Fermat's Last theorem short proof
in [Math]
|
From: bassam king karzeddin on 1 Mar 2007 09:22 > On 28 Feb, 07:06, bassam king karzeddin > <bas...(a)ahu.edu.jo> wrote: > > Hello Dear > > Hi, Doll > > > Now, I realised that somthing WRONG happeining > world wise in the holly > science-MATHEMATICS, AND, > PYTHAGOURS, FERMAT, ...,ARE ALL very , very, > VERY > ANGRY > > Well, prickly, at least. > Hello LOL You are the only one that is left with out an answer I wish to understand what do you really mean?? And what would you like to add?????? In your scale I know I loose the game, and all I have DONE in VAIN but it will be soon going to RAIN. Regards B.Karzeddin
From: Randy Poe on 1 Mar 2007 21:39 On Mar 1, 6:57 pm, 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) > > > > 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 > > > Why? > > > > (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), > > > Why? > > > If x = 4, y = 5, z = 7, for instance, then > > > x + y is divisible by 3, which is relatively prime to > > xyz. > > > > So > > > Not 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 > > > Dah~~~~! > > > Why are there so many replies to this simple argument! > > Hi H.S > > I got where you and may be Randy are lost Not lost. > > You assume an already a known counter example, No, I don't. But you started with "suppose x, y and z satisfy x^3 + y^3 + z^3 - 0". Hence, like you, I am starting with the assumption that x^3 + y^3 + z^3 = 0 and seeing where it leads. > but does that satisfies the equn.??, of course not, what I simply state that > if, x^3 + y^3 +z^3 = 0, implies the following eqn. > (x+y+z)^3 = (x+y)*(x+z)*(y+z), > > which of course not possible, as was proved by me directly, What you further assumed was that 3 does not divide xyz. You proved that under that assumption the above equation is not possible. Therefore this second assumption can not be true if x^3 + y^3 + z^3 = 0, which means that x^3 + y^3 + z^3 = 0 implies 3 divides xyz. It's your proof and your assumptions. Why are you balking at your own proof now? > What I have really done is changing the original eqn.form, that makes it a trillion time easier to handle by a number theorists, one already got it and run away-Das Gas-so dont let run away, and I really wanted you all to get it Um, no. I don't think number theorists would find "x^3 + y^3 + z^3 = 0" to be a trillion times easier to solve than x^3 + y^3 = z^3, which is exactly the same equation except for a change of variables from z to -z. Changing z to -z is not earth-shaking mathematics. - Randy
From: bassam king karzeddin on 1 Mar 2007 23:05 > On Feb 28, 10:27 am, bassam king karzeddin > <bas...(a)ahu.edu.jo> wrote: > > > On Feb 28, 4:32 am, bassam king karzeddin > > > <bas...(a)ahu.edu.jo> wrote: > > > > Dear Hisanobu Shinya > > > > > > HOW MANY TIMES SHOULD I REPEAT THAT I HAVE > PROVED > > > IT COMPLETLY IN MY POSTS HERE ONLY IN THIS THREAD > > > > > Repeating the claim is not a substitute for the > > > proof. > > > > > > Anad a part from other one 16 years back even > > > before I knew about it > > > > > > That doesn't mean I'm the only one, there may > be > > > many others, as yours I suppose > > > > > > The ISSUE is THERE IS SOMETHING WRONG WITH > > > MATHEMATICIANS > > > > IN POWER, AND THEY HAVE TO GO FOR SOMETHING > > > ELSE.... > > > > > No, all mathematicians require to believe a proof > > > exists > > > is to see that proof and verify it. You haven't > > > provided > > > a proof yet. > > > > > All you've provided is a proof of a result which > was > > > known long ago, namely that for any solution of > > > x^3 + y^3 = z^3, one of x, y, or z must be > divisible > > > by 3. > > > > Do you mean this is EXACTLY the only remaining > case, > > What case? > > I mean what I said. That x^3 + y^3 = z^3 implies one > of x, y or z is divisible by 3. > > FLT says "there are no integer solutions to x^n + y^n > = > z^n for n>2". > > Since you have not established this for any n at all, > including n=3, then I don't know what you mean by > "only remaining case". > > The "only remaining cases" in your proof that you > have > failed to address are n=3, n=4, n=5, n=6, n=7, ... > > Proofs for n=3 have been around a long time. Didn't > Fermat have one? > > But you DON'T have one. At least, not one that you've > shown yet. > > - Randy > Hello Randy Did you really think about the following Integer equn. A^p = p*B +C, as a hint - assuming (C=0),then you get A^P= P* B, Hence p must be a prime factor of A, other wise the eqn. can't have a solution, therefore a proof for the case when p doesn't divide xyz, 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 I know, you want Why gcd(p, N) = 1, IS THAT SO. If so, SEE Dr. Winter replies, or try it your self or wait It is not the main issue, isn't it ? and what about when p is not a factor of (x+y+z), which I PROVED IMPOSSIBLE REGIOUSLY The the main thing is that, there are very tiny ideas that can't be visible, exactly as an atom Integer-1 = Integer-2 + Integer-3, See, how many thousands of years it took this kind (only one man - First) to note the most important theorem that lies behind numbers, that are given to us in one hand only and making the five fingers And can you predict how many trillions of years when the first monkey will come to know about it alone Please Don't missunderstand me, (I do respect all), And I know that I have to explaine something else My Regards B.Karzeddin Al-Hussein Bin Talal University JORDAN
