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Analytical solution of basic equations set of atmospheric motion
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作者 施惟慧 沈春 王曰朋 《Applied Mathematics and Mechanics(English Edition)》 SCIE EI 2007年第3期385-394,共10页
On condition that the basic equations set of atmospheric motion possesses the best stability in the smooth function classes, the structure of solution space for local analytical solution is discussed, by which the thi... On condition that the basic equations set of atmospheric motion possesses the best stability in the smooth function classes, the structure of solution space for local analytical solution is discussed, by which the third-class initial value problem with typ- icality and application is analyzed. The calculational method and concrete expressions of analytical solution about the well-posed initial value problem of the third-kind are given in the analytic function classes. Near an appointed point, the relevant theoretical and computational problems about analytical solution of initial value problem are solved completely in the meaning of local solution. Moreover, for other type ofproblems for determining solution, the computational method and process of their stable analytical solution can be obtained in a similar way given in this paper. 展开更多
关键词 basic equations set of atmospheric motion structure of solution space analytical solution
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On Polynomials Solutions of Quadratic Diophantine Equations
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作者 Amara Chandoul 《Advances in Pure Mathematics》 2011年第4期155-159,共5页
Let P:=P(t) be a polynomial in Z[X]\{0,1} In this paper, we consider the number of polynomial solutions of Diophantine equation E:X2–(P2–P)Y2–(4P2–2)X+(4P2–4P)Y=0. We also obtain some formulas and recurrence rela... Let P:=P(t) be a polynomial in Z[X]\{0,1} In this paper, we consider the number of polynomial solutions of Diophantine equation E:X2–(P2–P)Y2–(4P2–2)X+(4P2–4P)Y=0. We also obtain some formulas and recurrence relations on the polynomial solution (Xn,Yn) of 展开更多
关键词 POLYNOMIAL solutionS Pell’s equation diophantine equation
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Non-Negative Integer Solutions of Two Diophantine Equations 2x + 9y = z2 and 5x + 9y = z2
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作者 Md. Al-Amin Khan Abdur Rashid Md. Sharif Uddin 《Journal of Applied Mathematics and Physics》 2016年第4期762-765,共4页
In this paper, we study two Diophantine equations of the type p<sup>x</sup> + 9<sup>y</sup> = z<sup>2</sup> , where p is a prime number. We find that the equation 2<sup>x</... In this paper, we study two Diophantine equations of the type p<sup>x</sup> + 9<sup>y</sup> = z<sup>2</sup> , where p is a prime number. We find that the equation 2<sup>x</sup> + 9<sup>y</sup> = z<sup>2</sup> has exactly two solutions (x, y, z) in non-negative integer i.e., {(3, 0, 3),(4, 1, 5)} but 5<sup>x</sup> + 9<sup>y</sup> = z<sup>2</sup> has no non-negative integer solution. 展开更多
关键词 Exponential diophantine equation Integer solutions
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Solution with Singularity of Order One for Singular Integral Equation with Hilbert Kernal 被引量:6
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作者 ZhongShou-guo ChenJing-song 《Wuhan University Journal of Natural Sciences》 CAS 2004年第1期6-12,共7页
