Our proofs are perfect and plain.
All_Pythagorean_triples.pdf4_Color_proof_and_FLT_proof.pdfNatural_numbers_and_two_each_methods_about_FLT_proof.pdf

   Our proofs are perfect and plain. Andrew Wiles`s FLT proof is a guess.
   Thanks.
Sincerely yours. Jae Yul Lee and You Jin Lee.
appendix pdf file
http://cafe.daum.net/leejaeyul5
4 Color Theorem proof of the regions on global surface
[1] 4 colors suffice for the distinguishing all regions about one region and the regions those share a common one region`s boundary line. Here, one region can contain all shapes many regions.
[Proof] The reason is this. 3 colors suffice for the distinguishing all regions those share a common one region`s boundary line.
[2] 3 colors suffice for the distinguishing all regions those share a common one region`s boundary line.
[Proof] The reason is this. When there are lined from a region`s inner one point to the points on one region`s boundary line those are met the adjacent regions boundary lines, all extension regions are the regions those share one common point. 3 colors suffice for the distinguishing all regions those share one common point.
[3] 3 colors suffice for the distinguishing all regions those share one common point.
[Proof] The reason is this. When one region is selected, 2 colors suffice for the distinguishing all regions those share a common one selected region`s boundary line.
Two each methods about FLT proof
Xn+Yn=Zn
A=Z-Y, B=Z-X
X=G(AB)1/n+A, Y=G(AB)1/n+B, Z=G(AB)1/n+A+B, X+Y-Z=G(AB)1/n
{G(AB)1/n+A}n+{G(AB)1/n+B}n={G(AB)1/n+A+B}n
When n=1, G=0 and when n=2, G=21/2>0 and when n=3, G=Function(A,B)>0.
X=(2AB)1/2+A, Y=(2AB)1/2+B, Z=(2AB)1/2+A+B
2AB=k2(k is 1,2,3...), XY=k(k+A)(k+2A)/2A=k(k+B)(k+2B)/2B
All Pythagorean triples cannot be the power numbers.
1st method about Fermat's Last Theorem proof
 Xn+Yn=Zn
 (Xn/2)2+(Yn/2)2=(Zn/2)2
a=Zn/2-Yn/2, b=Zn/2-Xn/2
{G(ab)1/2+a}2+{G(ab)1/2+b}2={G(ab)1/2+a+b}2
G=21/2>0, Xn/2=(2ab)1/2+a, Yn/2=(2ab)1/2+b, Zn/2=(2ab)1/2+a+b
(XY)n=ab{2a2+2b2+13ab+6(a+b)(2ab)1/2
The numbers (X,Y,Z) need to be co-prime and the number (n) is the odd and prime number.
2nd method about Fermat's Last Theorem proof
{G(AB)1/n+A}n+{G(AB)1/n+B}n={G(AB)1/n+A+B}n
When A=B, 2{G+A(n-2)/n}n={G+2A(n-2)/n}n. 
G={2(n-2)/n+…+21/n+1}{2A(n-2)}1/n=[{2(n-2)/n+…+21/n+1}n{2A(n-2)}]1/n
We make a numerical formula. {2(n-2)/n+…+21/n+1}[{2A(n-1)B}1/n+{2AB(n-1)}1/n]
q=2G(AB)1/n/{2(n-2)/n+…+21/n+1}[{2A(n-1)B}1/n+{2AB(n-1)}1/n]
G(AB)1/n=q{2(n-2)/n+…+21/n+1}[{2A(n-1)B}1/n+{2AB(n-1)}1/n]/2
when A=B, q must be 1.
   If the figure {G(AB)1/n} can be the natural number in some natural numbers (A,B), the formula {2G(AB)1/n} cannot have the figure {2(n-2)/n+…+21/n+1}. But the numerical formula that we made {2(n-2)/n+…+21/n+1}[{2A(n-1)B}1/n+{2AB(n-1)}1/n], have the figure {2(n-2)/n+…+21/n+1}. And when A=B, q=2G(AB)1/n/{2(n-2)/n+…+21/n+1}(21/nA). So, when A=B, the figure (q) cannot be 1. This is an apparent contradiction. Therefore, the figure {G(AB)1/n} cannot be the natural number in the natural numbers (A,B).
by 이재율 | 2008/05/11 22:42 | 트랙백
Our proof is perfect and plain

Our proof is perfect and plain. Andrew Wiles`s FLT proof is a guess and his proof is not plain.