From: bassam king karzeddin on 1 Mar 2007 23:27 Google the words now please "Fermat's last theorem short proof" "Fermat's last secret" "Angle division" ...... ...... And tell me please, Where are the others Where are the journals Didn't I really invade you, but to the light and freedom ISN'T GOOGLE OF OUR KIND?? ISN'T THIS MAKING THE HISTORY?? My Keen Regards B.Karzeddin
From: Randy Poe on 2 Mar 2007 10:15
On Mar 2, 9:05 am, bassam king karzeddin <bas...(a)ahu.edu.jo> wrote: > > On Feb 28, 10:27 am, bassam king karzeddin > > <bas...(a)ahu.edu.jo> wrote: > > > > On Feb 28, 4:32 am, bassam king karzeddin > > > > <bas...(a)ahu.edu.jo> wrote: > > > > > Dear Hisanobu Shinya > > > > > > HOW MANY TIMES SHOULD I REPEAT THAT I HAVE > > PROVED > > > > IT COMPLETLY IN MY POSTS HERE ONLY IN THIS THREAD > > > > > Repeating the claim is not a substitute for the > > > > proof. > > > > > > Anad a part from other one 16 years back even > > > > before I knew about it > > > > > > That doesn't mean I'm the only one, there may > > be > > > > many others, as yours I suppose > > > > > > The ISSUE is THERE IS SOMETHING WRONG WITH > > > > MATHEMATICIANS > > > > > IN POWER, AND THEY HAVE TO GO FOR SOMETHING > > > > ELSE.... > > > > > No, all mathematicians require to believe a proof > > > > exists > > > > is to see that proof and verify it. You haven't > > > > provided > > > > a proof yet. > > > > > All you've provided is a proof of a result which > > was > > > > known long ago, namely that for any solution of > > > > x^3 + y^3 = z^3, one of x, y, or z must be > > divisible > > > > by 3. > > > > Do you mean this is EXACTLY the only remaining > > case, > > > What case? > > > I mean what I said. That x^3 + y^3 = z^3 implies one > > of x, y or z is divisible by 3. > > > FLT says "there are no integer solutions to x^n + y^n > > = > > z^n for n>2". > > > Since you have not established this for any n at all, > > including n=3, then I don't know what you mean by > > "only remaining case". > > > The "only remaining cases" in your proof that you > > have > > failed to address are n=3, n=4, n=5, n=6, n=7, ... > > > Proofs for n=3 have been around a long time. Didn't > > Fermat have one? > > > But you DON'T have one. At least, not one that you've > > shown yet. > > Did you really think about the following Integer equn. Yes, I did. That's why I accept your proof that x^3 + y^3 + z^3 = 0 implies 3 divides xyz. > A^p = p*B +C, as a hint - assuming (C=0),then you get > > A^P= P* B, Hence p must be a prime factor of A, Yes. > other wise the eqn. can't have a solution, therefore a proof for > the case when p doesn't divide xyz, where Yes, this is why I'm saying "if you also assume 3 doesn't divide xyz..." > > A = x+y+z > B = (x+y)*(x+z)*(y+z)*N(x,y,z) > C = x^p + y^p + z^p, I don't know that you've convinced me of this result for general p. You claim it's true, and you claim it follows from the binomial theorem. Yet for some reason you won't walk through the argument. But let's say it's true. (That now makes 3 assumptions we have, that we are examining the consequences of... (1) Assume x, y, z solve x^p + y^p + z^p = 0. (2) Assume p does not divide xyz. (3) Assume (x+y+z)^p = p*B + C, where B and C are as given above. You just finished giving all three of these as assumptions. Why do you disagree when I say these are your assumptions? > and (x,y,z) are three (non-zero), distinct coprime integers Given those three assumptions, we have this: (x+y+z)^p = p*(x+y)(x+z)(y+z)*N(x,y,z) Right? That is a consequence of those three assumptions, and not in general true if those three assumptions are not true, OK? We're seeing where your 3 assumptions lead. Now the argument you omitted that I filled in is this: p must divide (x+y+z)^p, which means that p divides (x+y+z) and p^p divides (x+y+z). So we have x+y+z = p*k for some integer k x+y = p*k - z p does not divide x, y or z by assumption (2). Therefore it does not divide (x+y) = p*k - z, and similarly it does not divide (x+z) = p*m - y, or (y+z) = p*n - x. We know p^p divides the left hand side. Therefore p^p must divide the right hand side, which means p^(p-1) must divide N(x,y,z). Now here is where you failed to consider the general case. For p=3, we have N(x,y,z) = 1, and the three assumptions imply that 3^2 must divide 1. Clearly it does not. Hence the THREE ASSUMPTIONS lead to a contradiction. These three assumptions can not all be true for p=3. We know assumption (3) is true. Hence if (1) is true, (2) must be false. > > I know, you want Why gcd(p, N) = 1, IS THAT SO. > > If so, SEE Dr. Winter replies, or try it your self or wait > > It is not the main issue, isn't it ? I don't understand why you're arguing with me. All I'm doing is repeating your own proof back to you, unchanged. Are you arguing with yourself? What are you disagreeing with? - Randy |