In this paper we prove in a new way, the well known result, that Fermat’s equation a<sup>4</sup> + b<sup>4</sup> = c<sup>4</sup>, is not solvable in ℕ , when abc≠0 . To show this ...In this paper we prove in a new way, the well known result, that Fermat’s equation a<sup>4</sup> + b<sup>4</sup> = c<sup>4</sup>, is not solvable in ℕ , when abc≠0 . To show this result, it suffices to prove that: ( F 0 ): a 1 4 + ( 2 s b 1 ) 4 = c 1 4 , is not solvable in ℕ , (where a 1 , b 1 , c 1 ∈2ℕ+1 , pairwise primes, with necessarly 2≤s∈ℕ ). The key idea of our proof is to show that if (F<sub>0</sub>) holds, then there exist α 2 , β 2 , γ 2 ∈2ℕ+1 , such that ( F 1 ): α 2 4 + ( 2 s−1 β 2 ) 4 = γ 2 4 , holds too. From where, one conclude that it is not possible, because if we choose the quantity 2 ≤ s, as minimal in value among all the solutions of ( F 0 ) , then ( α 2 ,2 s−1 β 2 , γ 2 ) is also a solution of Fermat’s type, but with 2≤s−1<s , witch is absurd. To reach such a result, we suppose first that (F<sub>0</sub>) is solvable in ( a 1 ,2 s b 1 , c 1 ) , s ≥ 2 like above;afterwards, proceeding with “Pythagorician divisors”, we creat the notions of “Fermat’s b-absolute divisors”: ( d b , d ′ b ) which it uses hereafter. Then to conclude our proof, we establish the following main theorem: there is an equivalence between (i) and (ii): (i) (F<sub>0</sub>): a 1 4 + ( 2 s b 1 ) 4 = c 1 4 , is solvable in ℕ , with 2≤s∈ℕ , ( a 1 , b 1 , c 1 )∈ ( 2ℕ+1 ) 3 , coprime in pairs. (ii) ∃( a 1 , b 1 , c 1 )∈ ( 2ℕ+1 ) 3 , coprime in pairs, for wich: ∃( b ′ 2 , b 2 , b ″ 2 )∈ ( 2ℕ+1 ) 3 coprime in pairs, and 2≤s∈ℕ , checking b 1 = b ′ 2 b 2 b ″ 2 , and such that for notations: S=s−λ( s−1 ) , with λ∈{ 0,1 } defined by c 1 − a 1 2 ≡λ( mod2 ) , d b =gcd( 2 s b 1 , c 1 − a 1 )= 2 S b 2 and d ′ b = 2 s−S b ′ 2 = 2 s B 2 d b , where ( 2 s B 2 ) 2 =gcd( b 1 2 , c 1 2 − a 1 2 ) , the following system is checked: { c 1 − a 1 = d b 4 2 2+λ = 2 2−λ ( 2 S−1 b 2 ) 4 c 1 + a 1 = 2 1+λ d ′ b 4 = 2 1+λ ( 2 s−S b ′ 2 ) 4 c 1 2 + a 1 2 =2 b ″ 2 4;and this system implies: ( b 1−λ,2 4 ) 2 + ( 2 4s−3 b λ,2 4 ) 2 = ( b ″ 2 2 ) 2;where: ( b 1−λ,2 , b λ,2 , b ″ 2 )={ ( b ′ 2 , b 2 , b ″ 2 ) if λ=0 ( b 2 , b ′ 2 , b ″ 2 ) if λ=1;From where, it is quite easy to conclude, following the method explained above, and which thus closes, part I, of this article. .展开更多
The most interesting and famous problem that puzzled the mathematicians all around the world is much likely to be the Fermat’s Last Theorem. However, since the Theorem was proposed, people can’t find a way to solve ...The most interesting and famous problem that puzzled the mathematicians all around the world is much likely to be the Fermat’s Last Theorem. However, since the Theorem was proposed, people can’t find a way to solve the problem until Andrew Wiles proved the Fermat’s Last Theorem through a very difficult method called Modular elliptic curves in 1995. In this paper, I firstly constructed a geometric method to prove Fermat’s Last Theorem, and in this way we can easily get the conclusion below: If a and b are integer and?a = b, n ∈ Q and n > 1, the value of c satisfies the function an + bn = cn that can never be integer;if a, b and c are integer and a ≠ b, n is integer and n > 2, the function an + bn = cn cannot be established.展开更多
