1) Diophantine inequality
丢番图不等式
1.
For sufficiently large N, we also investigate the solvable of the Diophantine inequality |pc + Pkc - N|< ε in a prime p and a positive integer Pk with at most k prime factors.
本文证明了如果1丢番图不等式|P_1~c+P_2~c-N|
2) linear diophantine inequalitie
线性丢番图不等式
3) indeterminate(Diophantine) equation
不定(丢番图)方程
4) diophantus
丢番图
1.
In this paper,structured of generative function method and technique is introduced by finding the solution of a kind of diophantus equation,thus we introduce power series such as generative function to apply in combination probabicity.
通过求解一类丢番图方程解的个数,介绍了生成函数的构造方法和技巧,从而以幂级数作为生成函数,介绍了它在组合概率计算中的应用。
5) Diophantine equation
丢番图方程
1.
On the Diophantine equation x~p-1=Dy~n;
关于丢番图方程x~p-1=Dy~n
2.
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
3.
On the Diophantine equation(15n)~x+(112n)~y=(113n)~z;
关于丢番图方程(15n)~x+(112n)~y=(113n)~z
6) diophantine equations
丢番图方程
1.
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
2.
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。
3.
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
补充资料：丢番
1.放倒。
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