Fermat's Last theorem short proof
in [Math]
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From: bassam king karzeddin on 26 Feb 2007 01:53 > 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. > > Ah. OK. Glad somebody answered. OP refused to. > (Not sure how I got that arithmetic wrong.) > > OK, we'll take that as valid. Now let's look at the > rest > of the argument. > > >> 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) > > OK. So if a FLT counter-example exists, then there > will > be x, y >0, z<0 such that this identity holds. > > >> CASE-1 > >> If (p=3) implies N (x, y, z) = 1, > > Why does that follow? > > Perhaps there is something about N(x,y,z) we are not > being > told. I will ask OP for the third time to provide his > proof of this identity (which I will now accept as > true) > so I understand what N(x,y,z) is. > > It would be nice if he could answer the question I > just > asked as well. > > If for the third time he refuses, I'll abandon this > thread. > > - Randy > Hi Randy I will answer you Given two identical sets of things and a balance,and you are asked to put one set on only one side of the balance, so the second set is on the other side of the balance, then equilibrium state is attained, no matter if you keep rearranging them in deferent manners only on each side So, do the multiplication please, then all terms will be canceled, and you will get (0=0) A CUBE CAN'T BE TRIPLED, AN EQUATION IS BETTER THAN A CIVILIZATION I HOPE THAT CAN HELP MY REGARDS Bassam Karzeddin AL Hussein bin Talal University JORDAN
From: bassam king karzeddin on 26 Feb 2007 01:55 > 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. > > Ah. OK. Glad somebody answered. OP refused to. > (Not sure how I got that arithmetic wrong.) > > OK, we'll take that as valid. Now let's look at the > rest > of the argument. > > >> 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) > > OK. So if a FLT counter-example exists, then there > will > be x, y >0, z<0 such that this identity holds. > > >> CASE-1 > >> If (p=3) implies N (x, y, z) = 1, > > Why does that follow? > > Perhaps there is something about N(x,y,z) we are not > being > told. I will ask OP for the third time to provide his > proof of this identity (which I will now accept as > true) > so I understand what N(x,y,z) is. > > It would be nice if he could answer the question I > just > asked as well. > > If for the third time he refuses, I'll abandon this > thread. > > - Randy > Hi Randy I will answer you Given two identical sets of things and a balance,and you are asked to put one set on only one side of the balance, so the second set is on the other side of the balance, then equilibrium state is attained, no matter if you keep rearranging them in deferent manners only on each side So, do the multiplication please, then all terms will be canceled, and you will get (0=0) A CUBE CAN'T BE TRIPLED, AN EQUATION IS BETTER THAN A CIVILIZATION I HOPE THAT CAN HELP MY REGARDS Bassam Karzeddin AL Hussein bin Talal University JORDAN
From: Randy Poe on 26 Feb 2007 12:49 On Feb 26, 11:53 am, bassam king karzeddin <bas...(a)ahu.edu.jo> wrote: > > 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. > > > Ah. OK. Glad somebody answered. OP refused to. > > (Not sure how I got that arithmetic wrong.) > > > OK, we'll take that as valid. Now let's look at the > > rest > > of the argument. > > > >> 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) > > > OK. So if a FLT counter-example exists, then there > > will > > be x, y >0, z<0 such that this identity holds. > > > >> CASE-1 > > >> If (p=3) implies N (x, y, z) = 1, > > > Why does that follow? > > > Perhaps there is something about N(x,y,z) we are not > > being > > told. I will ask OP for the third time to provide his > > proof of this identity (which I will now accept as > > true) > > so I understand what N(x,y,z) is. > > > It would be nice if he could answer the question I > > just > > asked as well. > > > If for the third time he refuses, I'll abandon this > > thread. > > > - Randy > > Hi Randy > > I will answer you > > Given two identical sets of things and a balance,and you are asked to put one set on only one side of the balance, so the second set is on the other side of the balance, then equilibrium state is attained, no matter if you keep rearranging them in deferent manners only on each side > So, do the multiplication please, then all terms will be canceled, and you will get (0=0) > > A CUBE CAN'T BE TRIPLED, > AN EQUATION IS BETTER THAN A CIVILIZATION > > I HOPE THAT CAN HELP I now follow most of your argument. I'll just repeat my last statement: "what you've shown as far as I can tell is that if x^3 + y^3 + z^3 = 0 has a solution, then one of x, y or z must be divisible by 3. " 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 example implies that xyz = 0 mod 3. - Randy
From: bassam king karzeddin on 26 Feb 2007 05:29 > On Feb 26, 11:53 am, bassam king karzeddin > <bas...(a)ahu.edu.jo> wrote: > > > 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. > > > > > Ah. OK. Glad somebody answered. OP refused to. > > > (Not sure how I got that arithmetic wrong.) > > > > > OK, we'll take that as valid. Now let's look at > the > > > rest > > > of the argument. > > > > > >> 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) > > > > > OK. So if a FLT counter-example exists, then > there > > > will > > > be x, y >0, z<0 such that this identity holds. > > > > > >> CASE-1 > > > >> If (p=3) implies N (x, y, z) = 1, > > > > > Why does that follow? > > > > > Perhaps there is something about N(x,y,z) we are > not > > > being > > > told. I will ask OP for the third time to provide > his > > > proof of this identity (which I will now accept > as > > > true) > > > so I understand what N(x,y,z) is. > > > > > It would be nice if he could answer the question > I > > > just > > > asked as well. > > > > > If for the third time he refuses, I'll abandon > this > > > thread. > > > > > - Randy > > > > Hi Randy > > > > I will answer you > > > > Given two identical sets of