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Who Do solve Fermat in one page?

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Anonymous

 ∙ 12y ago
Updated: 8/20/2019

PIERRE DE FERMAT' S LAST THEOREM.

CASE SPECIAL N=3 AND.GENERAL CASE N>2. .

THE CONDITIONS.Z,X,Y,N ARE THE INTEGERS . Z*X*Y*N>0.N>2.

Z^3=/=X^3+Y^3 AND Z^N=/=X^N+Y^N.

SPECIAL CASE N=3.

WE HAVE

(X^2+Y^2)^2=X^4+Y^4+2X^2*Y^2.

BECAUSE

X*Y>0=>2X^2*Y^2>0.

SO

(X^2+Y^2)^2=/=X^4+Y^4.

CASE 1. IF

Z^2=X^2+Y^2

SO

(Z^2)^2=(X^2+Y^2)^2

BECAUSE

(X^+Y^2)^2=/=X^4+Y^4.

SO

(Z^2)^2=/=X^4+Y^4.

SO

Z^4=/=X^4+Y^4.

CASE 2. IF

Z^4=X^4+Y^4

BECAUSE

X^4+Y^4.=/= (X^2+Y^2.)^2

SO

Z^4=/=(X^2+Y^2.)^2

SO

(Z^2)^2=/=(X^2+Y^2.)^2

SO

Z^2=/=X^2+Y^2.

(1) AND (2)=> Z^4+Z^2=/=X^4+Y^4+X^2+Y^2.

SO

2Z^4+2Z^2=/=2X^4+2Y^4+2X^2+Y^2.

SO

(Z^4+Z^2+2Z^3+Z^4+Z^2-2Z^3)=/=(X^4+X^2+2X^3+X^4+X^2-2X^3)+)(Y^4+Y^2+2Y^3+Y^4+Y^2-2Y^3)

SO IF

(Z^4+Z^2+2Z^3)/4=(Z^4+Z^2+2Z^3)/4+(Z^4+Z^2+2Z^3)/4

=> (Z^4+Z^2-2Z^3)/4=/=(Z^4+Z^2-2Z^3)/4+(Z^4+Z^2-2Z^3/4)

AND

SO IF

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

=> (Z^4+Z^2+2Z^3)/4=/=(Z^4+Z^2+2Z^3)/4+(Z^4+Z^2+2Z^3)/4

BECAUSE

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

SO

Z^3=/=X^3+Y^3.

GENERAL CASE N>2.

Z^N=/=X^N+Y^N.

WE HAVE

[X^(N-1)/2+Y^(N-1)/2]^(N+1)/(N-1)=X^(N+1)/2+Y^(N+1)/2+ H.

BECAUSE X*Y>0=>H>0.

SO

[X^(N-1)/2+Y^(N-1)/2]^(N+1)/(N-1)=/= X^(N+1)/2+Y^(N+1)/2

CASE 1. IF

Z^(N-1)/2=X^(N-1)/2+Y^(N-1)/2

SO

[Z^(N-1)/2]^(N+1)/(N-1)=[X^(N-1)/2+Y^(N-1)/2 ]^(N+1)/(N-1).

BECAUSE

[X^(N-1)/2+Y^(N-1)/2 ]^(N+1)/(N-1)=/=X^(N+1)/2+Y(N+1)/2.

SO

[Z^(N-1)/2]^(N+1)/(N-1)=/=X^(N+1)/2+Y(N+1)/2.

SO

Z^(N+1)/2=/=X^(N+1)/2+Y^(N+1)/2.

CASE 2. IF

Z^(N+1)/2=X^(N+1)/2+Y^(N+1)/2

SO

[Z^(N+1)/2]^(N-1)/(N+1)=[X^(N+1)/2+Y^(N+1)/2 ]^(N-1)/(N+1)

BECAUSE

[X^(N+1)/2+Y^(N+1)/2](N-1)/(N+1)=/=X(N-1)/2+Y^(N-1)/2.

SO

[Z^(N+1)/2]^(N-1)/(N+1)=/=X(N-1)/2+Y^(N-1)/2.

SO

Z^(N-1)/2=/=X(N-1)/2+Y^(N-1)/2..