Considering Pythagorician divisors theory which leads to a new parameterization, for Pythagorician triplets ( a,b,c )∈ ℕ 3∗ , we give a new proof of the well-known problem of these particular squareless numbers n∈ ℕ...Considering Pythagorician divisors theory which leads to a new parameterization, for Pythagorician triplets ( a,b,c )∈ ℕ 3∗ , we give a new proof of the well-known problem of these particular squareless numbers n∈ ℕ ∗ , called congruent numbers, characterized by the fact that there exists a right-angled triangle with rational sides: ( A α ) 2 + ( B β ) 2 = ( C γ ) 2 , such that its area Δ= 1 2 A α B β =n;or in an equivalent way, to that of the existence of numbers U 2 , V 2 , W 2 ∈ ℚ 2∗ that are in an arithmetic progression of reason n;Problem equivalent to the existence of: ( a,b,c )∈ ℕ 3∗ prime in pairs, and f∈ ℕ ∗ , such that: ( a−b 2f ) 2 , ( c 2f ) 2 , ( a+b 2f ) 2 are in an arithmetic progression of reason n;And this problem is also equivalent to that of the existence of a non-trivial primitive integer right-angled triangle: a 2 + b 2 = c 2 , such that its area Δ= 1 2 ab=n f 2 , where f∈ ℕ ∗ , and this last equation can be written as follows, when using Pythagorician divisors: (1) Δ= 1 2 ab= 2 S−1 d e ¯ ( d+ 2 S−1 e ¯ )( d+ 2 S e ¯ )=n f 2;Where ( d, e ¯ )∈ ( 2ℕ+1 ) 2 such that gcd( d, e ¯ )=1 and S∈ ℕ ∗ , where 2 S−1 , d, e ¯ , d+ 2 S−1 e ¯ , d+ 2 S e ¯ , are pairwise prime quantities (these parameters are coming from Pythagorician divisors). When n=1 , it is the case of the famous impossible problem of the integer right-angled triangle area to be a square, solved by Fermat at his time, by his famous method of infinite descent. We propose in this article a new direct proof for the numbers n=1 (resp. n=2 ) to be non-congruent numbers, based on an particular induction method of resolution of Equation (1) (note that this method is efficient too for general case of prime numbers n=p≡a ( ( mod8 ) , gcd( a,8 )=1 ). To prove it, we use a classical proof by induction on k , that shows the non-solvability property of any of the following systems ( t=0 , corresponding to case n=1 (resp. t=1 , corresponding to case n=2 )): ( Ξ t,k ){ X 2 + 2 t ( 2 k Y ) 2 = Z 2 X 2 + 2 t+1 ( 2 k Y ) 2 = T 2 , where k∈ℕ;and solutions ( X,Y,Z,T )=( D k , E k , f k , f ′ k )∈ ( 2ℕ+1 ) 4 , are given in pairwise prime numbers.2020-Mathematics Subject Classification 11A05-11A07-11A41-11A51-11D09-11D25-11D41-11D72-11D79-11E25 .展开更多
