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March 15th, 2014, 11:57 PM   #1
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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=z-1, 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=z-2, 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^{(n-2)} 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
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March 16th, 2014, 03:16 AM   #2
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Re: What's wrong here ? FLT once again...

...of course is a joke !

Ciao
S.
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March 16th, 2014, 06:11 AM   #3
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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.
HallsofIvy is offline  
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 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.
Sorry I lindly ask you to read again.

- 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^{(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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