Tags Celebrity Relationships

Animals & Plants Arts & Entertainment Auto Beauty & Health Books and Literature Business Electronics Engineering & Technology Food & Drink History Hobbies Jobs & Education Law & Government Math People & Society Science Social Studies Sports Travel & Places

0

Were to go in n z to see whales?

Updated: 8/16/2019

Wiki User

∙ 15y ago

Best Answer

were do we go to see whales. were do we go to see whales.

Wiki User

∙ 15y ago

This answer is:

Add your answer:

Earn +20 pts

Q: Were to go in n z to see whales?

Write your answer...

Submit

Still have questions?

imp

Related questions

What does a n z skink need?

To be let go

How do you make a z in cursive?

See related links to learn how.

What is the name of the Gonzaga Theme Song?

Yes it's Go Gonzaga G-O-N-Z-A-G-A Go Gonzaga G-O-N-Z-A-G-A and it just repeats like that.

Who is the next villain after boo in Dragon Ball Z?

first of all its spelled buu n there is none after buu dragon ball z ends after buu is reborain as uub n him n goku go off n train

How do you beat 100 Doors Parallel Worlds Level 50?

This level requires some trial and error math to find the code.You have to find C, D, Z, and N so that:C+D=Z*N and Z+N=C*D.Plug in some numbers and you'll find that C = 3, D = 2, Z = 1, and N = 5.See related links for screenshots.

Can you tell how to submit an attempted proof of 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 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.

Who Do solve Fermat in one page?

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.

Who are some geniuses?

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 the 3 capital letters of the alphabet in block style with rotational symmetry?

H, I, N, O, S, X, Z H, I, N, O, S, X, Z H, I, N, O, S, X, Z H, I, N, O, S, X, Z

What is Distributions over the X N(ut)?

Z=N(0,1) and Z^2=X^2

What does A N Z mean on a scratch off lottery ticket?

What dose it mean A N Z and how much

Trending Questions

What is a gamete give example? Whose collection is larger the library or congress or the British library? What does the name gelita mean? Is naive a character trait? Which is the most famous football song in the world? Can a lawyer turn over an unpaid bill to a collection agency? What does 2 to the power of 2 equal? Can a convicted felon own a gun overseas? What economic changes have occurred in Slovakia since becoming an independent nation? How do you write 0.011 in words? What doesTax code K39 mean? How did Cleopatra lose the war and her power? How do you add your avatar to Yahoo Messenger for the Web? What middle school did Jensen Ackles go to? What is a 1906 5 franc worth? What happened to celies real father in the color purple? What do you mean by public private protected and friendly? What is fork oil capacity for a Yamaha virago 1100? What kind of number uses dots? ECGs trace electrical excitation to different parts of the heart?

Still have questions?

imp

Previously Viewed

Were to go in n z to see whales?

Resources

Top Categories

Product

Company

Copyright ©2024 Infospace Holdings LLC, A System1 Company. All Rights Reserved. The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Answers.