In this article, I consider the right triangle as the simplex in the Euclidean plane, and extend this definition to higher dimensions. The n-dimensional simplex has one hypotenuse and (n−1)legs (catheti). The (n−1)leg...In this article, I consider the right triangle as the simplex in the Euclidean plane, and extend this definition to higher dimensions. The n-dimensional simplex has one hypotenuse and (n−1)legs (catheti). The (n−1)legs define an orthogonal path of edges in the solid with perpendicular adjacent edges along the path. The length of the hypotenuse and the volume of the solid can be calculated without the Cayley-Menger determinant, by direct extension of the corresponding right triangle formulas. I give a proof of the existence of these shapes, describe the distribution of right angles in them, give an algebraic proof of the Coxeter trisection of a right tetrahedron into three smaller right tetrahedra, and generalize this construction to n-dimensional spaces. Finally, I investigate the connection between the Coxeter partition and the Hadwiger conjecture on the partition of the simplex into orthoschemes, which I call Pythagorean simplexes.展开更多
文摘In this paper we prove in a new way, the well known result, that Fermat’s equation a<sup>4</sup> + b<sup>4</sup> = c<sup>4</sup>, is not solvable in ℕ , when abc≠0 . To show this result, it suffices to prove that: ( F 0 ): a 1 4 + ( 2 s b 1 ) 4 = c 1 4 , is not solvable in ℕ , (where a 1 , b 1 , c 1 ∈2ℕ+1 , pairwise primes, with necessarly 2≤s∈ℕ ). The key idea of our proof is to show that if (F<sub>0</sub>) holds, then there exist α 2 , β 2 , γ 2 ∈2ℕ+1 , such that ( F 1 ): α 2 4 + ( 2 s−1 β 2 ) 4 = γ 2 4 , holds too. From where, one conclude that it is not possible, because if we choose the quantity 2 ≤ s, as minimal in value among all the solutions of ( F 0 ) , then ( α 2 ,2 s−1 β 2 , γ 2 ) is also a solution of Fermat’s type, but with 2≤s−1<s , witch is absurd. To reach such a result, we suppose first that (F<sub>0</sub>) is solvable in ( a 1 ,2 s b 1 , c 1 ) , s ≥ 2 like above;afterwards, proceeding with “Pythagorician divisors”, we creat the notions of “Fermat’s b-absolute divisors”: ( d b , d ′ b ) which it uses hereafter. Then to conclude our proof, we establish the following main theorem: there is an equivalence between (i) and (ii): (i) (F<sub>0</sub>): a 1 4 + ( 2 s b 1 ) 4 = c 1 4 , is solvable in ℕ , with 2≤s∈ℕ , ( a 1 , b 1 , c 1 )∈ ( 2ℕ+1 ) 3 , coprime in pairs. (ii) ∃( a 1 , b 1 , c 1 )∈ ( 2ℕ+1 ) 3 , coprime in pairs, for wich: ∃( b ′ 2 , b 2 , b ″ 2 )∈ ( 2ℕ+1 ) 3 coprime in pairs, and 2≤s∈ℕ , checking b 1 = b ′ 2 b 2 b ″ 2 , and such that for notations: S=s−λ( s−1 ) , with λ∈{ 0,1 } defined by c 1 − a 1 2 ≡λ( mod2 ) , d b =gcd( 2 s b 1 , c 1 − a 1 )= 2 S b 2 and d ′ b = 2 s−S b ′ 2 = 2 s B 2 d b , where ( 2 s B 2 ) 2 =gcd( b 1 2 , c 1 2 − a 1 2 ) , the following system is checked: { c 1 − a 1 = d b 4 2 2+λ = 2 2−λ ( 2 S−1 b 2 ) 4 c 1 + a 1 = 2 1+λ d ′ b 4 = 2 1+λ ( 2 s−S b ′ 2 ) 4 c 1 2 + a 1 2 =2 b ″ 2 4;and this system implies: ( b 1−λ,2 4 ) 2 + ( 2 4s−3 b λ,2 4 ) 2 = ( b ″ 2 2 ) 2;where: ( b 1−λ,2 , b λ,2 , b ″ 2 )={ ( b ′ 2 , b 2 , b ″ 2 ) if λ=0 ( b 2 , b ′ 2 , b ″ 2 ) if λ=1;From where, it is quite easy to conclude, following the method explained above, and which thus closes, part I, of this article. .
