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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) $A^n+B^n= z^n$ Where by the conditions: A2. 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^{(n-2)} we can ALWAYS rewrote the (1) to arrive at A^n so we can Alway fix A^n as: $A^2+B^2= z^2 = A^n + A^{(n-2)} * B^2 = A^{(n-2)} * z^2$ so we can ALWAYS have that A^n is: (5) $A^n= A^{(n-2)} * z^2 - A^{(n-2)} * B^2$ so now we can put this in the second equation of the system and try to see if it works: $A^n + D^n= E^n$ $A^n= E^n - D^n$ So from the (5) in the system (6) Fermat is asking if: $A^{(n-2)} * z^2 - A^{(n-2)} * B^2= E^n - D^n$ but since the first 2 term has a common factor we can rewrite: $z^2 - B^2= ( E^n - D^n)/(A^{(n-2)})$ 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 $A^2+ B^2= C^2$ which is not part of the conditions for FLT. What you have proved is the [b]IF[b] $A^2+ B^2= C^2$ then we cannot have $A^n+ B^n= C^n$ for any n greater than 2.
March 22nd, 2014, 10:12 AM   #4
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Re: What's wrong here ? FLT once again...

Quote:
 Originally Posted by HallsofIvy The obvious thing is that you are assuming that $A^2+ B^2= C^2$ which is not part of the conditions for FLT. What you have proved is the [b]IF[b] $A^2+ B^2= C^2$ then we cannot have $A^n+ B^n= C^n$ for any n greater than 2.

- I prove that

(1) $A^2= C^2-B^2$

Has a solution for any A integer, for so if Fermat is Right also the system

(6) $\begin{cases} A^2+B^2= z^2 \\ A^n+ D^n = E^n \end{cases}$

Must work... but finally it is the same to say:

$A^2= (E^n-D^n) / A^{(n-2)}$

so A must divide both E and D...

or A^{(n-2) has to divide perfectly (E^n-D^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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