The Pythagorean triples (a, b | c) of planar geometry which satisfy the equation a<sup>2</sup>+b<sup>2</sup>=c<sup>2</sup> with integers (a, b, c) are generalized to 3D-Pythagorean ...The Pythagorean triples (a, b | c) of planar geometry which satisfy the equation a<sup>2</sup>+b<sup>2</sup>=c<sup>2</sup> with integers (a, b, c) are generalized to 3D-Pythagorean quadruples (a, b, c | d) of spatial geometry which satisfy the equation a<sup>2</sup>+b<sup>2</sup>+c<sup>2</sup>=d<sup>2</sup> with integers (a, b, c, d). Rules for a parametrization of the numbers (a, b, c, d) are derived and a list of all possible nonequivalent cases without common divisors up to d<sup>2</sup> is established. The 3D-Pythagorean quadruples are then generalized to 4D-Pythagorean quintuples (a, b, c, d | e) which satisfy the equation a<sup>2</sup>+b<sup>2</sup>+c<sup>2</sup>+d<sup>2</sup>=e<sup>2</sup> and a parametrization is derived. Relations to the 4-square identity are discussed which leads also to the N-dimensional case. The initial 3D- and 4D-Pythagorean numbers are explicitly calculated up to d<sup>2</sup>, respectively, e<sup>2</sup>.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
Let a, b, c be relatively prime positive integers such that a^2＋ b^2= c^2. Jesmanowicz＇conjecture on Pythagorean numbers states that for any positive integer N, the Diophantine equation（aN）x＋（b N）y=（cN）zhas n...Let a, b, c be relatively prime positive integers such that a^2＋ b^2= c^2. Jesmanowicz＇conjecture on Pythagorean numbers states that for any positive integer N, the Diophantine equation（aN）x＋（b N）y=（cN）zhas no positive solution（x, y, z） other than x = y = z = 2. In this paper, we prove this conjecture for the case that a or b is a power of 2.展开更多
文摘The Pythagorean triples (a, b | c) of planar geometry which satisfy the equation a<sup>2</sup>+b<sup>2</sup>=c<sup>2</sup> with integers (a, b, c) are generalized to 3D-Pythagorean quadruples (a, b, c | d) of spatial geometry which satisfy the equation a<sup>2</sup>+b<sup>2</sup>+c<sup>2</sup>=d<sup>2</sup> with integers (a, b, c, d). Rules for a parametrization of the numbers (a, b, c, d) are derived and a list of all possible nonequivalent cases without common divisors up to d<sup>2</sup> is established. The 3D-Pythagorean quadruples are then generalized to 4D-Pythagorean quintuples (a, b, c, d | e) which satisfy the equation a<sup>2</sup>+b<sup>2</sup>+c<sup>2</sup>+d<sup>2</sup>=e<sup>2</sup> and a parametrization is derived. Relations to the 4-square identity are discussed which leads also to the N-dimensional case. The initial 3D- and 4D-Pythagorean numbers are explicitly calculated up to d<sup>2</sup>, respectively, e<sup>2</sup>.
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 .
基金Supported by the NSF of China(10901002)the Research Culture Funds of Anhui Normal University(2012xmpy009)+1 种基金The second author is supported by the NSF of China(11126173)Anhui Province Natural Science Foundation(1208085QA02)
基金Supported by Grant in Aid for JSPS Fellows(Grant No.25484)
文摘Let a, b, c be relatively prime positive integers such that a^2＋ b^2= c^2. Jesmanowicz＇conjecture on Pythagorean numbers states that for any positive integer N, the Diophantine equation（aN）x＋（b N）y=（cN）zhas no positive solution（x, y, z） other than x = y = z = 2. In this paper, we prove this conjecture for the case that a or b is a power of 2.