The basic sets of solutions in classH(orH *) for the characteristic equation and its adjoint equation with Hilbert kernel are given respectively. Thus the expressions of solutions and its solvable conditions are simpl... The basic sets of solutions in classH(orH *) for the characteristic equation and its adjoint equation with Hilbert kernel are given respectively. Thus the expressions of solutions and its solvable conditions are simplified. On this basis the solutions and the solvable conditions in classH 1 * as well as the generalized Noether theorem for the complete equation are obtained. Key words Hilbert kernel - solution with singularity of order one - basic set of solutions - Noether theorem - characteristic equation and its adjoint equation CLC number O 175.5 Foundation item: Supported by the National Natural Science Foundation of China (19971064) and Ziqiang Invention Foundation of Wuhan University (201990336)Biography: Zhong Shou-guo(1941-), male, Professor, research direction: singular integral equations and their applications. 展开更多
关键词 Hilbert kernel solution with singularity of order one basic set of solutions Noether theorem characteristic equation and its adjoint equation
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On Diophantine Equation X(X+1)(X+2)(X+3)=14Y(Y+1)(Y+2)(Y+3) 被引量:2
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作者 段辉明 郑继明 《Journal of Southwest Jiaotong University(English Edition)》 2009年第1期90-93,共4页
The Diophantine equation X( X + 1 ) ( X + 2 ) ( X + 3 ) = 14Y( Y + 1 ) ( Y + 2 ) ( Y + 3 ) still remains open. Using recurrence sequence, Maple software, Pell equation and quadraric residue, this pap... The Diophantine equation X( X + 1 ) ( X + 2 ) ( X + 3 ) = 14Y( Y + 1 ) ( Y + 2 ) ( Y + 3 ) still remains open. Using recurrence sequence, Maple software, Pell equation and quadraric residue, this paper proved it has only two positive integer solutions, i. e., (X,Y) = (5,2) ,(7,3). 展开更多
关键词 Integer solution diophantine equation Recurrent sequence Quadratic residue
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三次Diophantine方程x^(3)+1=pQy^(2)的整数解
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作者 沈秦豫 杨海 王成 《云南师范大学学报(自然科学版)》 2024年第5期17-21,共5页
利用同余式、Legendre符号、递归数列、Pell方程解的性质以及一些初等数论方法得到以下结论:当p、Q分别为6k+1和6k-1型奇素数或p、Q为2个互不相同的6k+1型奇素数时,丢番图方程x^(3)+1=pQy^(2)仅有整数解(x,y)=(-1,0).
关键词 丢番图方程 整数解 同余 LEGENDRE符号
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Heron Triangle and Diophantine Equation
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作者 YANG Shi-chun MA Chang- wei 《Chinese Quarterly Journal of Mathematics》 CSCD 北大核心 2005年第3期242-246,共5页
In this paper, we study the quantic Diophantine equation (1) with elementary geometry method, therefore all positive integer solutions of the equation (1) are obtained, and existence of Heron triangle whose median... In this paper, we study the quantic Diophantine equation (1) with elementary geometry method, therefore all positive integer solutions of the equation (1) are obtained, and existence of Heron triangle whose median lengths are all positive integer are discussed here. 展开更多
关键词 quantic diophantine equation positive integer solution Heron triangle MEDIAN
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The Odd Solutions of Equations Involving Euler-Like Function
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作者 Jiaxin Wu Zhongyan Shen 《Advances in Pure Mathematics》 2021年第5期440-446,共7页
<span style="white-space:nowrap;"><em><span style="white-space:nowrap;"><span style="white-space:nowrap;"><span style="white-space:nowrap;"><spa... <span style="white-space:nowrap;"><em><span style="white-space:nowrap;"><span style="white-space:nowrap;"><span style="white-space:nowrap;"><span style="white-space:nowrap;"><span style="white-space:nowrap;"><span style="white-space:nowrap;">φ</span></span></span></span></span></span><sub>e</sub></em>(<em>n</em>) </span>is a function similar to Euler function <em><span style="white-space:nowrap;"><span style="white-space:nowrap;"><span style="white-space:nowrap;"><span style="white-space:nowrap;"><span style="white-space:nowrap;">φ</span></span></span></span></span></em>(<em>n</em>). We discussed and obtained all the odd solutions of the equations <em style="white-space:normal;"><span style="white-space:nowrap;">φ<sub>e</sub></span></em><span style="white-space:nowrap;">(<em>xy</em>) </span><span style="white-space:nowrap;">= <em style="white-space:normal;"><span