Fermat's Last theorem short proof
in [Math]
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From: Randy Poe on 23 Feb 2007 09:55 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
From: Raymond Burhoe on 23 Feb 2007 00:11 > 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. Yes, I think that's what he's doing. That means there will be integers a, b, c, and d, where x + y = a^p z + y = b^p z + x = c^p X + y + z = pabcd, where d is a multiple of p for p > 3. Thus, p*N(x,y,z) = (pd)^p I think his assumption that d is coprime to p is based on wrongly extending what is true for p = 3 into higher values of p. > > > 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 >
From: Dan Cass on 23 Feb 2007 04:13 > 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 > Correction: (x+y)(x+z)(y+z) = 504, and then (248832 - 4392)/504 is 485 = 5*97 Based on a few Maple calcs, I think the identity (x+y+z)^p =p*(x+y)*(x+z)*(y+z)*N(x,y,z) + x^p+y^p+z^p gives an integer coefficient polynomial for N(x,y,z) whenever p is odd and at least 3. For p=3 we have N(x,y,z)=1, For p=5 I get N(x,y,z)=x^2+y^2+z^2+xy+xz+yz Note this correctly gives N(3,4,5) = 97. If this approach could even be pushed to a valid proof of FLT in the case n=3, it would be interesting...
From: bassam king karzeddin on 24 Feb 2007 23:35 > I'm not a mathematician, but maybe I can help: I'm also not a mathematician and this proof is ment mainly to nonmathematicians > > > 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. You have (x+y+z)^3 = 3*(x+y)*(x+z)*(y+z) for a counter example, you assume 3 is not a factor of xyz that implies 3 is not a factor of (x+y)(x+z)(y+z), and in your example (x+y+z) = 9*m, so aplly to the above eqn. after dividing both sides of the above eqn.by 3, then you get 3^5 on the left and no 3 factor on the right, hence your assumption is wrong The other part of your question is not that difficult, I think many have done the symbolic calculations and rearranged them in to a disirable shaps > > > > > 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
From: bassam king karzeddin on 24 Feb 2007 23:43
> > 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. > > Yes, I think that's what he's doing. That means > there will be integers a, b, c, and d, where > > x + y = a^p > > z + y = b^p > > z + x = c^p > > X + y + z = pabcd, where d is a multiple of p for p > > 3. > > Thus, p*N(x,y,z) = (pd)^p > > I think his assumption that d is coprime to p is > based on wrongly extending what is true for p = 3 > into higher values of p. Can you support your claim by anexample??? > > > > > > 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 > > |