Fermat's Last theorem short proof
in [Math]
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From: bassam king karzeddin on 2 Apr 2007 22:25 > On 2 abr, 13:47, bassam king karzeddin > <bas...(a)ahu.edu.jo> wrote: J. Antonio Perez M wrote > An arabian sibling of JHS, uh? Ok, enjoy the > family...:>) > Tonio > > Hello Tonio But I have the tools, and I have already equipped you with most necessary weapons to destruct the bad mathematics, hurry up and write your own paper, here is a free source, change little the script and make it your own, since here is not authorized mathematics, and you don't need to make a reference here, it is a very good chance for you all to make success, isn't it? Regards B.Karzeddin
From: Roman B. Binder on 9 Apr 2007 12:40 Hi, > > 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, * * * Why not ! See: 125 + 27 + 64 = 216 5^3 + 3^3 + 4^3 = 6^3 Did You really meant tripled as separated to three units or there is any extra condition not mentioned and secure, that such operation could not happen ??? * * * > AN EQUATION IS BETTER THAN A CIVILIZATION > > I HOPE THAT CAN HELP > > MY REGARDS > > Bassam Karzeddin > AL Hussein bin Talal University > JORDAN Regards Ro-Bin
From: AP on 10 Apr 2007 09:53 On Wed, 21 Feb 2007 09:51:51 EST, bassam king karzeddin <bassam(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) >P is odd prime number in fact N(x,y,z) is homogeneous in Z[X,Y,Z] and d�=p-3 Lam� (1839) and Wells (1986) have used this equation to prove Fermat for n=7 and this equation is used by ....James Harris >(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: bassam king karzeddin on 10 Apr 2007 06:10 > Hi, > > > > 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, > * * * > Why not ! See: 125 + 27 + 64 = 216 > 5^3 + 3^3 + 4^3 = 6^3 > Did You really meant tripled as separated to > three units or there is any extra condition > not mentioned and secure, that such operation could > not happen ??? > * * * > > AN EQUATION IS BETTER THAN A CIVILIZATION > > > > I HOPE THAT CAN HELP > > > > MY REGARDS > > > > Bassam Karzeddin > > AL Hussein bin Talal University > > JORDAN > > Regards > Ro-Bin What do I mean exactly (in integers) is the following: A cube can't be tripled (equal to three cubes), but can be a expressed as a sum of three cubes, similarly A square can't be doubeled (equal to two squares), but can be expressed as a sum of two squares And So on.... Regards Bassam Karzeddin Al Hussein bin Talal University JORDAN
From: Major Quaternion Dirt Quantum on 10 Apr 2007 23:49
I think he's referring to the classical "doubling of the hexahedron," although the tetragon is easily doubled with compasses. the "tripling" of the volume of the hexahedron, I've never seem as a problem, although it's trivial ( -ly impossible, using compasses ... or, wait .-) > What do I mean exactly (in integers) is the following: > > A cube can't be tripled (equal to three cubes), but can be a expressed as a sum of three cubes, > similarly > > A square can't be doubeled (equal to two squares), but can be expressed as a sum of two squares thus: let us assume that we cannot decide on which kind of proof is not which other kind; no big deal. so, we just plug the alleged proof into the formula -- given in the magazine -- and see what comes out, and *then* argue about what the **** it means. note that, already, the language you chose implies a sort of dualism, "in" directly; whereas I gather that most folks prefer "in" duction -- what ever duction is supposed to be. and before you get your panties all horribly wet & twisted, just remember, You asked for It! thus: if you want to quibble about inductive versus indirect, or what ever, go at it -- with all the powers thus: uh yeah; Borat wants you in Sudan, why, Baby?... Harry Potter wants you in Iran -- yeah, Baby; shag'US with a spoon? --DARFURIA CONSISTS OF ARABs & nonARABs; NEWS-ITEM: we are marching to Darfuria, Darfuria, Darfuria! Harry Potter IIX, ?Ordeal @ Oxford//Sudan ^ Aircraft Carrier! http://larouchepub.com/other/2007/3410caymans_hedges.html ALgoreTHEmovieFORpresident.COM: http://larouchepub.com/eirtoc/site_packages/2007/al_gore.html |