Fermat's Last theorem short proof
in [Math]
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From: bassam king karzeddin on 11 Mar 2007 06:12 > Trust me when I say that I really have little to > offer in terms of verifying your math. I will take > the formulas and convert them into a readable fashion > for interest, but from the outset one of your terms > immediately causes problems - and that is that there > is "an infinite sum" attached and that alone makes it > not a proof. It must be self-contained and computable > across the entire set of integers, not just the few > you have here. So every time you say that it is too > much work for you, imagine how little interest there > is from anyone else. > > Beal's Conjecture is on the Internet and you should > really take a shot at it - your method may work but I > cannot see that this is the correct f(x) > Joe Hello Joseph Parranto So, I have explained to you my formula, Could you please convert them into a readable fashion as you did promise , About the infinite sum, it is the main point. My Regards B.Karzeddin
From: bassam king karzeddin on 11 Mar 2007 06:18 > > 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 > t(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 > rs .........(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 > 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 Dear Roman B.Bider I will see if I can make use of your remarkable reply My Regards B.Karzeddin
From: bassam king karzeddin on 11 Mar 2007 06:29 > 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... I really didn't see Wile's proof, nor do I have intention since I'm not a profissional mathematician, nor I do have time. but is there any mention to the trinomial equation in his proof.? If so, is there any free web bage that shows the proof. Thanks B.Karzeddin > > 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: Joseph Parranto on 12 Mar 2007 13:21 Yes to both. The best is to find the original announcements in 199x. This forum has some emails from Wiles and others who give shorter explanations. It took Wiles 200 pages to do what you are trying to do in one line. Joe
From: Joseph Parranto on 12 Mar 2007 13:30
Sure... Case 1 has been known for about 100 years before you were born (the website)-- so you might consider that you are in good company and that you are taking credit for what rightfully belongs to others... Case 2 was not so easy and Wiles had to go out of his way to "create" the math to solve it - so unless you can show how you had it in about 1985, you are not likely to get much credit. Joe |