Địng lý cuối của PIERRE DE FERMAT.
(x,y,z,n) thuộc tập hợp ( N+ )^4..
n>2.
(a) thuộc tập hợp Z
F là hàm số của ( a.)
F(a)=[a(a+1)/2]^2
F(0)=0 и F(-1)=0.
Xét hai phương trình
F(z)=F(x)+F(y)
F(z-1)=F(x-1)+F(y-1)
Ta có một dãy suy luận
F(z)=F(x)+F(y) tương đương F(z-1)=F(x-1)+F(y-1)
F(z)=F(x)+F(y) suy ra F(z-1)=F(x-1)+F(y-1)
F(z-x-1)=F(x-x-1)+F(y-x-1) suy ra F(z-x-2)=F(x-x-2)+F(y-x-2)
ta có
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 )
cho
z=y
và
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)
cho
z=/=y.
do đó
F(z-x-1)=F(x-x-1)+F(y-x-1) không suy raF(z-x-2)=F(x-x-2)+F(y-x-2)
do đó
F(z)=F(x)+F(y) không suy ra F(z-1)=F(x-1)+F(y-1)
do đó
F(z)=F(x)+F(y) không tương đương F(z-1)=F(x-1)+F(y-1)
điều có thể xảy ra
[F(x)+F(y)] = F(z) и F(x-1)+F(y-1)]=/=F(z-1)
hay ngược lại
[F(x)+F(y)]-[F(x-1)+F(y-1)]=/=F(z)-F(z-1).
hoặc
F(x)-F(x-1)+F(y)-F(y-1)=/=F(z)-F(z-1).
ta có
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.
do đó
x^3+y^3=/=z^3.
n>2.tương tự
Ta có một dãy suy luận
F(z)=G(x)*F(x)+G(y)*F(y) tương đương 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) suy ra 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) suy ra G(z)*F(z-x-2)=G(x)*F(x-x-2)+G(y)*F(y-x-2)
ta có
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 )
cho
z=y.
và
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 do đó G(x)>0.
cho
z=/=y.
do đó
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)
do đó
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)
do đó
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)
điều có thể xảy ra
[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)
hay ngược lại
do đó
[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)].
hay
G(x)*[F(x) - F(x-1)] + G(y)*[F(y)-F(y-1)]=/=G(z)*[F(z)-F(z-1).]
ta có
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) ]
do đó
x^n+y^n=/=z^n
Hòa Bình.
Trần Tấn Cường.
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