SO

(1) AND (2)=> Z^(N+1)/2+Z^(N-1)/2=/=X^(N+1)/2+Y^(N+1)/2+X^(N-1)/2+Y^(N-1)/2.

SO

2[Z^(N+1)/2+Z^(N-1)/2]=/=2[X^(N+1)/2+Y^(N+1)/2]+2[X^(N-1)/2+Y^(N-1)/2.]

SO

[Z^(N+1)/2+Z^(N-1)/2+2Z^N ]+[Z^(N+1)/2+Z^(N-1)/2-2Z^N ]=/=[X^(N+1)/2+X^(N-1)/2+2X^N ]+[X^(N+1)/2+X^(N-1)/2-2X^N ]+[Y^(N+1)/2+Y^(N-1)/2+2Y^N ]+[Y^(N+1)/2+Y^(N-1)/2-2Y^N ]

SO IF

[Z^(N+1)/2+Z^(N-1)/2+2Z^N ]/4=[X^(N+1)/2+X^(N-1)/2+2X^N ] /4+ [Y^(N+1)/2+Y^(N-1)/2+2Y^N ]/4=>

[Z^(N+1)/2+Z^(N-1)/2-2Z^N ]/4=/=[X^(N+1)/2+X^(N-1)/2-2X^N ] /4+ [Y^(N+1)/2+Y^(N-1)/2-2Y^N ]/4

AND

IF

[Z^(N+1)/2+Z^(N-1)/2-2Z^N ]/4=[X^(N+1)/2+X^(N-1)/2-2X^N ] /4+ [Y^(N+1)/2+Y^(N-1)/2-2Y^N ]/4

=>

[Z^(N+1)/2+Z^(N-1)/2+2Z^N ]/4=/=[X^(N+1)/2+X^(N-1)/2+2X^N ]/4 + [Y^(N+1)/2+Y^(N-1)/2+2Y^N ]/4

BECAUSE

[Z^(N+1)/2+Z^(N-1)/2+2Z^N ] /4- [Z^(N+1)/2+Z^(N-1)/2-2Z^N ]/4=Z^N.

SO

Z^N=/=X^N+Y^N

HAPPY&PEACE.

Trantancuong. PIERRE DE FERMAT' S LAST THEOREM.

CASE SPECIAL N=3 AND.GENERAL CASE N>2. .

THE CONDITIONS.Z,X,Y,N ARE THE INTEGERS . Z*X*Y*N>0.N>2.

Z^3=/=X^3+Y^3 AND Z^N=/=X^N+Y^N.

SPECIAL CASE N=3.

WE HAVE

(X^2+Y^2)^2=X^4+Y^4+2X^2*Y^2.

BECAUSE

X*Y>0=>2X^2*Y^2>0.

SO

(X^2+Y^2)^2=/=X^4+Y^4.

CASE 1. IF

Z^2=X^2+Y^2

SO

(Z^2)^2=(X^2+Y^2)^2

BECAUSE

(X^+Y^2)^2=/=X^4+Y^4.

SO

(Z^2)^2=/=X^4+Y^4.

SO

Z^4=/=X^4+Y^4.

CASE 2. IF

Z^4=X^4+Y^4

BECAUSE

X^4+Y^4.=/= (X^2+Y^2.)^2

SO

Z^4=/=(X^2+Y^2.)^2

SO

(Z^2)^2=/=(X^2+Y^2.)^2

SO

Z^2=/=X^2+Y^2.

(1) AND (2)=> Z^4+Z^2=/=X^4+Y^4+X^2+Y^2.

SO

2Z^4+2Z^2=/=2X^4+2Y^4+2X^2+Y^2.

SO

(Z^4+Z^2+2Z^3+Z^4+Z^2-2Z^3)=/=(X^4+X^2+2X^3+X^4+X^2-2X^3)+)(Y^4+Y^2+2Y^3+Y^4+Y^2-2Y^3)

SO IF

(Z^4+Z^2+2Z^3)/4=(Z^4+Z^2+2Z^3)/4+(Z^4+Z^2+2Z^3)/4

=> (Z^4+Z^2-2Z^3)/4=/=(Z^4+Z^2-2Z^3)/4+(Z^4+Z^2-2Z^3/4)

AND

SO IF

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

=> (Z^4+Z^2+2Z^3)/4=/=(Z^4+Z^2+2Z^3)/4+(Z^4+Z^2+2Z^3)/4

BECAUSE

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

SO

Z^3=/=X^3+Y^3.

