leejaeyul5@yahoo.co.kr http://blog.empas.com/leejaeyul5 이재율(02-882-0830) [완전한 피타고라스 수 공식을 발견치 못하여, 불완전한 공식들을 사용 중이고, 페르마 정리를 증명치 못하여, 170 쪽 방대 난해한 비일반 추측 증명뿐이며, 4색문제를 증명치 못하여, 인간논리 아닌, 컴퓨터 의존 미확인 증명뿐입니다. 본인과 협조인 들은 오랜 연구 노력으로, 이들에 대한 간명한 증명을 발견하고, 이 논문들을 정부 교육 학계에 제안하고 있습니다.] 논문1. Pythagoras FLT proof : 피타고라스는 2580 년, 페르마는 400 년 전에 출생 하였습니다. 아마추어였으나 훌륭한 수학자로 인정받는 페르마는, 경이로운 증명 발견 사실을 분명히 기록하여 두었습니다. 대다수의 현대 수학자들이 그를 믿지 않고 있었으나, 우리는 그를 믿고 있으며, 분명하게 증명하여 제시합니다. 논문2. 4 color problem proof : 모든 나라들이 임의의 한 점에 접하도록 그려 보면, 이 나라들이 3가지 이하의 색으로 충분하게 구분이 됨으로서, 4색 문제가 증명이 될 수 있는 것입니다. 한 점을 공유하는 모든 나라가 3색으로 구분됨으로, 한 나라의 국경선 일부를 공유하는 주변의 모든 나라들도 역시 3색으로 충분하게 구분될 수가 있기 때문입니다. 참고. 우리 논문의 첨부 파일은 이재율-엠파스카페, 다음카페, 야후클럽, 네이버블로그, 네이버카페, 프리첼섬과 대한수학회 토론방 등에도 있습니다. Pythagorean numbers and Fermat s Last Theorem proof Abstract The Pythagorean Theorem and the Pythagorean numbers are well known things. In the equation X^2+Y^2=Z^2, the numbers of X, Y, Z can be the natural numbers known as the Pythagorean numbers. But anybody seemed not analyze the imperfect forms about the Pythagorean numbers carefully. In the equation X^n+Y^n=Z^n, when n be greater or equal 3, the equation can not have non zero integer solutions. The fact was also well known thing as the Fermat s Last Theorem. The Fermat had written what he found out the proof. But anybody could not find out his proof, so nobody could know his proof. We have considered A=Z-Y, B=Z-X in the equation and found out one new perfect form about the Pythagorean numbers and a new simple and plain proof about the Fermat s Last Theorem. _________________________________________________ Key Words and Phrases MSC : 11-A99 Number Theory Y+A=X+B=Z. 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=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 X+Y-Z=G(AB)^(1/n)= q[{2^(n-1)/n+…+2^(2/n)+2^(1/n)}/2][{A^(n-1)B}^(1/n)+{AB^(n-1)}^(1/n)] Sentence 1. Preface In the equation X^n+Y^n=Z^n, n be the natural numbers. When n=1, the equation be X+Y=Z. When n=2, the equation be X^2+Y^2=Z^2 and the numbers of X, Y, Z can become the natural numbers known as the Pythagorean numbers. But when n be greater or equal 3, the equation X^n+Y^n=Z^n can not have non zero integers solutions. That is known as the Fermat s Last Theorem. 2. General 2-1. It is generally acknowledged to be true what the equation X^n+Y^n=Z^n can not have non zero integer solutions and what the equation X^n+Y^n=Z^n can not have non zero natural number solutions are equivalent in meaning. 