style="white-space:nowrap;">φ<sub>e</sub></span></em><span style="white-space:nowrap;"><sub></sub>(<em>x</em>)</span></span><span style="white-space:nowrap;"> + </span><span style="white-space:nowrap;">2</span><em style="white-space:normal;"><span style="white-space:nowrap;"><em style="white-space:normal;"><span style="white-space:nowrap;">φ<sub>e</sub></span></em></span></em><span style="white-space:nowrap;"><span style="white-space:nowrap;"><sub></sub>(</span></span><em style="white-space:normal;"><span style="white-space:nowrap;"><em style="white-space:normal;"><span style="white-space:nowrap;">y)</span></em></span></em>, <em style="white-space:normal;"><span style="white-space:nowrap;">φ<sub>e</sub></span></em>(<em>xy</em>) = 2<em style="white-space:normal;"><span style="white-space:nowrap;">φ<sub>e</sub></span></em><sub></sub>(<em>x</em>) + 3<em style="white-space:normal;"><span style="white-space:nowrap;"><em style="white-space:normal;"><span style="white-space:nowrap;">φ<sub>e</sub></span></em></span></em><sub></sub>(<em style="white-space:normal;"><span style="white-space:nowrap;">y) </span></em>and <em style="white-space:normal;"><span style="white-space:nowrap;">φ<sub>e</sub></span></em><span style="white-space:normal;">(</span><em style="white-space:normal;">xyz</em><span style="white-space:normal;">) = </span><em style="white-space:normal;"><span style="white-space:nowrap;">φ<sub>e</sub></span></em><sub style="white-space:normal;"></sub><span style="white-space:normal;">(</span><em style="white-space:normal;">x</em><span style="white-space:normal;">) + </span><em style="white-space:normal;"><span style="white-space:nowrap;"><em style="white-space:normal;"><span style="white-space:nowrap;">φ<sub>e</sub></span></em></span></em><sub style="white-space:normal;"></sub><span style="white-space:normal;">(</span><em style="white-space:normal;"><span style="white-space:nowrap;">y)</span></em> <span style="white-space:normal;">+ </span><em style="white-space:normal;"><span style="white-space:nowrap;"><em style="white-space:normal;"><span style="white-space:nowrap;">φ<sub>e</sub></span></em></span></em><sub style="white-space:normal;"></sub><span style="white-space:normal;">(</span><em style="white-space:normal;"><span style="white-space:nowrap;">z)</span></em><span style="white-space:normal;"> </span>by using the methods and techniques of elementary number theory. 展开更多
关键词 Euler-Like Function diophantine equation Odd solutions
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A New Understanding on the Problem That the Quintic Equation Has No Radical Solutions
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作者 Xiaochun Mei 《Advances in Pure Mathematics》 2020年第9期508-539,共32页
It is proved in this paper that Abel’s and Galois’s proofs that the quintic equations have no radical solutions are invalid. Due to Abel’s and Galois’s work about two hundred years ago, it was generally accepted t... It is proved in this paper that Abel’s and Galois’s proofs that the quintic equations have no radical solutions are invalid. Due to Abel’s and Galois’s work about two hundred years ago, it was generally accepted that general quintic equations had no radical solutions. However, Tang Jianer <i><span style="font-family:Verdana;font-size:12px;">et</span></i><i><span style="font-size:12px;font-family:Verdana;"> al</span><span style="font-size:12px;font-family:Verdana;">.</span></i><span style="font-size:10pt;font-family:;" "=""><span style="font-family:Verdana;font-size:12px;"> recently prove that there are radical solutions for some quintic equations with special forms. The theories of Abel and Galois cannot explain these results. On the other hand, Gauss </span><i><span style="font-family:Verdana;font-size:12px;">et</span></i></span><i><span style="font-size:12px;font-family:Verdana;"> al</span><span style="font-size:12px;font-family:Verdana;">.