What is shortest solve about fermat?

Answer 1

To:

trantancuong21@Yahoo.com

PIERRE DE FERMAT's last Theorem.

(x,y,z,n) belong ( N+ )^4..

n>2.

(a) belong Z

F is function of ( a.)

F(a)=[a(a+1)/2]^2

F(0)=0 and F(-1)=0.

Consider two equations

F(z)=F(x)+F(y)

F(z-1)=F(x-1)+F(y-1)

We have a string inference

F(z)=F(x)+F(y) equivalent F(z-1)=F(x-1)+F(y-1)

F(z)=F(x)+F(y) infer F(z-1)=F(x-1)+F(y-1)

F(z-x-1)=F(x-x-1)+F(y-x-1) infer F(z-x-2)=F(x-x-2)+F(y-x-2)

we see

F(z-x-1)=F(x-x-1)+F(y-x-1 )

F(z-x-1)=F(-1)+F(y-x-1 )

F(z-x-1)=0+F(y-x-1 )

give

z=y

and

F(z-x-2)=F(x-x-2)+F(y-x-2)

F(z-x-2)=F(-2)+F(y-x-2)

F(z-x-2)=1+F(y-x-2)

give z=/=y.

So

F(z-x-1)=F(x-x-1)+F(y-x-1) don't infer F(z-x-2)=F(x-x-2)+F(y-x-2)

So

F(z)=F(x)+F(y) don't infer F(z-1)=F(x-1)+F(y-1)

So

F(z)=F(x)+F(y) is not equivalent F(z-1)=F(x-1)+F(y-1)

So have two cases.

[F(x)+F(y)] = F(z) and F(x-1)+F(y-1)]=/=F(z-1)

or vice versa

So

[F(x)+F(y)]-[F(x-1)+F(y-1)]=/=F(z)-F(z-1).

Or

F(x)-F(x-1)+F(y)-F(y-1)=/=F(z)-F(z-1).

We have

F(x)-F(x-1) =[x(x+1)/2]^2 - [(x-1)x/2]^2.

=(x^4+2x^3+x^2/4) - (x^4-2x^3+x^2/4).

=x^3.

F(y)-F(y-1) =y^3.

F(z)-F(z-1) =z^3.

So

x^3+y^3=/=z^3.

n>2. .Similar.

We have a string inference

G(z)*F(z)=G(x)*F(x)+G(y)*F(y) equivalent G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1)

G(z)*F(z)=G(x)*F(x)+G(y)*F(y) infer G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1)

G(z)*F(z-x-1)=G(x)*F(x-x-1)+G(y-x-1)*F(y) infer G(z)*F(z-x-2)=G(x)*F(x-x-2)+G(y)*F(y-x-2)

we see

G(z)*F(z-x-1)=G(x)*F(x-x-1)+G(y)*F(y-x-1 )

G(z)*F(z-x-1)=G(x)*F(-1)+G(y)*F(y-x-1 )

G(z)*F(z-x-1)=0+G(y)*F(y-x-1 )

give z=y.

and

G(z)*F(z-x-2)=G(x)*F(x-x-2)+G(y)*F(y-x-2)

G(z)*F(z-x-2)=G(x)*F(-2)+G(y)*F(y-x-2)

G(z)*F(z-x-2)=G(x)+G(y)*F(y-x-2)

x>0 infer G(x)>0.

give z=/=y.

So

G(z)*F(z-x-1)=G(x)*F(x-x-1)+G(y-x-1)*F(y) don't infer G(z)*F(z-x-2)=G(x)*F(x-x-2)+G(y)*F(y-x-2)

So

G(z)*F(z)=G(x)*F(x)+G(y)*F(y) don't infer G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1)

So

G(z)*F(z)=G(x)*F(x)+G(y)*F(y) is not equiivalent G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1)

So have two cases

[G(x)*F(x)+G(y)*F(y)]=G(z)*F(z) and [ G(x)*F(x-1)+G(y)*F(y-1)]=/=G(z-1)*F(z-1)

or vice versa.

So

[G(x)*F(x)+G(y)*F(y)] - [ G(x)*F(x-1)+G(y)*F(y-1)]=/=G(z)*[F(z)-F(z-1)].

Or

G(x)*[F(x) - F(x-1)] + G(y)*[F(y)-F(y-1)]=/=G(z)*[F(z)-F(z-1).]

We have

x^n=G(x)*[F(x)-F(x-1) ]

y^n=G(y)*[F(y)-F(y-1) ]

z^n=G(z)*[F(z)-F(z-1) ]

So

x^n+y^n=/=z^n

Happy&Peace.

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