GENERAL CASE N>2.

Z^N=/=X^N+Y^N.

WE HAVE

[X^(N-1)/2+Y^(N-1)/2]^(N+1)/(N-1)=X^(N+1)/2+Y^(N+1)/2+ H.

BECAUSE X*Y>0=>H>0.

SO

[X^(N-1)/2+Y^(N-1)/2]^(N+1)/(N-1)=/= X^(N+1)/2+Y^(N+1)/2

CASE 1. IF

Z^(N-1)/2=X^(N-1)/2+Y^(N-1)/2

SO

[Z^(N-1)/2]^(N+1)/(N-1)=[X^(N-1)/2+Y^(N-1)/2 ]^(N+1)/(N-1).

BECAUSE

[X^(N-1)/2+Y^(N-1)/2 ]^(N+1)/(N-1)=/=X^(N+1)/2+Y(N+1)/2.

SO

[Z^(N-1)/2]^(N+1)/(N-1)=/=X^(N+1)/2+Y(N+1)/2.

SO

Z^(N+1)/2=/=X^(N+1)/2+Y^(N+1)/2.

CASE 2. IF

Z^(N+1)/2=X^(N+1)/2+Y^(N+1)/2

SO

[Z^(N+1)/2]^(N-1)/(N+1)=[X^(N+1)/2+Y^(N+1)/2 ]^(N-1)/(N+1)

BECAUSE

[X^(N+1)/2+Y^(N+1)/2](N-1)/(N+1)=/=X(N-1)/2+Y^(N-1)/2.

SO

[Z^(N+1)/2]^(N-1)/(N+1)=/=X(N-1)/2+Y^(N-1)/2.

SO

Z^(N-1)/2=/=X(N-1)/2+Y^(N-1)/2..

SO

(1) AND (2)=> Z^(N+1)/2+Z^(N-1)/2=/=X^(N+1)/2+Y^(N+1)/2+X^(N-1)/2+Y^(N-1)/2.

SO

2[Z^(N+1)/2+Z^(N-1)/2]=/=2[X^(N+1)/2+Y^(N+1)/2]+2[X^(N-1)/2+Y^(N-1)/2.]

SO

[Z^(N+1)/2+Z^(N-1)/2+2Z^N ]+[Z^(N+1)/2+Z^(N-1)/2-2Z^N ]=/=[X^(N+1)/2+X^(N-1)/2+2X^N ]+[X^(N+1)/2+X^(N-1)/2-2X^N ]+[Y^(N+1)/2+Y^(N-1)/2+2Y^N ]+[Y^(N+1)/2+Y^(N-1)/2-2Y^N ]

SO IF

[Z^(N+1)/2+Z^(N-1)/2+2Z^N ]/4=[X^(N+1)/2+X^(N-1)/2+2X^N ] /4+ [Y^(N+1)/2+Y^(N-1)/2+2Y^N ]/4=>

[Z^(N+1)/2+Z^(N-1)/2-2Z^N ]/4=/=[X^(N+1)/2+X^(N-1)/2-2X^N ] /4+ [Y^(N+1)/2+Y^(N-1)/2-2Y^N ]/4

AND

IF

[Z^(N+1)/2+Z^(N-1)/2-2Z^N ]/4=[X^(N+1)/2+X^(N-1)/2-2X^N ] /4+ [Y^(N+1)/2+Y^(N-1)/2-2Y^N ]/4

=>

[Z^(N+1)/2+Z^(N-1)/2+2Z^N ]/4=/=[X^(N+1)/2+X^(N-1)/2+2X^N ]/4 + [Y^(N+1)/2+Y^(N-1)/2+2Y^N ]/4

BECAUSE

[Z^(N+1)/2+Z^(N-1)/2+2Z^N ] /4- [Z^(N+1)/2+Z^(N-1)/2-2Z^N ]/4=Z^N.

SO

Z^N=/=X^N+Y^N

HAPPY&PEACE.

Trantancuong.

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Wiki User

 ∙ 12y ago

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