things and a > balance,and you are asked to put one set on only one > side of the balance, so the second set is on the > other side of the balance, then equilibrium state is > attained, no matter if you keep rearranging them in > deferent manners only on each side > > So, do the multiplication please, then all terms > will be canceled, and you will get (0=0) > > > > A CUBE CAN'T BE TRIPLED, > > AN EQUATION IS BETTER THAN A CIVILIZATION > > > > I HOPE THAT CAN HELP > > I now follow most of your argument. I'll just repeat > my > last statement: > > "what you've shown as far as I can tell is that > if x^3 + y^3 + z^3 = 0 has a solution, then one of > x, y or z must be divisible by 3. " > > 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 example > implies that xyz = 0 mod 3. > > - Randy > Yes, and from the beginning I have shown you only a proof when p doesn't divide xyz, and in general the complete proof is so simple that an average mathematician can do Some have already got it as you see from the replies I wonder also why simply you deny me all my efforts to settle up this issue for ever. Is it because I give my knowledge for Free??? I can't write papers because mainly I don't have time, and I don't know how to write introduction and references, I also can't waite or tolerate refrees opinions especially when I'm a teacher and they are the students who are going to learn Here, many outstanding mathematicians used to have the final word in this topic but not any more, where they deliberately disappear now and may be working in secret, (the main code for success}- ask their first teacher whome I do respect to assure this fact, but the story it self didn't end.! and the final laugh will be for FERMAT if my proofs (which I will post completely whenever I get free time ) are stolen then I swear I will add and triple your WORLD THAT IS FULL OF buzzls for another few centuries, and the final word will be for the new born that is not a mind kind. I will give you another big hint a^p can't be equal to p*b^p, where (a, b) are distinct integers, p is prime number My Regards Bassam Karzeddin Al Hussein bin Talal University JORDAN
From: Michael Orion on 26 Feb 2007 05:38
> > On Feb 26, 11:53 am, bassam king karzeddin > > <bas...(a)ahu.edu.jo> wrote: > > > > 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. > > > > > > > Ah. OK. Glad somebody answered. OP refused to. > > > > (Not sure how I got that arithmetic wrong.) > > > > > > > OK, we'll take that as valid. Now let's look > at > > the > > > > rest > > > > of the argument. > > > > > > > >> 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) > > > > > > > OK. So if a FLT counter-example exists, then > > there > > > > will > > > > be x, y >0, z<0 such that this identity holds. > > > > > > > >> CASE-1 > > > > >> If (p=3) implies N (x, y, z) = 1, > > > > > > > Why does that follow? > > > > > > > Perhaps there is something about N(x,y,z) we > are > > not > > > > being > > > > told. I will ask OP for the third time to > provide > > his > > > > proof of this identity (which I will now > accept > > as > > > > true) > > > > so I understand what N(x,y,z) is. > > > > > > > It would be nice if he could answer the > question > > I > > > > just > > > > asked as well. > > > > > > > If for the third time he refuses, I'll abandon > > this > > > > thread. > > > > > > > - Randy > > > > > > Hi Randy > > > > > > I will answer you > > > > > > Given two identical sets of things and a > > balance,and you are asked to put one set on only > one > > side of the balance, so the second set is on the > > other side of the balance, then equilibrium state > is > > attained, no matter if you keep rearranging them > in > > deferent manners only on each side > > > So, do the multiplication please, then all terms > > will be canceled, and you will get (0=0) > > > > > > A CUBE CAN'T BE TRIPLED, > > > AN EQUATION IS BETTER THAN A CIVILIZATION > > > > > > I HOPE THAT CAN HELP > > > > I now follow most of your argument. I'll just > repeat > > my > > last statement: > > > > "what you've shown as far as I can tell is that > > if x^3 + y^3 + z^3 = 0 has a solution, then one of > > x, y or z must be divisible by 3. " > > > > 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 example > > implies that xyz = 0 mod 3. > > > > - Randy > > > > Yes, and from the beginning I have shown you only a > proof when > p doesn't divide xyz, and in general the complete > proof is so simple that an average mathematician can > do > > Some have already got it as you see from the replies > > > I wonder also why simply you deny me all my efforts > to settle up this issue for ever. > > Is it because I give my knowledge for Free??? > > I can't write papers because mainly I don't have > time, and I don't know how to write introduction and > references, I also can't waite or tolerate refrees > opinions especially when I'm a teacher and they are > the students who are going to learn > > Here, many outstanding mathematicians used to have > the final word in this topic but not any more, where > they deliberately disappear now and may be working in > secret, (the main code for success}- ask their first > teacher whome I do respect to assure this fact, but > the story it self didn't end.! and the final laugh > will be for FERMAT > > if my proofs (which I will post completely whenever I > get free time ) are stolen then > > I swear I will add and triple your WORLD THAT IS FULL > OF buzzls for another few centuries, and the final > word will be for the new born that is not a mind > kind. > > I will give you another big hint > > a^p can't be equal to p*b^p, where (a, b) are > distinct integers, p is prime number > > My Regards > > Bassam Karzeddin > Al Hussein bin Talal University > JORDAN I am laughing out loud. Just when Dr. Poe points out that you have proved only a very weak statement (that xyz is not divisible by 3) you let us all know that you have not posted the complete proof of FLT. So why did you not point out before that you did not post a complete proof? Still giggling, MO |