</span></i><span style="font-size:10pt;font-family:;" "=""><span style="font-family:Verdana;font-size:12px;"> proved the fundamental theorem of algebra. The theorem declared that there were </span><i><span style="font-family:Verdana;font-size:12px;">n</span></i><span style="font-family:Verdana;font-size:12px;"> solutions for the </span><i><span style="font-family:Verdana;font-size:12px;">n</span></i><span style="font-family:Verdana;font-size:12px;"> degree equations, including the radical and non-radical solutions. The theories of Abel and Galois contradicted with the fundamental theorem of algebra. Due to the reasons above, the proofs of Abel and Galois should be re-examined and re-evaluated. The author carefully analyzed the Abel’s original paper and found some serious mistakes. In order to prove that the general solution of algebraic equation</span></span><span style="font-size:10pt;font-family:;" "=""> </span><span style="font-size:12px;font-family:Verdana;">he proposed was effective for the cubic equation, Abel took the known solutions of cubic equation as a premise to calculate the parameters of his equation. Therefore, Abel’s proof is a logical circular argument and invalid. Besides, Abel confused the variables with the coefficients (constants) of algebraic equations. An expansion with 14 terms was written as 7 terms, 7 terms were missing.</span><span style="font-size:10pt;font-family:;" "=""> </span><span style="font-size:12px;font-family:Verdana;">We prefer to consider Galois’s theory as a hypothesis rather than a proof. Based on that permutation group </span><i><span style="font-size:12px;font-family:Verdana;">S</span></i><sub><span style="font-size:12px;font-family:Verdana;">5</span></sub><span style="font-size:12px;font-family:Verdana;"> had no true normal subgroup, Galois concluded that the quintic equations had no radical solutions, but these two problems had no inevitable logic connection actually. In order to prove the effectiveness of radical extension group of automorphism mapping for the cubic and quartic equations, in the Galois’s theory, some algebraic relations among the roots of equations were used to replace the root itself. This violated the original definition of automorphism mapping group, led to the confusion of concepts and arbitrariness. For the general cubic and quartic algebraic equations, the actual solving processes do not satisfy the tower structure of Galois’s solvable group. The resolvents of cubic and quartic equations are proved to have no symmetries of Galois’s soluble group actually. It is invalid to use the solvable group theory to judge whether the high degree equation has a radical solution. The conclusion of this paper is that there is only the </span><i><span style="font-size:10.0pt;font-family:;" "=""><span style="font-family:Verdana;font-size:12px;">S</span><sub><span style="font-family:Verdana;font-size:12px;">n</span></sub></span></i><span style="font-size:10pt;font-family:;" "=""><span style="font-family:Verdana;font-size:12px;"> symmetry for the </span><i><span style="font-family:Verdana;font-size:12px;">n</span></i><span style="font-family:Verdana;font-size:12px;"> degree algebraic equations. The symmetry of Galois’s solvable group does not exist. Mathematicians should get rid of the constraints of Abel and Galois’s theories, keep looking for the radical solutions of high degree equations.</span></span> 展开更多
关键词 Quintic equation Gauss basic Theorem of Algebra Radical solution Abel’s Theory Galois’s Theory Solvable Group Lagrange’s Resolvents
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On the Diophantine Equation y^2= px(x^2+ 2)
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作者 WANG Xiao-ying 《Chinese Quarterly Journal of Mathematics》 CSCD 2009年第4期499-503,共5页
