Fermat's Last theorem short proof
in [Math]
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From: Hisanobu Shinya on 28 Feb 2007 12:23 > 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. I am not even sure if he proved the so-called first case n = 3. When I gave a counterexample (x, y, z) to his claim that if 3 is not a divisor of x*y*z, then (x + y)(x + z)(y + z) is not divisible by 3, he started talking about the triangle stuff, which says that any counterexample to Fermat's Last Theorem for the case n = 3 must form a triangle. Are you familiar with the triangle idea? > > - Randy > HS
From: Randy Poe on 28 Feb 2007 23:14 On Feb 28, 10:23 pm, Hisanobu Shinya <eprinthshi...(a)yahoo.com> 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. > > I am not even sure if he proved the so-called first case n = 3. When I gave a counterexample (x, y, z) to his claim that if 3 is not a divisor of x*y*z, then (x + y)(x + z)(y + z) is not divisible by 3, he started talking about the triangle stuff, which says that any counterexample to Fermat's Last Theorem for the case n = 3 must form a triangle. Are you familiar with the triangle idea? No, I"m not. But I think I did convince myself the argument for the result I paraphrased above was correct, i.e. that if 3 divides x*y*z AND x^3 + y^3 + z^3 = 0, then you get a contradiction. Hence both of those can not be true and x^3 + y^3 + z^3 = 0 implies 3 divides x*y*z. - Randy
From: Hisanobu Shinya on 28 Feb 2007 14:18 > On Feb 28, 10:23 pm, Hisanobu Shinya > <eprinthshi...(a)yahoo.com> 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. > > > > I am not even sure if he proved the so-called first > case n = 3. When I gave a counterexample (x, y, z) to > his claim that if 3 is not a divisor of x*y*z, then > (x + y)(x + z)(y + z) is not divisible by 3, he > started talking about the triangle stuff, which says > that any counterexample to Fermat's Last Theorem for > the case n = 3 must form a triangle. Are you familiar > with the triangle idea? > > No, I"m not. > > But I think I did convince myself the argument for > the result > I paraphrased above was correct, i.e. that if 3 > divides > x*y*z AND x^3 + y^3 + z^3 = 0, then you get a > contradiction. The following does not make sense: > Hence both of those can not be true and x^3 + y^3 + > z^3 = 0 > implies 3 divides x*y*z. Let A = "3 divides xyz" and B = "x^3 + y^3 + z^3 = 0". I understand that you said: If A and B are true, then we get a contradiction. Therefore, [your "hence..." part] both of A and B can not be true, and so if B is true, then A is true. How does "A is true" follow? Perhaps what you really mean was If A and B are ture, then we get a contradiction. Hence A could not be true (because the argument assumes B is true) i.e., 3 is NOT a divisor of xyz? Sorry. I am not sure about which is the so-called first case. It doesn't matter, does it. > > - Randy > HS
From: bassam king karzeddin on 1 Mar 2007 00:08 > > Dear ALL > > > > Yes, there are few things I kept them hidden, > > > > First- something very interesting about N (x, y, > z), > > I did not mention, since that wasn't needed to > prove > > the first case- when (p) is not a factor of > (x*y*z)- > > and that is too simple to prove > > > > p, (x+y), (x+z), (y+z), N (x, y, z) are all > coprime > > pair wise > > > > Second- and according to "ONT" or "odd number > > theorem" > > > > Gcd {(x+y), (x^p+y^p) / (x+y)} = p^k, where (k=0), > if > > p is not a factor of (x+y), and (k=1), if p is a > > prime factor of (x+y) > > > > And that was due to a very valuable note from a > > member in sci.math QUASI whom we do miss and I do > > appreciate. > > > > On a comment to a previous thread of mine Fermat > > Last Secret > > Proof > > Let p be a prime factor of (x+y), assume (x=a Mod > > p), then (y= -a Mod p), implies > > (x^p+y^p) / (x+y) = p*a^(p-1) Mod p, which is a > > multiple of (p^1), the rest follows directly > > > > Third- if you whish to see the prime factorization > of > > (x^p+y^p), then assume (z=0), in the general > equation > > I provided in the beginning of this thread > > > > Now, still the proof is too short, but can you get > > the complete picture?? > > > > IF YOU CAN'T I WILL... > > > > My Regards > > 27,TH FEB.2007 > > Bassam Karzeddin > > AL Hussein Bin Talal University > > JORDAN > > If you do the steps you will simply arrive to the > following equation > > (x+y)^(p-1) = p*x*y*N(x,y) + (x^p + y^p) / (x+y) > > Now, assuming (p) is prime factor of (x+y), then > accordingly you get (1 Mod p) on the left hand side > of the equation and (0 Mod p) on the right hand side > of the same above equation because it will be a > multiple of (p) > > A contradiction implies that is not possible with all > integer numbers, and hence no counter example > exists. > > Similarly you may apply assuming either (x=o), or > (y=0) > Yes, There is another flaw in this proof I found,(for p divides xyz) No, this is obviously not correct argument to conclude, And my proof is not a proof But it still easy provable, I have to check my old notes to see how I surely did prove it, so you have to waite and the proof is certainly there. > My Regards > Bassam Karzeddin > AL Hussein bin Talal University > JORDAN Yes, There is also a flaw in this proof
From: Randy Poe on 1 Mar 2007 11:43
On Mar 1, 12:18 am, Hisanobu Shinya <eprinthshi...(a)yahoo.com> wrote: > > On Feb 28, 10:23 pm, Hisanobu Shinya > > <eprinthshi...(a)yahoo.com> 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. > > > > I am not even sure if he proved the so-called first > > case n = 3. When I gave a counterexample (x, y, z) to > > his claim that if 3 is not a divisor of x*y*z, then > > (x + y)(x + z)(y + z) is not divisible by 3, he > > started talking about the triangle stuff, which says > > that any counterexample to Fermat's Last Theorem for > > the case n = 3 must form a triangle. Are you familiar > > with the triangle idea? > > > No, I"m not. > > > But I think I did convince myself the argument for > > the result > > I paraphrased above was correct, i.e. that if 3 > > divides > > x*y*z AND x^3 + y^3 + z^3 = 0, then you get a > > contradiction. > > The following does not make sense: > > > Hence both of those can not be true and x^3 + y^3 + > > z^3 = 0 > > implies 3 divides x*y*z. Suppose P and Q can't both be true. So if you know P is true, then Q is not true. If you know Q is true, then P is not true. Oh wait. You're quite correct. The statement was "if you know 3 does NOT divide xyz, then x^3 + y^3 + z^3 = 0 leads to a contradiction". So "3 does not divide xyz" and "x^3 + y^3 + z^3 = 0" are contradictory, and therefore x^3 + y^3 + z^3 implies 3 divides xyz. > How does "A is true" follow? Perhaps what you really mean was > > If A and B are ture, then we get a contradiction. Hence A could not be true (because the argument assumes B is true) i.e., 3 is NOT a divisor of xyz? Yes, more or less. See my correction. A is "3 is not a divisor of xyz", and B is "x^3 + y^3 + z^3 = 0". A and B can't both be true. So B implies "not A", i.e. 3 IS a divisor of xyz. - Randy |