基本解释：丢番图方程组
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On the Diophantine equation x~p-1=Dy~n;关于丢番图方程x~p-1=Dy~n
On the solution of the Diophantine equations x~2-2p=y~n;关于丢番图方程x~2-2p=y~n的解
On the Diophantine equation(15n)~x+(112n)~y=(113n)~z;关于丢番图方程(15n)~x+(112n)~y=(113n)~z
On the Diophantine equations x~4±y~6=z~2 and x~2+y~4=z~6;关于丢番图方程x~4±y~6=z~2与x~2+y~4=z~6
When p is a odd prime and p ≠1 (mod 8), we get all solutions of diophantine equations ( x(x+1)(2x+1)=2p~ky~(2n) ) with elementary theory of number.若p为奇素数,且p≠1(mod8)时,本文给出了丢番图方程x(x+1)(2x+1)=2pky2n的所有正整数解,并给出了Lucas猜想的一个简单证明。
With the help of the elementary theory of number and Fermat method of infinite descent,some necessary conditions have been proved provided that the Diophantine equations x 4+mx 2y 2+ny 4=z 2 has positive Integer solutions that fit (x,y) =1 m.利用数论方法及Fermat无穷递降法 ,证明了丢番图方程x4 +mx2 y2 +ny4 =z2 在 (m ,n) =(± 6,-3 ) ,(6,3 ) ,(± 3 ,3 ) ,(-12 ,2 4) ,(± 12 ,-2 4) ,(± 6,15 ) ,(-6,-15 ) ,(3 ,6)仅有平凡整数解 ,并且获得了方程在 (-6,3 ) ,(12 ,2 4) ,(3 ,-6) ,(-6,3 3 )时的无穷多组正整数解的通解公式 ,从而完善了Aubry等人的结
Let p>3 be a prime integer prime,when the elementary grade method and the Diophantus Equation theories are used.设p>3为素数,证明了丢番图方程x6-y6=2pz2无正整数解,证明了丢番图方程x6+y6=2pz2在p 1(mod24)时无正整数解,同时获得了方程在p≡1(mod24)时有正整数解的计算公式。
In this paper two theorems are given by using matrixvector description of polynomial multiplication, which are useful to resolve the Diophantus equation.采用多项式乘积的矩阵－向量表示方法，证明了对求解丢番图方程极为有用的定理１和定理２，从丢番图方程的基本解法着手，给出了各种设计要求下的极点配置算法。
For combinatorics, linear Diophantine equations,counting integer points in polytope,Frobenins problemes etc.组合数学方面,可对线性丢番图方程组整解数目、多面体内整点计数、Frobenius问题等相关问题进行研究。
With the rapid development of computer science in many fields, the peoblem of linear Diophantine equations has attracted more attention again.许多领域中的经典问题都可以归纳为求解线性丢番图方程组。
ABS Algorithms for Solving Linear Diophantine Equations and Inequations;求解线性丢番图方程组及不等式组的ABS算法
On the Diophantine Equation System a~2+b~2=c~3 and a~x+b~y=c~z联立丢番图方程组a~2+b~2=c~3和a~x+b~y=c~z
Sequential and Parallel Algorithms for Solving Linear Diophantine Equations求解线性丢番图方程(组)的串、并行算法
An elementary proof of the Diophantine equation (The equation abbreviated) is given by using recursion sequence method.运用递推序列法，给出组合数丢番图方程(方程式略)的一个初等解法。
On the Diophantine Equation of a~x-b~(2y)=46~2;丢番图方程a~x-b~(2y)=46~2的一解
Diophantine Equation x~3-y~6=pz~2 and Tijdeman Corijecture;丢番图方程x~3-y~6=pz~2与Tijdeman猜想
On Integer Solution of A Diophantine Equation;关于丢番图方程x~2-py~4=1
On the Diophantine Equation x~3±p~(3n)=Dy~2;关于丢番图方程x~3+p~(3n)=Dy~2的讨论
About the Diophantine Equation X + mY4= Z;关于丢番图方程x~4+my~4=z~4
On the Diophantine Equation x~3 ±5~6 =Dy~2;关于丢番图方程x~3±5~6=Dy~2
On the Diophantine Equations X~5 ± X~3 = DY~3;关于丢番图方程X~5±X~3=Dy~3
On the Diophantine Equation x~4+mx~2y~2+ny~4=z~2;关于丢番图方程x~4+mx~2y~2+ny~4=z~2
On the Diophantine Equations x~2±y~4=z~3;关于丢番图方程x~2±y~4=z~3
On the Diophantine Equations x(x+1)=Dy~4;关于丢番图方程x(x+1)=Dy~4
On the Diophantine Equation (x~m-1)(x~(mn) -1)=y~2;关于丢番图方程(x~m-1)(x~(mn)-1)=y~2
On the Diophantine Equations x_4±y_6=z_2;关于丢番图方程x~4±y~6=z~2
On the Diophantine Equations x~3±y~6=z~2;关于丢番图方程x~3±y~6=z~2的解
On the Sum of Equal Powers and Diophantine Equation S_5(x) =y~n;关于幂和丢番图方程S_5(x)=y~n