For any fixed odd prime p, let N(p) denote the number of positive integer solutions (x, y) of the equation y^2 = px(x^2 + 2). In this paper, using some properties of binary quartic Diophantine equations, we pro... For any fixed odd prime p, let N(p) denote the number of positive integer solutions (x, y) of the equation y^2 = px(x^2 + 2). In this paper, using some properties of binary quartic Diophantine equations, we prove that ifp ≡ 5 or 7(mod 8), then N(p) = 0; ifp ≡ 1(mod 8), then N(p) 〈 1; if p〉 3 andp ≡ 3(rood 8), then N(p) ≤ 2. 展开更多
关键词 cubic and quartic diophantine equation number of solutions upper bound 2000 MR Subject Classification: 11D25
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A Variant of Fermat’s Diophantine Equation
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作者 Serdar Beji 《Advances in Pure Mathematics》 2021年第12期929-936,共8页
A variant of Fermat’s last Diophantine equation is proposed by adjusting the number of terms in accord with the power of terms and a theorem describing the solubility conditions is stated. Numerically obtained primit... A variant of Fermat’s last Diophantine equation is proposed by adjusting the number of terms in accord with the power of terms and a theorem describing the solubility conditions is stated. Numerically obtained primitive solutions are presented for several cases with number of terms equal to or greater than powers. Further, geometric representations of solutions for the second and third power equations are devised by recasting the general equation in a form with rational solutions less than unity. Finally, it is suggested to consider negative and complex integers in seeking solutions to Diophantine forms in general. 展开更多
关键词 Variant of Fermat’s Last equation Positive Integer solutions of New Fermat-Type equations Geometric Representations for solutions of New diophantine equations
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On a Matrix Equation AX+XB=C over a Skew Field 被引量:1
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作者 王卿文 秦建国 《Chinese Quarterly Journal of Mathematics》 CSCD 1993年第3期97-102,共6页
In this paper we study a matrix equation AX+BX=C(I)over an arbitrary skew field,and give a consistency criterion of(I)and an explicit expression of general solutions of(I).A convenient,simple and practical method of s... In this paper we study a matrix equation AX+BX=C(I)over an arbitrary skew field,and give a consistency criterion of(I)and an explicit expression of general solutions of(I).A convenient,simple and practical method of solving(I)is also given.As a particular case,we also give a simple method of finding a system of fundamental solutions of a homogeneous system of right linear equations over a skew field. 展开更多
关键词 matrix equation over a skew field fundamental system of solutions basic solution matrix subdirect product homogeneous system of right linear equations
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广义Erdös-Straus猜想的互异正整数解的存在性
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作者 尤利华 李佳姻 袁平之 《数学理论与应用》 2024年第2期65-79,共15页
本文研究当n>k≥2且t≥2时方程k/n=1/x_(1)+1/x_(2)+…+1/x_(t)的互异正整数解,证明若方程有正整数解,则至少有一互异正整数解;当k=5,t=3时,除了n≡1,5041,6301,8821,13861,15121(mod 16380)外方程有一互异正整数解;当n≥3,t=4时,除... 本文研究当n>k≥2且t≥2时方程k/n=1/x_(1)+1/x_(2)+…+1/x_(t)的互异正整数解,证明若方程有正整数解,则至少有一互异正整数解;当k=5,t=3时,除了n≡1,5041,6301,8821,13861,15121(mod 16380)外方程有一互异正整数解;当n≥3,t=4时,除了n≡1,81901(mod 163800)外方程有一互异正整数解;并进一步指出对于任意的n(>k),当t≥k≥2时,方程至少有一互异正整数解. 展开更多
关键词 不定方程 正整数解 互异 Erd?s-Straus 猜想
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不定方程∑_(i=1)^(m)x_(i)^(2)=∑_(j=1)^(n)y_(j)^(2)(m>n,m,n∈N_(+))的整数解
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作者 杨雅琴 《高师理科学刊》 2024年第6期1-5,共5页
给出不定方程∑_(i=1)^(m)x_(i)^(2)=∑_(j=1)^(n)y_(j)^(2)(m>n,m,n∈N_(+))的一种参数求解方法,以及该不定方程的一种拆分求解方法.
关键词 参数法 不定方程 整数解 拆分方法
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关于Diophantine方程(20n)~x+(21n)~y=(29n)~z(英文) 被引量:11
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作者 程智 孙翠芳 杜先能 《应用数学》 CSCD 北大核心 2013年第1期129-133,共5页
设a,b,c是满足条件a2+b2=c2的两两互素的正整数.Jesmanowicz于1956年猜想对于任意给定的正整数n,方程(an)x+(bn)y=(cn)z仅有解(x,y,z)=(2,2,2).本文证明了方程(20n)x+(21n)y=(29n)z有唯一解(x,y,z)=(2,2,2).
