Fermat's Last theorem short proof
in [Math]
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From: Gerry Myerson on 19 Mar 2007 18:33 In article <1174322686.872938.90260(a)l75g2000hse.googlegroups.com>, "Randy Poe" <poespam-trap(a)yahoo.com> wrote: > On Mar 19, 6:22 am, bassam king karzeddin <bas...(a)ahu.edu.jo> wrote: > > Dear All > > > > I don't know if this question had been asked before ! > > > > Is Infinity is odd or even or WHAT? > > Neither odd nor even. Those are properties of integers, > and "infinity" is not an integer. Never even is a palindrome. oo is a palindrome. Therefore, oo is never even. -- Gerry Myerson (gerry(a)maths.mq.edi.ai) (i -> u for email)
From: David Harden on 19 Mar 2007 19:36 On Feb 23, 2:13 pm, Dan Cass <d...(a)sjfc.edu> wrote: > > 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. It works in general. A very easy proof uses the fact that Z[x,y,z] is a unique factorization domain: Since p, x+y, x+z and y+z are distinct primes in this ring, showing D(x,y,z) = (x+y+z)^p - (x^p+y^p+z^p) is a multiple of their product just amounts to showing it is a multiple of each of them. First of all, the binomial theorem definitely gives D(x,y,z) as a multiple of p in this ring. Next, it suffices to show that D is a multiple of x+y: D is symmetric in x,y and z and therefore is a multiple of x+z and y+z if it is a multiple of x+y. Write Z[x,y,z]= (Z[y,z])[x]. Since x+y is a monic polynomial in x, division by x+y can be applied to write D(x,y,z) = Q(x,y,z)*(x+y) + R(y,z). We need to show that R(y,z)=0. Since R(y,z) is independent of x, it suffices to show that R(y,z)=0 when x = -y: R(y,z) = D(-y,y,z) = (-y + y + z)^p - ((-y)^p + y^p + z^p) = -(1+(-1)^p)y^p. Since p is odd, this last expression equals 0. ---- David To send me email, move the r from the beginning to the end of the part before the @ and insert "alum." at the beginning of the part after the @.
From: David Harden on 19 Mar 2007 19:45 On Feb 27, 6:49 am, bassam king karzeddin <bas...(a)ahu.edu.jo> wrote: > > In article > > <11886980.1172521796572.JavaMail.jaka...(a)nitrogen.math > > forum.org> bassam king karzeddin <bas...(a)ahu.edu.jo> > > writes: > > > Randy Poe: > > > > I agree that if p=3 and (x,y,z) is a FLT counter > > > > example, then assuming xyz <> 0 mod 3 leads to > > > > a contradiction. Hence (x,y,z) FLT counter > > r example > > > > implies that xyz = 0 mod 3. > > > > Yes, and from the beginning I have shown you only > > y a proof when > > > p doesn't divide xyz, and in general the complete > > e proof is so simple > > > that an average mathematician can do > > > For a general p? I do not know as I do not follow it > > very closely, but > > Sophie Germain proved that if p and 2p+1 are both > > prime that > > x^p + y^p + z^p = 0 > > implies that one of x, y or z is divisible by p (p = > > 3 and p = 5 both > > fall into this special case, and since that time such > > primes are called > > Sophie Germain primes). From that time on FLT is > > split in two cases: > > (1): None of x, y and z is divisible by p (the easy > > sy case) > > (2): One of x, y and z is divisible by p (the > > he difficult case). > > Sophie Germain further proved Case 1 for all p less > > than 100 and Legendre > > extended it to all numbers less than 197. > > I didn't get that > Do you main sir that proof is based on numerical test > we know that primes are infinit, if so then it is not a proof unless I'm mistaken No, he doesn't mean that the proof is based on a numerical test. This is why no one credits Germain or Legendre or anyone before Wiles with a proof. There are infinitely many primes, so that alone would not suffice. Wiles' proof uses individual checks only (if I remember correctly) for p=3, 5 and 7. It uses a different argument (involving lots of algebraic geometry) for the remaining odd primes. ---- David
From: bassam king karzeddin on 20 Mar 2007 02:58 > On Feb 23, 2:13 pm, Dan Cass <d...(a)sjfc.edu> wrote: > > > 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. > > It works in general. A very easy proof uses the fact > that > Z[x,y,z] is a unique factorization domain: > > Since p, x+y, x+z and y+z are distinct primes in this > ring, showing > D(x,y,z) = (x+y+z)^p - (x^p+y^p+z^p) is a multiple of > their product > just amounts to showing it is a multiple of each of > them. > > First of all, the binomial theorem definitely gives > D(x,y,z) as a multiple of p in this ring. > > Next, it suffices to show that D is a multiple of > x+y: > D is symmetric in x,y and z and therefore is a > multiple of x+z and y+z > if it is a multiple of x+y. > > Write Z[x,y,z]= (Z[y,z])[x]. Since x+y is a monic > polynomial in x, > division by x+y can be applied to write D(x,y,z) = > Q(x,y,z)*(x+y) + > R(y,z). > We need to show that R(y,z)=0. > Since R(y,z) is independent of x, it suffices to show > that R(y,z)=0 > when x = -y: > R(y,z) = D(-y,y,z) = (-y + y + z)^p - ((-y)^p + y^p + > z^p) = > -(1+(-1)^p)y^p. > Since p is odd, this last expression equals 0. > > ---- David > > To send me email, move the r from the beginning to > the end of the part > before the @ and insert "alum." at the beginning of > the part after the > @. > And finally comes David Harden with a very simple proof in the language mathematicians understand, and put it bravely on the net-against the rules of JOURNALS. where he is not plying a game... but no comments from the experts or big heads!! Did you surrender? then rise up the White Flag. Watch out the talks are explosive and -indirectly- about FLT each of his own method naturally, HERE AND EVERY WHERE, except of course the mathematicians and Doctors in my University, - they really still don't know what I'm talking about. The question is open now, could Fermat had really a much simpler proof??!! Regards B.Karzeddin
From: bassam king karzeddin on 20 Mar 2007 03:38
> In article > <1174322686.872938.90260(a)l75g2000hse.googlegroups.com> > , > "Randy Poe" <poespam-trap(a)yahoo.com> wrote: > > > On Mar 19, 6:22 am, bassam king karzeddin > <bas...(a)ahu.edu.jo> wrote: > > > Dear All > > > > > > I don't know if this question had been asked > before ! > > > > > > Is Infinity is odd or even or WHAT? > > > > Neither odd nor even. Those are properties of > integers, > > and "infinity" is not an integer. > > Never even is a palindrome. > oo is a palindrome. > Therefore, oo is never even. > > -- > Gerry Myerson (gerry(a)maths.mq.edi.ai) (i -> u for > email) Hi Gerry Myerson It will take me too long to understand this paindrome concept. My Regards B.Karzeddin |