2-1-1. When n be even numbers, what the equation X^n+Y^n=Z^n can not have non zero integer solutions and what the equation X^n+Y^n=Z^n can not have non zero natural number solutions are equivalent in meaning. When n be even numbers and U, V, W are natural numbers, (-U)^n+V^n=W^n U^n+V^n=W^n 2-1-2. When n be odd numbers, what the equation X^n+Y^n=Z^n can not have non zero integer solutions and what the equation X^n+Y^n=Z^n can not have non zero natural number solutions are equivalent in meaning. When n be odd numbers and U, V, W are natural numbers, (-U)^n+V^n=W^n -U^n+V^n=W^n W^n+U^n=V^n 2-2. Therefore what the equation X^n+Y^n=Z^n can not have non zero integer solutions be what the equation X^n+Y^n=Z^n can not have non zero natural number solutions. 3. Introduction 3-1. In the equation X^n+Y^n=Z^n, the mutual relation forms between A, B and X, Y, Z are A=Z-Y, B=Z-X, Y+A=X+B=Z, X-A=Y-B=Z-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, etc. In the equation X^n+Y^n=Z^n the numbers of X, Y, Z be regarded as non zero natural numbers. Y+A=X+B=Z The numbers of A, B must become non zero natural numbers. A=Z-Y, B=Z-X Therefore X-A=Y-B=Z-A-B=X+Y-Z So (X-A)/(AB)^(1/n)=(Y-B)/(AB)^(1/n)=(Z-A-B)/(AB)^(1/n) This be G. G=(X-A)/(AB)^(1/n)=(Y-B)/(AB)^(1/n)=(Z-A-B)/(AB)^(1/n) And X=G(AB)^(1/n)+A, Y=G(AB)^(1/n)+B, Z=G(AB)^(1/n)+A+B Therefore X+Y-Z=G(AB)^(1/n) and {G(AB)^(1/n)+A}^n+{G(AB)^(1/n)+B}^n={G(AB)^(1/n)+A+B}^n 350 years ago, the Fermat wrote what the proof was beautiful and wonderful thing. We believe in the Fermat. 3-2. When n=1, G=0 and when n=2, G=G=2^(1/2) but when n=3, G=F(A,B) 3-2-1. When n=1, G=0 G(AB)^(1/n)=0 (0+A)+(0+B)=(0+A+B) X=A, Y=B, Z=A+B 3-2-2. When n=2, G=2^(1/2) G(AB)^(1/n)=(2AB)^(1/2) {(2AB)^(1/2)+A}^2+{(2AB)^(1/2)+B}^2={(2AB)^(1/2)+A+B}^2 X=(2AB)^(1/2)+A, Y=(2AB)^(1/2)+B, Z=(2AB)^(1/2)+A+B 4. Pythagorean numbers form 4-1. This be one new perfect form about the Pythagorean numbers. X=(2AB)^(1/2)+A, Y=(2AB)^(1/2)+B, Z=(2AB)^(1/2)+A+B 4-1-1. A=Z-Y, B=Z-X 4-1-2. X-A=Y-B=Z-A-B=X+Y-Z=(2AB)^(1/2) 4-2. These were 3 imperfect forms about the Pythagorean numbers. First X=2A+1, Y=2A^2+2A, Z=2A^2+2A+1 Second X=4A, Y=4A^2-1, Z=4A^2+1 Third X=A^2-B^2, Y=2AB, Z=A^2+B^2 4-2-1. The inducement of the forms be uncertain. 4-2-2. The mutual relation forms between A, B and X, Y, Z are uncertain. 4-2-3. The irrational numbers of A, B are needed for some numbers of X, Y, Z as the Pythagorean numbers. 4-2-4. The changed forms between X and Y are needed for some numbers of X, Y, Z as the Pythagorean numbers. 5. Fermat s Last Theorem proof 5-1. In the equation X^n+Y^n=Z^n, when n be greater or equal 3, the equation be {G(AB)^(1/n)+A}^n+{G(AB)^(1/n)+B}^n={G(AB)^(1/n)+A+B}^n When n be greater or equal 3 X^n+Y^n=Z^n A=Z-Y, B=Z-X {G(AB)^(1/n)+A}^n+{G(AB)^(1/n)+B}^n={G(AB)^(1/n)+A+B}^n G=F(A,B) One G=F(A,B) has the positive numbers but (n-1) each G=F(A,B) have the non positive numbers. 