Fermat's Last theorem short proof
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From: bassam king karzeddin on 20 Feb 2007 23:51 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
From: Randy Poe on 21 Feb 2007 10:52 On Feb 21, 9:51 am, 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) Are you claiming this is true in general? Counterexample: x=3, y=4, z=5, p=5. (x+y+z)^p = 248832 (x^p + y^p + z^p) = 4392 (x+y)(x+z)(y+z) = 560 => p*N(x,y,z) = (248832 - 4392)/560 = 436.5 > 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) How can this be if x, y, z, p > 0? - Randy
From: bassam king karzeddin on 22 Feb 2007 20:56 > How can this be if x, y, z, p > 0? > > - Randy > Hi Randy I may have forgotten to answer your second 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, 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) My Regards B.Karzeddin
From: bassam king karzeddin on 22 Feb 2007 21:22 Hello all Why not come out of your holes and say a word of truth? strangely, If some one makes a mistake then many outstanding mathematicians will appear immediately correcting or harassing the OP I will be grateful and happier if my proof is wrong, or worthless or known before I'm also in my own way to make a much simpler and shorter proof that you may teach your kids before going to school Here, I will not stop, and my aim is far a head of FLT. My Regards B.Karzeddin
From: Raymond Burhoe on 22 Feb 2007 23:32
I'm not a mathematician, but maybe I can help: > We have the following general equation which any > mathematician can prove(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 odd integer function in terms of > (x, y, z) and is prime to (p) > P is odd prime number > (x, y, z) are three distinct(none zero) co prime > integers? > > Assuming a counter example (x, y, z) exists such that > (x^p+y^p+z^p=0) > > Then we have the following equation > > > (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),because > All prime factors of (x+y) devide z, > All prime factors of (x+z) devide y > All prime factors of (y+z) devide x > So the above equation does not have solution , > because 3 must be a prime factor of both sides of the > above equation > (That is by dividing both sides by 3, you get > (3)^(3k-1),on the left hand and no (3) factor on the > right hand of the above same equation,( where k is > positive integer) , and that implies our wrong > assumptiono of the existance of a counter example > which of course is impossible > I think proof is completed for (p=3, and 3 is not a > factor of (x*y*z) P = 3 is a special case, where (x + y + z) is a multiple of 3 but not a multiple of 9 whenever x*y*z is a nonmultiple of 3. It can be proved that any prime factor of x*y*z that is coprime to (x + y)(z + y)(z + x) must have the form 2pj + 1, so that it follows from the FLT equation that (x + y + z) = 0 (mod p^2). If you want anyone to believe your proof, you're going to have to prove your claim that N(x,y,z) is coprime to p when p > 4. > > The same arguments applies for any odd prime power p > , > (that is only needed to prove FLT > where (p) doesn't divide (x*y*z), as the following > If x^p+y^p+z^p=0, implies > > (x+y+z)^p =p* (x+y)*(x+z)*(y+z)*N(x,y,z) > > Assuming (p) does not divide (x*y*z), then (p) does > not divide (x+y)*(x+z)*(y+z),because > All prime factors of (x+y) devide z, > All prime factors of (x+z) devide y > All prime factors of (y+z) devide x > > And we have gcd(p,N(x,y,z))=1 > So the above equation does not have solution , > because p must be a prime factor of both sides of the > above equation > (That is by dividing both sides by p, you get > (p)^(p*k-1),on the left hand and no (p) factor on the > right hand of the above same equation,( where k is > positive integer) , and that implies our wrong > assumptiono of the existance of a counter example > which of course is impossible > > I think proof is completed for (p) any odd prime > number, where (p) doesn't divide (x*y*z) > > The same arguments applies for any odd prime power p > , > (that is only needed to prove FLT > > > My question to the specialists, 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 > > > Message was edited by: bassam king karzeddin |