 Thanks.

Natural numbers and two each methods about Fermat's Last Theorem proof
[Jae Yul Lee and You Jin Lee]
Abstract
   The natural numbers (X,Y,Z) are known as the Pythagorean triples in the equation X2+Y2=Z2. When the number (n) is greater or equal 3 in the equation Xn+Yn=Zn, the equation cannot have non zero integer solutions. The truth is known as the Fermat's Last Theorem. The Fermat had written that he had found out the proof, but nobody could see his proof. We have considered the natural numbers (A=Z-Y) and (B=Z-X) in the equation Xn+Yn=Zn and we have found out two each methods of the Fermat's Last Theorem proof.
Key Words and Phrases
MSC : 11-A99  Number Theory
Xn+Yn=Zn
A=Z-Y, B=Z-X
X=G(AB)1/n+A, Y=G(AB)1/n+B, Z=G(AB)1/n+A+B, X+Y-Z=G(AB)1/n
{G(AB)1/n+A}n+{G(AB)1/n+B}n={G(AB)1/n+A+B}n
When n=1, G=0 and when n=2, G=21/2>0 and when n=3, G=Function(A,B)>0.
X=(2AB)1/2+A, Y=(2AB)1/2+B, Z=(2AB)1/2+A+B
2AB=k2(k is 1,2,3...)
XY=k(k+A)(k+2A)/2A=k(k+B)(k+2B)/2B
All Pythagorean triples cannot be the power numbers.
(Xn/2)2+(Yn/2)2=(Zn/2)2
a=Zn/2-Yn/2, b=Zn/2-Xn/2
(XY)n=ab{2a2+2b2+13ab+6(a+b)(2ab)1/2}
When A=B, G={2(n-2)/n+…+21/n+1}{2A(n-2)}1/n.
We make a numerical formula with the figure {2(n-2)/n+…+21/n+1}.
{2(n-2)/n+…+21/n+1}[{2A(n-1)B}1/n+{2AB(n-1)}1/n]
q=2G(AB)1/n/{2(n-2)/n+…+21/n+1}[{2A(n-1)B}1/n+{2AB(n-1)}1/n]
G(AB)1/n=q{2(n-2)/n+…+21/n+1}[{2A(n-1)B}1/n+{2AB(n-1)}1/n]
When A=B, q=1.
q=2G(AB)1/n/{2(n-2)/n+…+21/n+1}(21/nA)
When the numbers (A,B) are the natural numbers,
the figure {G(AB)}1/n cannot be the natural number.
Sentence
1. Preface
   What the equation Xn+Yn=Zn cannot have non zero integer solutions and what the equation Xn+Yn=Zn cannot have the natural number solutions are equivalent in meaning.
   When n=0, the equation is X0+Y0=Z0. This equation means nothing.
   When n=1, the equation is X+Y=Z. Every natural numbers (X,Y) make some natural number (Z).
   When n=2, the equation is X2+Y2=Z2. The natural numbers (X,Y,Z) are the Pythagorean triples in the equation X2+Y2=Z2.
   When the number (n) is greater or equal 3, the equation Xn+Yn=Zn cannot have non zero integer solutions. This truth is known as the Fermat's Last Theorem.
   In the equation Xn+Yn=Zn, when the numbers (X,Y,Z) are the natural numbers, the numbers (A=Z-Y) and (B=Z-X) are also the natural numbers.
2. General
2-1. What the equation Xn+Yn=Zn cannot have non zero integer solutions and what the equation Xn+Y=Zn cannot have the natural number solutions are equivalent in meaning.
[Explanation] 
   When the number (n) is the even number,
(-U)n+Vn=Wn
Un+Vn=Wn.
   When the number (n) is the odd number,
(-U)n+Vn=Wn
-Un+Vn=Wn
Wn+Un=Vn.
2-2. When the equation Xn+Yn=Zn can have some natural number solutions, the natural numbers (X,Y,Z) need to be co-prime.
[Explanation] 
U=QX, V=QY, W=QZ.
And
the natural numbers (X,Y,Z) are co-prime.
