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A New Proof for Congruent Number’s Problem via Pythagorician Divisors
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作者 Léopold Dèkpassi Keuméan François Emmanuel Tanoé 《Advances in Pure Mathematics》 2024年第4期283-302,共20页
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
关键词 Prime Numbers-Diophantine Equations of Degree 2 & 4 Factorization Greater Common Divisor Pythagoras Equation pythagorician Triplets Congruent Numbers Inductive Demonstration Method Infinite Descent BSD Conjecture
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Pythagorician Divisors and Applications to Some Diophantine Equations
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作者 François Emmanuel Tanoé Prosper Kouadio Kimou 《Advances in Pure Mathematics》 2023年第2期35-70,共36页
We consider the Pythagoras equation X<sup>2</sup> +Y<sup>2</sup> = Z<sup>2</sup>, and for any solution of the type (a,b = 2<sup>s</sup>b<sub>1 </sub>≠0,c) ... We consider the Pythagoras equation X<sup>2</sup> +Y<sup>2</sup> = Z<sup>2</sup>, and for any solution of the type (a,b = 2<sup>s</sup>b<sub>1 </sub>≠0,c) ∈ N<sup>*3</sup>, s ≥ 2, b<sub>1</sub>odd, (a,b,c) ≡ (±1,0,1)(mod 4), c > a , c > b, and gcd(a,b,c) = 1, we then prove the Pythagorician divisors Theorem, which results in the following: , where (d,d′′) (resp. (e,e<sup>n</sup>)) are unique particular divisors of a and b, such that a = dd′′ (resp. b = ee′′ ), these divisors are called: Pythagorician divisors from a, (resp. from b). Let’s put λ ∈{0,1}, defined by: and S = s -λ (s -1). Then such that . Moreover the map is a bijection. We apply this new tool to obtain a new classification of the primitive, positive and non-trivial solutions of the Pythagoras equations: a<sup>2</sup> + b<sup>2</sup> = c<sup>2</sup> via the Pythagorician parameters (d,e,S ). We obtain for (d, e) fixed, the equivalence class of any Pythagorician solution (a,b,c), checking , namely: . We also update the solutions of some Diophantine equations of degree 2, already known, but very important for the resolution of other equations. With this tool of Pythagorean divisors, we have obtained (in another paper) new recurrent methods to solve Fermat’s equation: a<sup>4</sup> + b<sup>4 </sup>= c<sup>4</sup>, other than usual infinite descent method;and to solve congruent numbers problem. We believe that this tool can bring new arguments, for Diophantine resolution, of the general equations of Fermat: a<sup>2p</sup> + b<sup>2p</sup> = c<sup>2p</sup> and a<sup>p</sup> + b<sup>p</sup> = c<sup>p</sup>. MSC2020-Mathematical Sciences Classification System: 11A05-11A51-11D25-11D41-11D72. 展开更多
关键词 Pythagoras Equation pythagorician Triplets Diophantine Equations of Degree 2 Factorisation-Gcd-Fermat’s Equations
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Fermat and Pythagoras Divisors for a New Explicit Proof of Fermat’s Theorem:a4 + b4 = c4. Part I
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作者 Prosper Kouadio Kimou François Emmanuel Tanoé Kouassi Vincent Kouakou 《Advances in Pure Mathematics》 2024年第4期303-319,共17页
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
关键词 Factorisation in Greatest Common Divisor Pythagoras Equation pythagorician Triplets Fermat's Equations pythagorician Divisors Fermat's Divisors Diophantine Equations of Degree 2 4-Integral Closure of in
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