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
对于不定方程multiply from k=1 to n(k^2+1)=a·m^2,J.Cillerelo证明了当a=1时,当且仅当n=3方程有解.证明了当a=5和7时,此方程无解;当a=17时,方程只有唯一解;还证明了一般情形,当a满足(a,17×101×1297×739 601)=1且...对于不定方程multiply from k=1 to n(k^2+1)=a·m^2,J.Cillerelo证明了当a=1时,当且仅当n=3方程有解.证明了当a=5和7时,此方程无解;当a=17时,方程只有唯一解;还证明了一般情形,当a满足(a,17×101×1297×739 601)=1且a的最大素因子p(a)≤2×738 740时,当n>3,方程无解.展开更多
A(n,k)=sum from m=1 to k sum r=1 to m sum j=0 to [k/m]-1 (tm,r,j (k)×nj×s(r,m)×ζmnr,ζm=e2πi/m,s(r,m)={1,gcd(r,m)=1 0,其他)为丢番图方程sum i=1 to k (ixi=n)的非负整数解的个数.虽然用解线性方程组的方法...A(n,k)=sum from m=1 to k sum r=1 to m sum j=0 to [k/m]-1 (tm,r,j (k)×nj×s(r,m)×ζmnr,ζm=e2πi/m,s(r,m)={1,gcd(r,m)=1 0,其他)为丢番图方程sum i=1 to k (ixi=n)的非负整数解的个数.虽然用解线性方程组的方法可求得A(n,k)的所有系数,然而,该求解过程却非常耗时.本文利用方程(1-x)(1-x2)...(1-xk)=0的相异根的幂可能存在的相等关系,即取适当的正整数g使某些相异根的g次幂相等来实现同类项系数的合并以降低方程的维数,达到提高方程求解速度的目的.展开更多
若丢番图方程multiply from i=1 to i(x_i)=sum from i=1 to k(x_i)仅有唯一解,则正整数k称为Schnizel数.本文给出了k为Schinzel数的充要条件,并证明了:对于k≤500,000,除了k=2,3,4,6,24,114.174,444外无其它Schinzel数.
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 .
文摘对于不定方程multiply from k=1 to n(k^2+1)=a·m^2,J.Cillerelo证明了当a=1时,当且仅当n=3方程有解.证明了当a=5和7时,此方程无解;当a=17时,方程只有唯一解;还证明了一般情形,当a满足(a,17×101×1297×739 601)=1且a的最大素因子p(a)≤2×738 740时,当n>3,方程无解.
文摘A(n,k)=sum from m=1 to k sum r=1 to m sum j=0 to [k/m]-1 (tm,r,j (k)×nj×s(r,m)×ζmnr,ζm=e2πi/m,s(r,m)={1,gcd(r,m)=1 0,其他)为丢番图方程sum i=1 to k (ixi=n)的非负整数解的个数.虽然用解线性方程组的方法可求得A(n,k)的所有系数,然而,该求解过程却非常耗时.本文利用方程(1-x)(1-x2)...(1-xk)=0的相异根的幂可能存在的相等关系,即取适当的正整数g使某些相异根的g次幂相等来实现同类项系数的合并以降低方程的维数,达到提高方程求解速度的目的.
文摘若丢番图方程multiply from i=1 to i(x_i)=sum from i=1 to k(x_i)仅有唯一解,则正整数k称为Schnizel数.本文给出了k为Schinzel数的充要条件,并证明了:对于k≤500,000,除了k=2,3,4,6,24,114.174,444外无其它Schinzel数.