Un+Vn=Wn
(QX)n+(QY)n=(QZ)n
Qn(Xn+Yn=Zn)
When
the number (Q) is the natural number,
the number (Qn) is also the natural number.
3. Introduction
3-1. When A=Z-Y and B=Z-X in the equation Xn+Yn=Zn, we can find the equation {G(AB)1/n+A}n+{G(AB)1/n+B}n={G(AB)1/n+A+B}n.
[Explanation]
 Xn+Yn=Zn
A=Z-Y, B=Z-X
Y+A=X+B=Z
X-A=Y-B=Z-A-B=X+Y-Z
And
(X-A)/(AB)1/n=(Y-B)/(AB)1/n=(Z-A-B)/(AB)1/n=(X+Y-Z)/(AB)1/n
This is G.
G=(X-A)/(AB)1n=(Y-B)/(AB)1/n=(Z-A-B)/(AB)1/n=(X+Y-Z)/(AB)1/n
So,
X=G(AB)1/n+A, Y=G(AB)1/n+B, Z=G(AB)1/n+A+B, X+Y-Z=G(AB)1/n
Therefore,
{G(AB)1/n+A}n+{G(AB)1/n+B}n={G(AB)1/n+A+B}n
3-2. When n=1, G=0 and when n=2, G=21/2>0 and when n=3, G=Function(A,B)>0.
[Explanation] 
   When n=1, G=0.
(X+Y)=Z
X+Y-Z=G(AB)=0, G=0.  
{G(AB)+A}+{G(AB)+B}={G(AB)+A+B}
G(AB)=0
G=0
   When n=2, G=21/2>0.
(X2+Y2)>Z2
X+Y-Z=G(AB)1/2>0, G>0.
{G(AB)1/2+A}2+{G(AB)1/2+B}2={G(AB)1/2+A+B}2
G2(AB)=2AB
G=21/2>0
   When n=3, G=Function(A,B)>0.
(X3+Y3)>Z3
X+Y-Z=G(AB)1/3>0, G>0.
{G(AB)1/3+A}3+{G(AB)1/3+B}3={G(AB)1/3+A+B}3
G=Function(A,B)>0
Generally,
we cannot find the solutions to the figure {G(AB)1/3}.
4. 1st method about Fermat's Last Theorem proof
4-1. Transformation
 Xn+Yn=Zn
 (Xn/2)2+(Yn/2)2=(Zn/2)2
a=Zn/2-Yn/2, b=Zn/2-Xn/2
{G(ab)1/2+a}2+{G(ab)1/2+b}2={G(ab)1/2+a+b}2
G=21/2>0
Xn/2=(2ab)1/2+a, Yn/2=(2ab)1/2+b, Zn/2=(2ab)1/2+a+b
4-2. When the number (n) is the even number in the equation Xn+Yn=Zn, the numbers (X,Y,Z) cannot be the natural numbers.
4-2-1. All Pythagorean triples.
[Explanation]
Xn+Yn=Zn
A=Z-Y, B=Z-X
X=(2AB)1/2+A, Y=(2AB)1/2+B, Z=(2AB)1/2+A+B
   When the numbers (X,Y,Z) are the natural numbers,
the numbers (A,B) are also the natural numbers.
Therefore, 
when 2AB=k2(k is 1,2,3...),
the natural numbers (X,Y,Z) are all Pythagorean triples.
4-2-2. All Pythagorean triples cannot be the power numbers.
[Explanation]
X=(2AB)1/2+A, Y=(2AB)1/2+B, Z=(2AB)1/2+A+B
2AB=k2(k is 1,2,3...)
X=k+A=k(k+2B)/2B, Y=k+B=k(k+2A)/2A
And
XY=k(k+A)(k+2A)/2A=k(k+B)(k+2B)/2B
   So,
the figure (XY) cannot be the power number in co-prime (A,B).
Therefore, 
all Pythagorean triples cannot be the power numbers.
4-2-3. When the number (n) is the even number, the numbers (Xm,Ym,Zm) cannot be the natural numbers. So, the numbers (X,Y,Z) cannot be the natural numbers.
[Explanation]
Xn+Yn=Zn
(Xn/2)2+(Yn/2)2=(Zn/2)2
 When n=2m,
(Xm)2+(Ym)2=(Zm)2.
