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What is the newest Proof about Fermat's last theorem?
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.
What are diagonals in a rhombus?
The equation for diagonals is n(n - 3)/2, and when 4 is replaced for n, this yields 4(4 - 3)/2 = 2(1) = 2.
What is the pattern 2 4 16 3 9 81 4 16 256 5 25 625?
n, n^2, n^2^2, n+1, (n+1)^2, (n+1)^2^2, n+2, ...
Is 6 -4 2 -4 -4 2 4 6 and 2 6 a function?
n-2,4,6-number b-b,c,d-ball coin disc i-a,a,a-name
What is the closed form of a geometric sequence with ratio between consecutive terms 4 and initial term 8?
It is G(n) = 8*4^(n-1) or 2*4^n for n = 1, 2, 3, ...
