x) = x^n +a*x^m + b = 0 Then, at least One real root can be easily obtained as a direct application of my following formula, that is too simple to program. SOLUTION If, b^(n-m) / (abs (a))^n < (m^m)*(n-m)^(n-m) / n^n, then x = - (b/a)^(1/m) sum_{i=0}^infty u_i, Where (u_0 = 1), (u_1 = - r / m), and r = (b^(n-m) / a^n)^(1 / m) And for (i > 1) u_i = ((-r)^i / i! *m^i)product_{j=1}^{i-1)(i*n - j*m +1) Similarly If b^(n-m) / (abs (a))^n > (m^m)*(n-m)^(n-m) / n^n, then x = - (b)^(1/n) sum_{i=0}^infty u_i, Where (u_0 = 1), (u_1 = - r / n), and r = (a^n / b^(n-m))^(1 / n) And for (i > 1) u_i = ((-r)^i / i! n^i product_{j=1}^{i-1) (i*m - j*n +1) I wont show you also how all coefficients of a general finite degree equation would be dancing for a root, because it is too much to tolerate Thanking you for more information My Regards Bassam Karzeddin Al Hussein Bin Talal University JORDAN |