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.
Trần Tấn Cường.