5-2. When A=B, G=F(A)=F(B). This be one positive irrational numbers G. G=F(A)={2^(n-1)/n+…+2^(2/n)+2^(1/n)}A^(n-2)/n In {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 (n-1) each G=F(A) have the non positive irrational numbers but one G=F(A) has the positive irrational numbers. This be one positive irrational numbers G. G=F(A)={2^(n-1)/n+…+2^(2/n)+2^(1/n)}A^(n-2)/n 5-3. One positive numbers G=F(A,B) and G(AB)^(1/n)=F(A,B)(AB)^(1/n) We can divide and multiply by 2{2^(n-1)/n+…+2^(2/n)+2^(1/n)}{A^(n-2)/n+B^(n-2)/n} to one positive numbers G=F(A,B) Therefore q=2F(A,B)/{2^(n-1)/n+…+2^(2/n)+2^(1/n)}{A^(n-2)/n+B^(n-2)/n} G=q[{2^(n-1)/n+…+2^(2/n)+2^(1/n)}/2]{A^(n-2)/n+B^(n-2)/n} When A=B, q must become 1. And X+Y-Z=G(AB)^(1/n)= q[{2^(n-1)/n+…+2^(2/n)+2^(1/n)}/2][{A^(n-1)B}^(1/n)+{AB^(n-1)}^(1/n)] 5-3-1. If the numbers of X, Y, Z be non zero natural numbers, the numbers of A, B must become non zero natural numbers. And in all the non zero natural numbers A, B, G(AB)^(1/n)=q[{2^(n-1)/n+…+2^(2/n)+2^(1/n)}/2][{A^(n-1)B}^(1/n)+{AB^(n-1)}^(1/n)] must become the irrational numbers. So the numbers of X=G(AB)^(1/n)+A, Y=G(AB)^(1/n)+B, Z=G(AB)^(1/n)+A+B must become the irrational numbers. That be an apparent contradiction. Therefore the numbers of X, Y, Z can not become non zero natural numbers. 5-3-2. And if the numbers of X, Y, Z become non zero natural numbers by the non positive numbers G, then another non zero natural numbers A, B must come into existence. And another positive numbers G must come into existence. So the numbers of X=G(AB)^(1/n)+A, Y=G(AB)^(1/n)+B, Z=G(AB)^(1/n)+A+B become the irrational numbers by all the non zero natural numbers A, B and another positive numbers G. That be an apparent contradiction. Therefore the numbers of X, Y, Z can not become non zero natural numbers by the non positive numbers G. 5-4. Therefore when n be greater or equal 3, the equation X^n+Y^n=Z^n can not have non zero integer solutions. 6. Conclusion Consequently, in the equation X^n+Y^n=Z^n when n=1 and when n=2, the numbers of X, Y, Z can become non zero integers. But when n be greater or equal 3, the equation X^n+Y^n=Z^n can not have non zero integer solutions. End. References ParkBeumSoo. (2003). Yes we have no neutron. The armchair universe. Alexander Keewatin Dewdney : eclio. Parkyoungsuk. (1998). Out of there mind. Dennis Shasha. Cathy Lazere. Howard eves. Leewooyoung. (2003). Citation and References Mathematics : The science of patterns. Fermat s Last Theorem. Keith Devlin. Barry Cipra. (1996). What s happening in Mathematical Sciences. American Mathematical Society. Leejaeyul. (2005). appendix about FLT. X^n+Y^n=Z^n {X^(n/2)}^2+{Y^(n/2)}^2={Z^(n/2)}^2 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 X^(n/2)Y^(n/2)=3AB+(A+B)(2AB)^1/2 (XY)^n=2A^3B+2AB^3+13(AB)^2+6AB(A+B)(2AB)^1/2