Re: WHAT THE MODERN MATHEMATHICIEN ARE SHY TO TALK ABOUT

Hello,
1) z+1 is divisible by 2 as many times as k+1 is divisible by 2
"as long as (z+1) or k+1 is not divisible by 2 as many times
as [u^(n-1)]/2".What counts is the fact that if(k+1) is divisible by 4
that makes (z+1) divisible by 4
There is a step in the proof which shows that k+1 is divisible by 4
because it should be divisible by 2 more times than (n-1) is
divisible by2.Therfore k=1 must be disible at least by 4 and that makes
z+1 divisible by 4 at least because z-k=a*u^(n-1)=b=k

2)Z=r*z*u where r=1 or r=n when Z is divisible by n
Go to the poster "to *** Winter about FLT 5/18/05 where you
can find a better editing of the proof
Best Wishes
george ghiata

> There is no way to follow your "argument". However, I
> will try to make
> stupid comments:
>
> > > > > > > > >We know that b+2*k=z
> > > > > > > > >Therfore z-k=b+k.
>
> This I can understand. Is this the main step in your
> derivation?
>
> > > > > > > > >Therfore z+1 is divisible by2 as many
> > > times
> > > > as
> > > > > k+1 is divisible by 2
>
> This is where I've lost you once again. And that
> would be why? Pure
> magic?
>
> > > > > when Z is divisible by n
> > > > > > > > >Therfore z-k=b+k.
>
> What is the realtionship between Z and z? Father and
> son?
>
> george ghiata wrote:
> > >One more statement has been ommited and add-up to
> > second part
> > george ghiata
> > Here it is posted a corected Second part
> > > george ghiata
> > >
> > > > Now we going to see the elementary genaral
> proof
> > > of
> > > > FLT.
> > > > george ghiata
> > > >
> > > > > Here is another exemple to suport my point
> what
> > > > this
> > > > > message
> > > > > and title is about besides the exemple given
> > > with
> > > > > Pell's equation
> > > > > The text below shows the proof of FLT for
> > > n=4*L+1
> > > > > I have not seen or heard about this kind the
> > > > proof.
> > > > > Since this proof exists, we have to think
> twice
> > > > > before we say the great
> > > > > mathematicien could not had missed a
> elementary
> > > > proof
> > > > > for long time.
> > > > > That is because now it looks possible that it
> > > > might
> > > > > exist a similar one
> > > > > for=n*L+3.
> > > > > And guess what? Yes ,there is one.
> > > > > Next time I will present it.
> > > > > george ghiata
> > > > > > >Hello,
> > > > > > >Here is Archimedes Proof of Fermat Last
> > > theorem
> > > > > for n=4*L+1
> > > > > > >george Ghiata
> > > > > > > >
> > > > > > > > >To Alf and Andrew,
> > > > > > > > >A simplfied Fermat last theorm
> > > > > proof:Archimedes FLT proof when n=4*L+1
> > > > > > > > >X^n+Y^n=Z^n is Imposible if Z,X,YAre
> > > > Integers
> > > > > and n=prime>2
> > > > > > > > >We take Z=even number
> > > > > > > > >X+Y-Z=B
> > > > > > > > >X=B+Q
> > > > > > > > >Y=B+P
> > > > > > > > >2*B+Q+P=s*u^n wher s=1 or n^(n-1) if Z
> is
> > > > > divisible by n
> > > > > > > > >B+Q+P=r*z*u wher r is 1 or n if Z is
> > > > divisible
> > > > > by n
> > > > > > > > >From above we get that B=b*u wher
> u=even
> > > > > Therefore Q+P=2*u*k where k
> > > > > > > > >is =odd number.
> > > > > > > > > Therefore too we get that z= -
> > > > > b+v*u^(n-1).Therfore b=odd numbre and
> > > > > v=1 or n^(n-2) when Z is divisible by n
> > > > > > > > >From what has been shown by now we get
> > > that
> > > > > the equation :
> > > > > > > > >
> (B+Q+)^n+(B+P)^n-(B+Q+P)^n=0
> > > > > > > > >can be written as:
> > > > > > > > > EQ A: B^n+M + G*H*(u^n)=0 where H
> is
> > > > 2^t
> > > > > which comes from
> > > > > > > > > 2*(n-1)=2^t*d where d is
> =odd
> > > > and
> > > > > G=odd number and
> > > > > > > > >M is comming from following:
> > > > > > > > >Now:EQ.C : (B+Q)^n+(B+P)^n - (Q+P)^n=
> > > B^n+
> > > > > n*B*(Q+P)+B^n+M
> > > > > > > > >From above we get from EQ A : that
> B^n+M
> > > is
> > > > > divisible by 2 as many times as
> > > > > > > > > (2^t)*u^n is divisible by 2
> > > > > > > > >From EQ C we get F =[ (B+Q)^n+(B+P)^n
> -
> > > > > (Q+P)^n - B^n-[B^(n-1)]*n*(Q+P) is divisible
> by
> > > 2
> > > > as
> > > > > many times as
> > > > > > > > > (2^t)*u^n is divisible by
> 2
> > > > > > > > > Therefore F=Z^n- B^n -
> > > > > (Q+P)^n-n*[B^(n-1)]*(Q+P) =(z*u)^n-
