关于方程ax+by=cz的Terai-Jesmanowicz猜想

设m是正整数,证明了:(A)如果b是奇素数,且a=m3-3m,b=3m2-1,c=m2+1,那么丢番图方程ax+by=cz(1)仅有正整数解(x,y,z)=(2,2,3);(B)如果b是奇素数,且a=m|m4-10m2+5|,b=5m4-10m2+1,c=m2+1,那么丢番图方程(1)仅有正整数解(x,y,z)=(2,2,5).     

关于丢番图方程(an)x+(bn)y=(cn)z

设n,a,b,c是正整数,gcd(a,b,c)=1,a,b≥3,且丢番图方程a~x+b~y=c~z只有正整数解(x,y,z)=(1,1,1).证明了若(x,y,z)是丢番图方程(an)~x+(bn)~y=(cn)~z的正整数解且(x,y,z)≠(1,1,1),则yzz或xzy.还证明了当(a,b,c)=(3,5,8),(5,8,13),(8,13,21),(13,21,34)时,丢番图方程(an)~x+(bn)~y=(cn)~z只有正整数解(x,y,z)=(1,1,1).     

关于Diophantine方程a~x+b~y=c~z的Terai猜想

设m是正偶数.证明了(A)若b是奇素数,且a=m|m~6-21m~4+35m~2-7|,b=|7m~6-35m~4+21m~2-1|,c=m~2+1,则Diophantine方程G:a~x+b~y=c~z仅有正整数解(x,y,z)=(2,2,7);(B)若m2863,且a=m|m~8-36m~6+126m~4-84m~2+9|,b=|9m~8-84m~6+126m~4-36m~2+1|,c=m~2+1,则Diophantine方程G仅有正整数解(x,y,z)=(2,2,9);(C)若a,b,c适合a=m|∑_(i=0)~((r-1)/2)(-1)~i(_(2i)~r)m~(r-2i-1)|,b=|∑_(i=0)~((r-1)/2)(-1)~i(_(2i+1)~r)m~(r-2i-1)|,c=m~2+1,r≡1(mod4),2|x,2|y,且b为奇素数或m145r(log r),则方程G仅有解(x,y,z)=(2,2,r).     

指数丢番图方程a~x+b~y=c~z

设a=|m(m~4-10m~2+5)|,b=5m~4-10m~2+1,c=m~2+1,其中m是正偶数。利用Bilu,Hanrot和Voutier关于本原素除子的深刻结果,证明了指数丢番图方程a~x+b~y=c~z仅有正整数解(x,y,z)=(2,2,5)。     

关于三项纯指数Diophantine方程ax+by=cz

设a、b、c是互素的正整数.本文证明了:当a b2l-1=c2,b≡5(mod 12),c是适合c≡-1(mod b2l)的奇素数,其中l是正整数时,方程ax by=cz仅有正整数解(x,y,z)=(1,2l-1,2).     

丢番图方程(an)^(x)+(bn)^(y)=(cn)^(z)

设a,b,c,n均是大于1的整数且a+b=c^(2),gcd(a,b,c)=1.得到了一些关于丢番图方程(an)^(x)+(bn)^(y)=(cn)^(z)正整数解(x,y,z)的结论.     

关于丢番图方程b~x+(2~α)~y=(b+2~α)~z

结合初等和高等的方法研究丢番图方程b~x+2~(αy)=(6+2~α)~z,α≥3正整数解的问题,并得到了如下结论:1.若b为平方数,则方程只有正整数解(x,y,z)=(1,1,1);若b+2~α为平方数,则x=1.2.若x1,则2■z.3.方程3~x+(2~(2k+1))~y=(3+2~(2k+1))~z,k■2 (mod 6),2k+1∈N,2k+11只有正整数解(x,y,z)=(1,1,1).     

Diophantine方程组a~2+b~2=c~r和x~2+b~y=c~z的一点注记

设r是大于1的正奇数,a,b,c是满足a~2+b~2=c~r的互素正整数.证明了:当r(?)5(mod8),c>10~(12)r~4且b是奇素数的方幂时,方程x~2+b~y=c~z仅有正整数解(x,y,z)=(a,2,r).     

