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
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
Abstract This paper deals with how to perturb a given set of polynomials so as to include a common linear factor. An algorithm is derived for determining such a set of perturbation polynomials which are subject to cer...Abstract This paper deals with how to perturb a given set of polynomials so as to include a common linear factor. An algorithm is derived for determining such a set of perturbation polynomials which are subject to certain constrains at the endpoints of a prescribed parametric interval and minimized in a certain sense. This result can be combined with subdivision technique to obtain a continuous piecewise approximation to a rational curve.展开更多
Denote by a non-trivial primitive solution of Fermat’s equation (p prime).We introduce, for the first time, what we call Fermat principal divisors of the triple defined as follows. , and . We show that it is possible...Denote by a non-trivial primitive solution of Fermat’s equation (p prime).We introduce, for the first time, what we call Fermat principal divisors of the triple defined as follows. , and . We show that it is possible to express a,b and c as function of the Fermat principal divisors. Denote by the set of possible non-trivial solutions of the Diophantine equation . And, let<sub></sub><sub></sub> (p prime). We prove that, in the first case of Fermat’s theorem, one has . In the second case of Fermat’s theorem, we show that , ,. Furthermore, we have implemented a python program to calculate the Fermat divisors of Pythagoreans triples. The results of this program, confirm the model used. We now have an effective tool to directly process Diophantine equations and that of Fermat. .展开更多
提出了一种RS码的快速盲识别方法。该方法基于RS码的等效二进制分组码的循环移位特性,通过欧几里德算法计算循环移位前后码字的最大公约式,根据最大公约式指数的相关性来估计码长,并快速剔除含错码字,进而利用伽罗华域的傅里叶变换(Galo...提出了一种RS码的快速盲识别方法。该方法基于RS码的等效二进制分组码的循环移位特性,通过欧几里德算法计算循环移位前后码字的最大公约式,根据最大公约式指数的相关性来估计码长,并快速剔除含错码字,进而利用伽罗华域的傅里叶变换(Galois Field Fourier Transform,GFFT)实现RS码的本原多项式和生成多项式的识别。仿真结果表明,该算法复杂度低,计算量小,在误码率为10-3的情况下,对RS码的识别概率高于90%。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Abstract This paper deals with how to perturb a given set of polynomials so as to include a common linear factor. An algorithm is derived for determining such a set of perturbation polynomials which are subject to certain constrains at the endpoints of a prescribed parametric interval and minimized in a certain sense. This result can be combined with subdivision technique to obtain a continuous piecewise approximation to a rational curve.
文摘Denote by a non-trivial primitive solution of Fermat’s equation (p prime).We introduce, for the first time, what we call Fermat principal divisors of the triple defined as follows. , and . We show that it is possible to express a,b and c as function of the Fermat principal divisors. Denote by the set of possible non-trivial solutions of the Diophantine equation . And, let<sub></sub><sub></sub> (p prime). We prove that, in the first case of Fermat’s theorem, one has . In the second case of Fermat’s theorem, we show that , ,. Furthermore, we have implemented a python program to calculate the Fermat divisors of Pythagoreans triples. The results of this program, confirm the model used. We now have an effective tool to directly process Diophantine equations and that of Fermat. .
文摘提出了一种RS码的快速盲识别方法。该方法基于RS码的等效二进制分组码的循环移位特性,通过欧几里德算法计算循环移位前后码字的最大公约式,根据最大公约式指数的相关性来估计码长,并快速剔除含错码字,进而利用伽罗华域的傅里叶变换(Galois Field Fourier Transform,GFFT)实现RS码的本原多项式和生成多项式的识别。仿真结果表明,该算法复杂度低,计算量小,在误码率为10-3的情况下,对RS码的识别概率高于90%。