关键词 JESMANOWICZ猜想 diophantine方程 正整数解
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关于Diophantine方程x^3±1=pqy^2 被引量:22
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作者 管训贵 杜先存 《安徽大学学报(自然科学版)》 CAS 北大核心 2014年第1期29-35,共7页
关于Diophantine方程x3±1=Dy2至今仍未解决.论文利用同余式、平方剩余、Pell方程解的性质、递归序列证明:(1)p≡1(mod 12)为素数,q=12s2+1(s是正奇数)为素数,(p q)=-1时,Diophantine方程x3±1=pqy2仅有整数解(x,y)=(1,0);(2)p... 关于Diophantine方程x3±1=Dy2至今仍未解决.论文利用同余式、平方剩余、Pell方程解的性质、递归序列证明:(1)p≡1(mod 12)为素数,q=12s2+1(s是正奇数)为素数,(p q)=-1时,Diophantine方程x3±1=pqy2仅有整数解(x,y)=(1,0);(2)p≡1(mod 24)为素数,q=12s2+1(s是正奇数)为素数,(p q)=-1时,Diophantine方程x3±1=pqy2仅有整数解(x,y)=(-1,0). 展开更多
关键词 diophantine方程 奇素数 整数解 同余式 平方剩余 递归序列
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关于Diophantine方程x^3±1=3Dy^2 被引量:26
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作者 杜先存 吴丛博 赵金娥 《沈阳大学学报(自然科学版)》 CAS 2013年第1期84-86,共3页
设D是奇素数,运用同余式、平方剩余、递归序列、Maple程序等初等方法得出了当D=27t2+1(t∈Z+)时,Diophantine方程x3±1=3 Dy2无正整数解的一个充分条件.
关键词 diophantine方程 奇素数 同余 平方剩余 递归序列 正整数解
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关于Diophantine方程x^3-1=13qy^2的整数解 被引量:4
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作者 杜先存 管训贵 万飞 《南京师大学报(自然科学版)》 CAS CSCD 北大核心 2015年第4期103-105,共3页
设D=multiply from i=1 to s p_i(s≥2),p_i=1(mod 6)(1≤i≤s)为不同的奇素数.关于Diophantine方程x^3-1=Dy^2的初等解法至今仍未解决.主要利用同余式、平方剩余、Pell方程的解的性质、递归序列,证明了q≡7(mod 24)为奇素数.(q/13)=-1... 设D=multiply from i=1 to s p_i(s≥2),p_i=1(mod 6)(1≤i≤s)为不同的奇素数.关于Diophantine方程x^3-1=Dy^2的初等解法至今仍未解决.主要利用同余式、平方剩余、Pell方程的解的性质、递归序列,证明了q≡7(mod 24)为奇素数.(q/13)=-1时,Diophantine方程x^3-1=13qy^2仅有整数解(x,y)=(1,0). 展开更多
关键词 diophantine方程 奇素数 整数解 同余式 平方剩余 递归序列
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关于三次Diophantine方程x^3+1=2p_1p_2Qy^2的可解性 被引量:3
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作者 杨海 候静 付瑞琴 《中山大学学报(自然科学版)》 CAS CSCD 北大核心 2017年第5期30-33,共4页
设p_1,p_2是适合_p1≡p_2≡1(mod 6)以及(p_1/p_2)=-1的奇素数,其中(p_1/p_2)是Legendre符号。设Q是至少有两个不同素因数且每个素因数q都满足q≡5(mod 6)的无平方因子正整数。运用初等数论方法证明了:如果p_1≡1(mod 8),p_2≡5(mod 8),... 设p_1,p_2是适合_p1≡p_2≡1(mod 6)以及(p_1/p_2)=-1的奇素数,其中(p_1/p_2)是Legendre符号。设Q是至少有两个不同素因数且每个素因数q都满足q≡5(mod 6)的无平方因子正整数。运用初等数论方法证明了:如果p_1≡1(mod 8),p_2≡5(mod 8),Q≡1(mod 4),那么方程x^3+1=2p_1p_2Qy^2无正整数解(x,y)。 展开更多
关键词 三次diophantine方程 正整数解 同余条件
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关于指数Diophantine方程x^3-1=2py^2 被引量:5
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作者 万飞 杜先存 《西南民族大学学报(自然科学版)》 CAS 2012年第5期736-738,共3页
设p是6k+1型的奇素数,运用Pell方程px2-3y2=1的最小解、同余式、平方剩余、勒让德符号的性质等初等方法证明了当p=3n(n+1)+1≡1,7(mod8)(n为单数)为奇素数,且2n+1为奇素数时,指数Diophantine方程x3-1=2py2无正整数解.
关键词 指数diophantine方程 奇素数 同余 最小解 正整数解 勒让德符号
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