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Can you tell how to submit an attempted proof of Fermat's Last Theorem?
Wiki User
∙ 11y ago
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.
Wiki User
∙ 11y ago
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