关于丢番图方程X~2+4Y~4=pZ~4

设p为素数,p=4A~2+1+2|A,A∈N~*.运用二次和四次丢番图方程的结果证明了方程G:X~2+4Y~4=pZ~4,gcd(X,Y,Z)=1,除开正整数解(X,Y,Z)=(1,A,1)外,当A≡1(mod4)时,至多还有正整数解(X,Y,Z)满足X=|p(a~2-b~2)~2-4(A(a~2-b~2)±ab)~2|,Y~2=A(a~2-b~2)~2±2ab(a~2-b~2)-4a~2b~2A,Z=a~2+b~2;当A≡3(mod4)时,至多还有正整数解(X,Y,Z)满足X=|4a~2b~2A-(4abA±(a~2-b~2))~2|,Y~2=4a~2b~2A±2ab(a~2-b~2)-A(a~2-b~2)~2,Z=a~2+b~2.这里a,b∈N~*并且ab,gcd(a,b)=1,2|(a+b).同时具体给出了p=5时方程G的全部正整数解.     

一类3p~2阶4度1-正则图

设p为大于3的素数,群G=和H=(其中r(?)1(mod p~2),r~3≡1(mod p~2),3|(p-1))是两类3p~2阶非交换群.通过研究Cayley图的正规性,完成了对G和H的所有4度Cayley图的分类,并得到了一类新的4度1-正则图.     

丢番图方程X~2-(a~2+1)Y~4=3-4a

设a≥2是正整数.本文证明了:当a=2时,方程X~2一(a~2+1)Y~4=3-4a仅有正整数解(X,Y)=(20,3);当a=3时,该方程仅有2组互素的正整数解(X,Y)=(1,1)和(79,5);当a≥4且4a+1非平方数时,该方程最多有4组互素的正整数解(X,Y);当a≥4且4a+1为平方数时,该方程最多有5组互素的正整数解(X,Y).     

关于丢番图方程X~2-(a~2+1)Y~4=8-6a

设a是正整数.本文证明了:当a=1时,方程X~2-(a~2+1)Y~4=8~6a仅有正整数解(X,Y)=(2,1);当a=2时,该方程仅有正整数解(X,Y)=(1,1);当a=3时,该方程无正整数解(X,Y);当a=4时,该方程仅有2组互素的正整数解(X,Y)=(1,1)和(103,5);当a≥5且6a+1非平方数时,该方程最多有3组互素的正整数解(X,Y);当a≥5且6a+1为平方数时,该方程最多有4组互素的正整数解(X,Y).     

4p~2阶4度半传递图

如果图X的全自同构群Aut(X)作用在其顶点集V(X)和边集E(X)上都是传递的,但作用在弧集Arc(X)上非传递,则称X是半传递图.研究了4p~2(p3且p≡-1(mod4))阶4度半传递图,确定了4p~2阶4度半传递图的连通性及其自同构群的阶.     

关于指数Diophantine方程x~2=D~(2m)-D~mp~n+p~(2n)

设D 1是正整数,p是适合p?D的素数.本文研究了指数Diophantine方程x~2=D~(2m)-D~mp~n+p~(2n)的满足m 1的正整数解.根据Diophantine方程的性质,结合已有的结论,运用初等方法确定了方程满足m 1的所有正整数解(D,p,x,m,n).这个结果修正并完整解决了文献[4]的猜想.     

一类p~3阶群的Burnside环之增广商群

群环理论将群论和环论有机地结合了起来,是代数学中的重要分支之一,其中增广理想和增广商群是群环理论中的一个经典课题.设G有限群,分别记的Burnside环及其增广理想为Ω(G)和Δ(G).本文对任意正整数n,具体构造了Δ~n(I_p)作为自由交换群的一组基,并确定了商群Δ~n(I_p)/Δ~(n+1)(I_p)的结构,其中I_p=〈a,b|a~(p~2)=b~p=1,b~(-1)ab=a~(p+1)〉,p为奇素数.     

