Talk:Pythagorean triple
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2 methods of FLT proof and Pythagorean triples X^n+Y^n=Z^n Fermat had made a proof that the equation cannot have nonzero natural number solution in the even number n that is greater or equal 4. Therefore we need to make a proof that the equation cannot have nonzero natural number solution in the odd and prime number n. Y+A=X+B=Z A=Z-Y, B=Z-X X-A=Y-B=Z-A-B=X+Y-Z G=(X-A)/(AB)^(1/n)=(Y-B)/(AB)^(1/n)=(Z-A-B)/(AB)^(1/n)=(X+Y-Z)/(AB)^(1/n) X=G(AB)^(1/n)+A, Y=G(AB)^(1/n)+B, Z=G(AB)^(1/n)+A+B {G(AB)^(1/n)+A}^n+{G(AB)^(1/n)+B}^n={G(AB)^(1/n)+A+B}^n When n=1, G=0, when n=2, G=2^(1/2), when n>2, G=function(A,B) is the positive real number. X=(2AB)^(1/2)+A, Y=(2AB)^(1/2)+B, Z=(2AB)^(1/2)+A+B (X,Y,Z) are the irrational numbers or Pythagorean triples in natural number (A,B). We can translate the upper form into this. AB=2k^2,B=2k^2/A, k=hA, B=2Ah^2 X=A(2h+1), Y=2Ah(h+1), Z=A(2h^2+2h+1) XY=2A^2h(2h+1)(h+1) can not be the power numbers in natural number (A,h).
- * * * * 1st method of FLT proof * * * * *
G(AB)^(1/n) is the irrational number in natural number (A,B), so (X,Y,Z) are the irrational numbers. {G(AB)^(1/n)+A}^n+{G(AB)^(1/n)+B}^n={G(AB)^(1/n)+A+B}^n When A=B, 2{GA^(2/n)+A}^n={GA^(2/n)+2A}^n {2^(1/n)-1}GA^(2/n)={2-2^(1/n)}A G=[2^{(n-1)/n}+…+2^(2/n)+2^(1/n)]A^{(n-2)/n} At least (n-2) each terms are the irrational numbers in (n-1) each terms. We make new form with [2^{(n-1)/n}+…+2^(2/n)+2^(1/n)]. [2^{(n-1)/n}+…+2^(2/n)+2^(1/n)][{A^(n-1)B}^(1/n)+{AB^(n-1)}^(1/n)]/2 This form is the irrational number in natural number (A,B). [{2^{(n-1)A^(n-1)B}^(1/n)+…+{2^2A^(n-1)B}^(1/n)+{2A^(n-1)B}^(1/n)+ {2^{(n-1)AB^(n-1)}^(1/n)+…+{2^2AB^(n-1)}^(1/n)+{2AB^(n-1)}^(1/n)}/2 At least 2(n-2) each terms are the irrational numbers in 2(n-1) each terms. We divide and multiply by [2^{(n-1)/n}+…+2^(2/n)+2^(1/n)][{A^(n-1)B}^(1/n)+{AB^(n-1)}^(1/n)]/2 to G(AB)^(1/n), and we can get two forms. And when A=B, q=1. G(AB)^(1/n)=q[2^{(n-1)/n}+…+2^(2/n)+2^(1/n)][{A^(n-1)B}^(1/n)+{AB^(n-1)}^(1/n)]/2 q=2G(AB)^(1/n)/[2^{(n-1)/n}+…+2^(2/n)+2^(1/n)][{A^(n-1)B}^(1/n)+{AB^(n-1)}^(1/n)] If G(AB)^(1/n) is the natural number (N) in some (a,b), when A=B, q can not be 1. That is an apparent contradiction. G(ab)^(1/n)=N can not have [2^{(n-1)/n}+…+2^(2/n)+2^(1/n)], so G(AB)^(1/n) can not have [2^{(n-1)/n}+…+2^(2/n)+2^(1/n)]. 1st method. end.
- * * * * 2nd method of FLT proof * * * * *
X^n+Y^n=Z^n {X^(n/2)}^2+{Y^(n/2)}^2={Z^(n/2)}^2 When n=2, we can display {X^(n/2),Y^(n/2),Z^(n/2)} with (a,b). a=Z^(n/2)-Y^(n/2), b=Z^(n/2)-X^(n/2) X^(n/2)=(2ab)^(1/2)+a, Y^(n/2)=(2ab)^(1/2)+b, Z^(n/2)=(2ab)^(1/2)+a+b When n is the prime number and (X,Y,Z) is co prime, the ab is the irrational number. ab=Z^n-(YZ)^(n/2)-(XZ)^(n/2)+(XY)^(n/2) We multiply X^(n/2) and Y^(n/2). (XY)^n=2a^3b+2ab^3+13(ab)^2+6ab(a+b)(2ab)^(1/2) If (X,Y,Z) is the natural number, (XY)^n is the natural number, but 2a^3b+2ab^3+13(ab)^2+6ab(a+b)(2ab)^(1/2) is the irrational number. That is an apparent contradiction. Therefore (X,Y,Z) must be the irrational number. 2nd method. end.