文摘The most interesting and famous problem that puzzled the mathematicians all around the world is much likely to be the Fermat’s Last Theorem. However, since the Theorem was proposed, people can’t find a way to solve the problem until Andrew Wiles proved the Fermat’s Last Theorem through a very difficult method called Modular elliptic curves in 1995. In this paper, I firstly constructed a geometric method to prove Fermat’s Last Theorem, and in this way we can easily get the conclusion below: If a and b are integer and?a = b, n ∈ Q and n > 1, the value of c satisfies the function an + bn = cn that can never be integer;if a, b and c are integer and a ≠ b, n is integer and n > 2, the function an + bn = cn cannot be established.
文摘Considering Pythagorician divisors theory which leads to a new parameterization, for Pythagorician triplets ( a,b,c )∈ ℕ 3∗ , we give a new proof of the well-known problem of these particular squareless numbers n∈ ℕ ∗ , called congruent numbers, characterized by the fact that there exists a right-angled triangle with rational sides: ( A α ) 2 + ( B β ) 2 = ( C γ ) 2 , such that its area Δ= 1 2 A α B β =n;or in an equivalent way, to that of the existence of numbers U 2 , V 2 , W 2 ∈ ℚ 2∗ that are in an arithmetic progression of reason n;Problem equivalent to the existence of: ( a,b,c )∈ ℕ 3∗ prime in pairs, and f∈ ℕ ∗ , such that: ( a−b 2f ) 2 , ( c 2f ) 2 , ( a+b 2f ) 2 are in an arithmetic progression of reason n;And this problem is also equivalent to that of the existence of a non-trivial primitive integer right-angled triangle: a 2 + b 2 = c 2 , such that its area Δ= 1 2 ab=n f 2 , where f∈ ℕ ∗ , and this last equation can be written as follows, when using Pythagorician divisors: (1) Δ= 1 2 ab= 2 S−1 d e ¯ ( d+ 2 S−1 e ¯ )( d+ 2 S e ¯ )=n f 2;Where ( d, e ¯ )∈ ( 2ℕ+1 ) 2 such that gcd( d, e ¯ )=1 and S∈ ℕ ∗ , where 2 S−1 , d, e ¯ , d+ 2 S−1 e ¯ , d+ 2 S e ¯ , are pairwise prime quantities (these parameters are coming from Pythagorician divisors). When n=1 , it is the case of the famous impossible problem of the integer right-angled triangle area to be a square, solved by Fermat at his time, by his famous method of infinite descent. We propose in this article a new direct proof for the numbers n=1 (resp. n=2 ) to be non-congruent numbers, based on an particular induction method of resolution of Equation (1) (note that this method is efficient too for general case of prime numbers n=p≡a ( ( mod8 ) , gcd( a,8 )=1 ). To prove it, we use a classical proof by induction on k , that shows the non-solvability property of any of the following systems ( t=0 , corresponding to case n=1 (resp. t=1 , corresponding to case n=2 )): ( Ξ t,k ){ X 2 + 2 t ( 2 k Y ) 2 = Z 2 X 2 + 2 t+1 ( 2 k Y ) 2 = T 2 , where k∈ℕ;and solutions ( X,Y,Z,T )=( D k , E k , f k , f ′ k )∈ ( 2ℕ+1 ) 4 , are given in pairwise prime numbers.2020-Mathematics Subject Classification 11A05-11A07-11A41-11A51-11D09-11D25-11D41-11D72-11D79-11E25 .
文摘In this article, I consider the right triangle as the simplex in the Euclidean plane, and extend this definition to higher dimensions. The n-dimensional simplex has one hypotenuse and (n−1)legs (catheti). The (n−1)legs define an orthogonal path of edges in the solid with perpendicular adjacent edges along the path. The length of the hypotenuse and the volume of the solid can be calculated without the Cayley-Menger determinant, by direct extension of the corresponding right triangle formulas. I give a proof of the existence of these shapes, describe the distribution of right angles in them, give an algebraic proof of the Coxeter trisection of a right tetrahedron into three smaller right tetrahedra, and generalize this construction to n-dimensional spaces. Finally, I investigate the connection between the Coxeter partition and the Hadwiger conjecture on the partition of the simplex into orthoschemes, which I call Pythagorean simplexes.
