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What is new solve about Fermat?

Updated: 8/20/2019
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11y ago

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trantancuong21@Yahoo.com

Последнее Пьер де Ферма теоремы. .

(x,y,z,n) принадлежать( N+ )^4..

n>2.

(a) принадлежать Z

F является функцией( a.)

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

F(0)=0 и F(-1)=0.

Рассмотрим два уравнения

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

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

непрерывный дедуктивного рассуждения

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

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

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

мы видим,

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 )

давать

z=y

и

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)

давать

z=/=y.

так

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

так

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

так

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

Таким образом, возможны два случая.

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

или наоборот

так

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

или

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

у нас есть

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.

так

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

n>2. аналогичный

непрерывный дедуктивного рассуждения

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

мы видим,

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 )

давать

z=y.

и

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 выводить G(x)>0.

давать

z=/=y.

так

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

так

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

так

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

Таким образом, возможны два случая.

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

или наоборот.

так

[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)].

или

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

у нас есть

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) ]

так

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

Счастливые и мира.

Trần Tấn Cường.

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