Fermat's Last theorem short proof
in [Math]
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From: bassam king karzeddin on 2 Mar 2007 05:22 > 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 > O.K Randy Here, is the same previous equation but in positive (non zero) distinct coprime integers (x<y<z) Then it implies that, the following integer eqn. doesn't have any solun.in positive integers (x+y-z)^p = p*(x+y)*(z-x)*(z-y)*N(x,y,z) Where, p is odd prime number, N(x,y,z) is odd integer function in terms of (x,y,z) I shall give another BIG hint to you all (x,y,z) forms a triangle where The largest angle Is between (PI/2, PI/3), where (PI) is 180 degrees only one integer of (x,y,z) is even x^2 + y^2 > z^2 > x^2 + y^2 - x*y I wish I could offer a prize for that What is my aim is to make the whole world know about such a scandle My Regards B.Karzeddin
From: bassam king karzeddin on 3 Mar 2007 04:40 Dear All Think about the following equn.- Series soln. as another hint for any function f(x) over the real variable x, of the following form: f(x) = x^n + x^m + 1 = 0 Where, (n&m) are two odd positive distinct integers,and (n>m) x = -[1-(1/n)+(2m-n+1)/(2!*n^2) -(3m-2n+1)*(3m-n+1)/(3!*n^3) +(4m-3n+1)*(4m-2n+1)*(4m-n+1)/(4!*n^4) -(5m-n+1)*(5m-2n+1)*(5m-3n+1)*(5m-4n+1)/(5!*n^5)+...] The general term is SO obvious Regards B.Karzeddin Al Hussein Bin Talal University JORDAN
From: bassam king karzeddin on 3 Mar 2007 23:27 Dear All Think about the following Trinomial Equation - Series solution. As another hint For any function f (x) over the real variable x, of the following form: f (x) = x^n + x^m + 1 = 0 Where, (n & m) are two odd positive distinct integers, and (n>m) x = -[1-(1/n)+(2m-n+1)/(2!*n^2) -(3m-2n+1)*(3m-n+1)/(3!*n^3) +(4m-3n+1)*(4m-2n+1)*(4m-n+1)/(4!*n^4) -(5m-n+1)*(5m-2n+1)*(5m-3n+1)*(5m-4n+1)/(5!*n^5)+...] The general term is SO obvious then Is that too disgusting, or brain blasting, I know most of you are in love with the quadratic equation mainly, and you keep rotating about it for thousands of years. Are you ashamed to only confirm the result which many of you surely did, but in secret, Some others are much more dangerous they may steal it and present it in their work in a slightly different forms, and make their success When he saw it, and that is courage yes Dr.Robert one day assured it My words are directed towords true mathematicians which I will depend on, and others are not allowed, with my respect to ALL. If you see some thing you most likely don't understand, then you should ask your teachers, and if they still cant understand it, then they have to ask their Brain masters about it, and again if your Brain masters or JOURNALS still can't answer, then you have to go to the one who originated the problem, That is how we can protect each others ideas, and you can protect and save your children and coming generations from puzzles that WILL COST THEM A LOT. But, if you are lazy, careless, so ignore it. My question now IS OPEN for the JOURNALS, who SAW my formula in 1991,yes in 1991, and wasn't suitable for them , which I'm going to reveal their names soon. From their replies that I do HAVE, with signatures and exact dates DO YOU KNOW THAT IS ONLY A REMIDY TO YOUR RIDDLE PUZZLE, Regards B.Karzeddin Al Hussein Bin Talal University JORDAN
From: bassam king karzeddin on 4 Mar 2007 01:25 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), doesn't have any solution Believe it, the rest is DONKEY TYPE WORK, I have left it for them, since they have a better language and deeper throats, and more over it will not be accepted from me since I don't have there DIRTY talents And I know that I may have to explain something else My Regards B.Karzeddin Al-Hussein Bin Talal University JORDAN
From: bassam king karzeddin on 4 Mar 2007 23:13
> On Feb 23, 6:56 am, bassam king karzeddin > <bas...(a)ahu.edu.jo> wrote: > > > How can this be if x, y, z, p > 0? > > > > > - Randy > > > > Hi Randy > > I may have forgotten to answer your second question > > What about my first question? > > > > > For a purpose of FLT I have defined (x, y, z) as > integer numbers (I mean positive and negative > numbers), so if > > > > x^p+y^p+z^p=0, then obviously not all of them > (x,y,z)are positive, > > OK. And after I posted I realized that perhaps you > were > just writing FLT in a different form. > > If x^p + y^p = z^p for p odd, then x^p + y^p + (-z)^p > = 0. > > > but in general one can see and prove (using the > general binomial theorem) that the following identity > holds true always > > > > (x+y+z)^n = n*(x+y)*(x+z)*(y+z)*f(x,y,z) + x^n +y^n > +z^n > > > > where n is odd positive integer > > (x,y,z) belong to C, complex numbers > > f(x,y,z) is function in terms of (x,y,z) > > Then what do you make of my counterexample? What are > the > properties of f(x,y,z)? I thought in your > first post you said it was an integer. But I gave > a counterexample where n*f(x,y,z) is not an integer. > > Can you sketch out how you get this identity from the > binomial theorem? > > - Randy > Yes Randy That can't be generalized to n where n is odd integer, the generalization would be something looks different from the identity I have already established for n when it is odd prime only. Thanks for noting that My Regards B.Karzeddin |