> > > > > [b*u]^n-(2*u*k)^n- n*(b^n)*k*2*(u^n)
> > > > > > > > > can be written as :
> > > > > > > > > F= (u^n)*{ z^n -
> b^n
> > > -
> > > > > (2*k)^n - n*2*k*(b^n)}
> > > > > > > > >Since z=- b+v*u^(n-1) we get that F
> > > > > =(u^n)*{2*(b^n)*(n*k+1) -d*u^(n-1) -2*k)^n}
> > > > > > > > >But F is divisible by 2 as many times
> > > as
> > > > > (2^t)*u^n is divisible by 2
> > > > > > > > >Therfore We get that (k+1)=2^m where
> > > > m>(t-1)
> > > > > and (n-1)=j*2^(t-1) wher j is odd number
> > > > > > > > >We know that b+2*k=z and
> > > > > 2*b+2*k=h*u*(n-1)where h=1 or n^(n-2) when Z
> is
> > > > > divisible by n
> > > > > > > > >Therfore z-k=b+k.
> > > > > > > > >Therfore z+1 is divisible by2 as many
> > > times
> > > > as
> > > > > k+1 is divisible by 2
> > > > > > > > >But when the EQuation
> > > > > (B+Q+P)^n=(B+Q)^n+(B+P)^n=(z*u)^n is divided
> by
> > > > > (2*B+Q+P)
> > > > > > > > >we get :
> > > > > > > > > H= [(n-1)*u*V
> +n*Q^(n-1)}=r*z^n
> > > > > > > > >If r=1 then we get:
> > > > > > > > > [(n-1)*u*V
> > > > > +(n-1)*Q^(n-1)+Q^(n-1) +1=z^n+1
> > > > > > > > >We know that (z+1)=(2^m)*c where
> m>(t-1)
> > > > and
> > > > > [2^(t-1)]*a =(n-1)
> > > > > > > > >Therfore n=4*g+3
> > > > > > > > >If r=n then we get :
> > > > > > > > >
> > > > > [(n-1)*u*V+(n-1)*Q^(n-1)
> > > > > [(n-1)*u*V+(n-1)*Q^(n-1)
> > > > > -1) +Q^(n-1)+1=(n-1)*z^n+z^n+1
> > > > > > > > > We see that the right side
> of
> > > EQ.
> > > > is
> > > > > divisible by 2^(t-1).
> > > > > > > > >If (n-1) is divisible by 2 only then
> the
> > > > left
> > > > > side is divisible by 4
> > > > > > > > >If (n-1) is divisible by 4 then The
> left
> > > > side
> > > > > is divisible by 2 only and the right side by
> 4
> > > > > > > > >Conclusion:
> > > > > > > > > If n=4*L+1 then X^n
> +Y^n
> > > > =Z^n
> > > > > is impossible
> > > > > > > > >created by george ghiata-may 11 -05
> > > > > > > > > Second Part of The general proof of
> > > > FLT.:n=4*L+3
> > > > We use the information from the case n=4*L+1
> > Corection:From First Part we get that
> > if Z is divisible by 2 then Z is not divisible
> by n
> >
> > > Let's take (B+Q)^n+(B+P)^n=(B+Q+P)^n
> > > > We multiply the Eq by (2*u)^n and get:
> > > >
> (2*B+Q+P+Q-P)^n+(2*B+Q+P+P-Q)^n=(2*B+Q+P+Q+P)^n
> > > > After we develop the parantheses and move
> > > everything
> > > > g to
> > > > the left of Eq. and divide by (2*u)^n we get :
> > > > EQ.S1: (u^n)*C
> > > > )*C +n*(Q-u*k)^(n-1)-n*[(k*u)^(n-1)]/(2) -k^n=0
> > > Corection:
> > > EQ.S1 should be read:
> > > EQ.S1= [2^(n-3)]*(u^n)*C +(t^n)*[2^(n-1)]*E+
> > > +n*(Q-u*k)^(n-1)-n*[(k*u)^(n-1)]/2-k^n
> > >
> > > > We see that D=[n*(Q-u*k)^(n-1)]-k^n must be
> > > divisible
> > > > by 2 as many times
> > > > as [u^(n-1)]/2 is divisible by 2.
> > > > Now we take X^n+Y^n=Z^n and after division by
> > > > (X+Y)=u^n we change it
> > > > to: EQJ: [(X+Y)*E +n*X^(n-1)]=z^n
> > > > We know that (z-k)=[u^(n-1)]/2
> > > > Now we change the EQJ to
> > > > EQ.S :
> > > > :
> [(X+Y)*E+n*(n-1)*(b+k)*A+n*(Q-u*k)^(n-1)-z^n=0
> > > Corection:Above in the EQ.S we should read
> > > n*(n-1)*(b+k)*u*A instead of
> > > n*(n-1)*(b+k)*A
> > >
> > > > where A is a odd number.We see thatT=
> > > > [n*(Q-u*k)^(n-1)]-z^n is divisible by 2 as
> > > > many times as (n-1)*(b+k) is divisible by 2
> > > Corection:in the above line we should read
> > > (n-1)*(b+k)*u instead of
> > > (n-1)*(b+k)
> > > > Now we substract and add up in the EQ.S1 the
> > > > Q.S1 the value z^n and then
> > > > we divide by [u^(n-1)]/2=(b+k) and then cancel
> > > > n*k^(n-1)and get
> > > > EQ.S2. Since n=4*L+3 we get that the EQ.S2 is
> > > > divisible by 2 and
> > > > not by 4.Therefore the EQ.S2 is impossible.
> > > Corection: In the above line about EQ.S2 we
> > > should read:
> > > Since n=4*L+3 we get that the EQ.S2 is
> > > divisible by 2 as many times as (u) is divisible
> > > ble by 2
> > > (because {n*(Q-u*k)^(n-1)-z^n}/(b+k) must be
> > > e divisible
> > > by 2*u conform to the EQ.S
> > >
> > > > Therfore Fermat last theorem is impossible for
> > > > n=4*L+3 too.
> > > > Therfore Fermat Last theorem is true.
> > > > Created by George Ghiata-AA
>
.

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