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March 15th, 2014, 11:57 PM  #1 
Banned Camp Joined: Dec 2012 Posts: 1,028 Thanks: 24  What's wrong here ? FLT once again...
Still on the FLT: I probably find another little piece of the puzzle (but I fore sure make some mistake since this trick seems working !): PART1) What is well known (sorry if I make some mistake, pls find wher ! ): (1) Where by the conditions: A<B<z ; A,B,z integers First think non written by Fermat, but obvius is: MCD( A,B,z ) = 1 since: if there is a common factor k, than: (2) can be immediately rewritten as: (1) So we can reduce the (2) till arrive to the (1) where the odd/even table tell us that: 0) A = odd ; B = odd ; z= odd is impossible (even = odd is false) 00) A=even ; B= odd ; z = even is not possible by definition where MCD(A,z)=1 , and here is fore sure MCD(A,z)=2 a) A = even ; B = Even ; z= Even can be for sure reduced at the (1) dividing by 2 till possible so we can only have: b) A=odd ; B= even ; C = odd PART. 2) Several years ago I found that the case n=2: Is soo special since For any A integer there is a solution for the (1) (1) infact: the case A, B=z1, z can solve the (1) for any A odd (from 1 to infinite) in the form: a) if A=odd you just keep: and that comes from the identity: (3) b) if A=even you just keep A, B=z2, z can solve the (1) for any A even (from 2 to infinite) in the form: and: that comes from the identity: (4) The identity (3) and (4) are abvious so not necessary to prove nothing. What I'm not able to discover immediately is the conclusion to Fermat: Since we prove that the (1) has ALWAYS a solution for ANY A, so FLT becomes a system of 2 equations: where we need now to introduce D and E (insthead of B and z that we already use in the (1)) to have the system: (6) where: A^2+B^2 = z^2 is for sure true for ANY "A" integer and A^n + D^n = E^n has to be discover if possible, for n>2. We know by the initial conditions that MCD(A,E,D)=1 PART 3) Now the too simple trick (where I for sure make some mistake, but I'm not able to discover it !): Using a common factor A^{(n2)} we can ALWAYS rewrote the (1) to arrive at A^n so we can Alway fix A^n as: so we can ALWAYS have that A^n is: (5) so now we can put this in the second equation of the system and try to see if it works: So from the (5) in the system (6) Fermat is asking if: but since the first 2 term has a common factor we can rewrite: Where the left is for sure AN INTEGER, the right is fore sure A NON INTEGER for any n<>2, by the initial conditions MCD(A,E,D)=1 So Fermat is Right for any n>3 Where are my errors ?????? Thanks Ciao Stefano 
March 16th, 2014, 03:16 AM  #2 
Banned Camp Joined: Dec 2012 Posts: 1,028 Thanks: 24  Re: What's wrong here ? FLT once again...
...of course is a joke ! Ciao S. 
March 16th, 2014, 06:11 AM  #3 
Math Team Joined: Sep 2007 Posts: 2,409 Thanks: 6  Re: What's wrong here ? FLT once again...
The obvious thing is that you are assuming that which is not part of the conditions for FLT. What you have proved is the [b]IF[b] then we cannot have for any n greater than 2. 
March 22nd, 2014, 10:12 AM  #4  
Banned Camp Joined: Dec 2012 Posts: 1,028 Thanks: 24  Re: What's wrong here ? FLT once again... Quote:
 I prove that (1) Has a solution for any A integer, for so if Fermat is Right also the system (6) Must work... but finally it is the same to say: so A must divide both E and D... or A^{(n2) has to divide perfectly (E^nD^n)... that is again Fermat's problem... in anothe way. So I prove nothing.... once again. But If we consider that since A<>B<>C and that A,B,C are in relation (for each A) with the (1) and we know that A, B, C has an unique factorization than: The only possible case is that A,B or A,C or B,C has a common factor. For so is necessary to go on this to prove something.... Ciao Stefano  

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