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PIERRE DE FERMAT' S LAST THEOREM.

CASE SPECIAL N=3 .

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

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

Happy&Peace.

Trantancuong.

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More answers

The equation a^n+b^n=c^n, n>2 , where n is a positive integer , has no non trivial integer solution (a,b,c).

First by Andrew Wiles-Very difficult proof

Now simple proofs are available;A simple and short analytical proof of Fermat's last theorem, CNMSEM,Vol.2,No.3 March 2011 pg.57-63

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12y ago
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Andrew Wiles did.

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13y ago
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