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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.
Andrew Wiles
Fermat's Last Theorem
long time.
Pierre de Fermat. The problem was called Fermat's Last Theorem
Fermat Prize was created in 1989.
Yes, the famous Fermat's Last Theorem, a conjecture by Fermat, that an equation of the form an + bn = cn has no integer solution, for n > 2. This was conjectured by Fermat in 1637, but it was only proved in 1995.Yes, the famous Fermat's Last Theorem, a conjecture by Fermat, that an equation of the form an + bn = cn has no integer solution, for n > 2. This was conjectured by Fermat in 1637, but it was only proved in 1995.Yes, the famous Fermat's Last Theorem, a conjecture by Fermat, that an equation of the form an + bn = cn has no integer solution, for n > 2. This was conjectured by Fermat in 1637, but it was only proved in 1995.Yes, the famous Fermat's Last Theorem, a conjecture by Fermat, that an equation of the form an + bn = cn has no integer solution, for n > 2. This was conjectured by Fermat in 1637, but it was only proved in 1995.
who meny juseph have fermat
It was 1647 not 1847 and by Fermat himself.
Fermat's Room was created on 2007-10-07.
A Fermat Prime refers to a proof that the mathematician Fermat discovered. It refers to a integer that is subject to an equation and the predictable result. Below is a webpage that explains it with examples.
Pierre De Fermat is famous for Fermat's Last Theorem, which states that an+bn=cn will never be true as long as n>2
Pierre de Fermat was born on August 17, 1601.