   The natural numbers (Xm,Ym,Zm) need to be the Pythagorean triples,
but all Pythagorean triples cannot be the power numbers.
So, 
the numbers (Xm,Ym,Zm) cannot be the natural numbers.
Therefore,
the numbers (X,Y,Z) cannot be the natural numbers.
4-3. When the number (n) is the odd number in the equation Xn+Yn=Zn, the numbers (X,Y,Z) cannot be the natural numbers.
4-3-1. The number (n) needs to be the odd and prime number.
[Explanation]
Xr+Yr=Zr
When 
r=nop..(n,o,p.. are the prime numbers.),
(Xop..)n+(Yop..)n=(Zop..)n.
When the numbers (X,Y,Z) are co-prime,
the numbers (Xop..,Yop..,Zop..) are also co-prime.
Therefore,
when the numbers (X,Y,Z) are co-prime,
 the number (n) needs to be the odd and prime number.
4-3-2. The figure (XY)n is the natural number in the natural numbers (X,Y,Z), but the figure [ab{2a2+2b2+13ab+6(a+b)(2ab)1/2}] cannot be the natural number in the natural numbers (X,Y,Z). This is an apparent contradiction. Therefore, the numbers (X,Y,Z) cannot be the natural numbers, when the number (n) is the odd number.
[Explanation]
Xn+Yn=Zn
(Xn/2)2+(Yn/2)2=(Zn/2)2
a=Zn/2-Yn/2, b=Zn/2-Xn/2
Xn/2=(2ab)1/2+a, Yn/2=(2ab)1/2+b, Zn/2=(2ab)1/2+a+b
(XY)n/2={(2ab)1/2+a}{(2ab)1/2+b}
Here,
the natural numbers (X,Y,Z) are co-prime
and the number (n) is the odd and prime number.
(XY)n=ab{2a2+2b2+13ab+6(a+b)(2ab)1/2}
=(Zn/2-Xn/2)(Zn/2-Yn/2){2Xn+2Yn+17Zn-17(ZX)n/2-17(ZY)n/2+13(XY)n/2+6(21/2)(2Zn/2-Xn/2-Yn/2)(Zn/2-Xn/2)1/2(Zn/2-Yn/2)1/2}
The figure (XY)n is the natural number in the natural numbers (X,Y,Z).
But 
the figure [ab{2a2+2b2+13ab+6(a+b)(2ab)1/2}]
cannot be the natural number in the natural numbers (X,Y,Z).
This is an apparent contradiction.
Therefore, 
the numbers (X,Y,Z) cannot be the natural numbers.
4-4. Therefore, the equation Xn+Yn=Zn cannot have the natural number solutions in the even number (n) and in the odd number (n).
5. 2nd method about Fermat's Last Theorem proof
5-1. Transformation
Xn+Yn=Zn
A=Z-Y, B=Z-X
X+Y-Z=G(AB)1/n>0
{G(AB)1/n+A}n+{G(AB)1/n+B}n={G(AB)1/n+A+B}n
5-2. In the equation {G(AB)1/n+A}n+{G(AB)1/n+B}n={G(AB)1/n+A+B}n, when the numbers (A,B) are the natural numbers, the figure {G(AB)1/n} cannot be the natural number.
5-2-1. When the numbers (A,B) are the natural numbers, one each figure {G(AB)1/n} is the positive number in the equation {G(AB)1/n+A}n+{G(AB)1/n+B}n={G(AB)1/n+A+B}n.
[Explanation] 
{G(AB)1/n+A}n+{G(AB)1/n+B}n={G(AB)1/n+A+B}n
   Generally,
we cannot solve this equation,
but 
we see that we can have n each solutions
to the figure {G(AB)1/n} in this equation.
When
the numbers (A,B) are the natural numbers,
one each figure {G(AB)1/n} is the positive number
and
(n-1) each figures {G(AB)1/n} cannot be the positive numbers.
5-2-2. The figure [G={2(n-2)/n+…+21/n+1}{2A(n-2)}1/n>0] cannot be the natural number in the equation {G(AB)1/n+A}n+{G(AB)1/n+B}n={G(AB)1/n+A+B}n , when the number (A) is the natural number and (A=B).
[Explanation]
{G(AB)1/n+A}n+{G(AB)1/n+B}n={G(AB)1/n+A+B}n