关于本原商高数的Miyazaki猜想

设a,b,c是适合a=2~(2r)-n~2,b=2~(r+1)n,c=2~(2r)+n~2的正整数,其中r是正整数,n是奇素数.运用初等数论方法讨论了指数Diophantine方程c~x+b~y=a~z.证明了:当2~r=n+1时,方程仅有正整数解(x,y,z)=(1,1,2);否则,方程无解。上述结果部分地证实了有关本原商高数的Miyazaki猜想。     

一类指数丢番图方程的解

设m为正整数,且a=8m~3+3m,b=3m~2+1,c=4m~2+1.本文同时使用两个代数数的对数线性型下界估计,两个有理数方幂之差的P-adic赋值的下界估计的一些结果,以及二次域的类数与本原素因子的深刻结论,证明了,当m为任意正整数时,指数丢番图方程a~x+b~y=c~z仅有正整数解(x,y,z)=(2,1,3).     

EI^n代数与^＊EI^n代数

为了更好地解决模糊概念的表示问题，文[1].[2]引入了AFS．代数和AFS．结构。为了讨论AFS．结构的拓扑性质，本文在[1],[2]的基础上，进一步讨论了EI^n代数和^*EI^n代数的性质与结构。     

Une question sur les équations xm ? ym = Rzn (A Question on the Equations xm ? ym = Rzn)

Let m and n be integers at least two and R be a nonzero natural number. In this paper, we study the problem of the determination of the proper solutions of the Diophantine equation x ^m – y ^m equals; Rz ⁿ . We raise a question concerning the existence of any proper nontrivial solution of this equation, in case some precise conditions are satisfied by the triple (m, n, R). We prove some results about it.     

Rational proper holomorphic maps from B n into B 2 n

In this paper, we will investigate rational proper holomorphic maps from Bⁿ into B²ⁿ, when n4.Mathematics Subject Classification (2000): 32H35Partially supported by Grant-in-Aid for Scientific Research (C) no. 14540195 from Japan Society for the Promotion of Science, 2004.Acknowledgement The author thanks the referee for helpful comments and suggestions.     

Z／（2∧e）上本原序列不同压缩映射的导出序列

设f(x）是Z/（2∧e）上n次强本原多项式,对形如xe-1 η（x0,…,xe-2）的二个e元布尔函数φ（xo,…,xe-1）和ψ（x0,…,xe-1）及二条序列a,b∈G（f（x)）e,若φ（a0,…,ae-1）=ψ（b0,…,be-1）,给出了函数φ（x0,…,xe-1）和ψ（x0,…,xe-1）之间的关系与序列a和b之间的关系,所给出的结论进一步说明了导出的二元序列具有良好的密码性质。     

The diophantine equation x3 + 3y3 = 2n

It is proved that the equation of the title has a finite number of integral solutions (x, y, n) and necessary conditions are given for (x, y, n) in order that it can be a solution (Theorem 2). It is also proved that for a given odd x₀ there is at most one integral solution (y, n), n ≥ 3, to x₀³ + 3y³ = 2ⁿ and for a given odd y₀ there is at most one integral solution (x, n), n ≥ 3, to x³ + 3y₀³ = 2ⁿ.     

AVERAGE σ-K WIDTH OF CLASS OF Lp（R^n） IN Lq（R^n）

ＡＶＥＲＡＧＥ　σ－Ｋ　ＷＩＤＴＨ　ＯＦ　ＣＬＡＳＳ　ＯＦ　Ｌ_ｐ（Ｒ~ｎ）ＩＮ　Ｌ_ｑ（Ｒ~ｎ）ＡＶＥＲＡＧＥσ－ＫＷＩＤＴＨＯＦＣＬＡＳＳＯＦＬ_ｐ（Ｒ~ｎ）ＩＮＬ_ｑ（Ｒ~ｎ）￥ＬＩＵＹＯＮＧＰＩＮＧＡｂｓｔｒａｃｔ：Ｔｈｅａｖｅｒａｇｅσ－Ｋｗｉｄｔ...     

Proof of the impossibility of the fermat equation Xp + Yp = Zp for special values of p and of the more general equation bXn + cYn = dZn

This paper continues the search to determine for what exponents n Fermat's Last Theorem is true. The main theorem and Corollary 1 consider the set of prime exponents p for which mp + 1 is prime for certain even integers m and prove the truth of FLT in Case 1 for such primes p. The remaining theorems prove the inequality of the more general Fermat equation bXⁿ + cYⁿ = dZⁿ.