关于丢番图方程(na)~x+(nb)~y=(nc)~z（英文）

设(a,b,c)为本原的商高数组,满足a~2+b~2=c~2且2|b.1956年,Jesmanowicz猜想:对任给的正整数n,丢番图方程(na)~x+(nb)~y=(nc)~z仅有正整数解x=y=z=2.令P(n)表示n的所有不同素因子乘积.对商高数组(a,b,c)=(p~(2r)-4,4p~r,p~(2r)+4),其中p为大于3的素数且p■1(mod 8),本文证明在条件P(a)|n或者P(n)a下,Jesmanowicz猜想成立.     

求等差数前列n项和的一个公式

设a_0,a_2,…,a_n,a_(n+1),…为等差数列,其公差为d,则有公式 (?)a_i~3=(a_n·a_(n+1))~2+(a_1a_0)~2/4d 下面给出证明。给定n个等式。 (a_n~2+da_n)~2-(a_n~2-da_n)~3=4da_n~3; (a_(n-1)~2+da_(n-1))-(a_(n-1)~2-da_(n-1))~2=4da_(n-1)~3; (a_(n-2)~2+da_(n-2))~2-(a_(n-2)~3 2-da_(n-2))~2=4da_(n-2)~3,…, (a_3~2+da_3)~2-(a_3~2-da_3)~2=4da_3~3,     

以静窥动 动中求静

我们在解决某些几何题时,可以把某一儿何图形看成是另一图形运动的结果。从这一思想出发,常能获得较为新颖或较好的解法。例1 证明:椭圆x~2/a~2 y~2/b~2=1的内接三角形的面积的最大值为3 3~(1/2)ab/4。证:我们把椭圆x~2/a~2 y~2/b~2=1看成是由圆:X~2 Y~2=a~2经均匀压缩变换 x=X y=bY/a 运动而得到的。设A(x_1,y_1),B(x_2,y_2),C(x_3,y_3)一是椭圆内接三角形三个顶点,它们在圆X~2 Y~2=α~2上的对应点为A′     

Menon Difference Sets in Groups of Order 4p~2q~(2b)

In this paper we study the existence of abelian Menon difference sets with parameter (4p2q2b,wp2q2b-pqb,p2q2b-pqb)in G=H×Cp×Cp or H×Cp2 and get a necessary condition for it.     

不动点集为RP(4)∪ P(4,2n-1)的对合

设(M,T)是一个光滑闭流形上的对合,不动点集为F=RP(4)UP(4,2n-1),则它的每一个对合(M,T)必协边(RP(4)×RP(4),twist)和(P(4,2n),T')之一.     

关于指数型超椭圆方程x^2=p^2m-p^m＋n＋1

证明了指数型超椭圆方程x^2=p^2m-p^m＋n＋1无解（x,p,m,n）,其中x,m,n∈N^＋,m〉n〉1,p∈P．上述结果部分解决了组合论中关于可逆Abel差集的Ma猜想．     

The Number of Triangulations of the Cyclic Polytope C (n , n -4)

We show that the exact number of triangulations of the standard cyclic polytope C(n,n-4) is (n+4)2 ^(n-4)/2 -n if n is even and \left((3n+11)/2\right)2 ^(n-5)/2 -n if n is odd. These formulas were previously conjectured by the second author. Our techniques are based on Gale duality and the concept of virtual chamber. They further provide formulas for the number of triangulations which use a specific simplex. We also compute the maximum number of regular triangulations among all the realizations of the oriented matroid of C(n,n-4) . Received October 24, 2000, and in revised form July 8, 2001. Online publication November 7, 2001.     

Quantum superdeterminants for OSP q (1-2n)

It is shown that there exists a quantum superdeterminant sdet _q T for the quantum super group OSP _q (1|2n). It is also shown that the quantum superdeterminant sdet _q T is a group-like element and central, and that the square of sdet _q T for OSP _q (1|2n) is equal to 1.     

The exponential Diophantine equation x2 + (3n 2 + 1) y = (4n 2 + 1) z

Let n be a positive integer. In this paper, using the results on the existence of primitive divisors of Lucas numbers and some properties of quadratic and exponential diophantine equations, we prove that if n ≡ 3 (mod 6), then the equation x ² + (3n ² + 1) ^y = (4n ² + 1) ^z has only the positive integer solutions (x, y, z) = (n, 1, 1) and (8n ³ + 3n, 1, 3).     

On the Exponential Diophantine Equation $$(a^{n}-2)(b^{n}-2)=x^{2}$$

Mathematical Notes - In this paper, we deal with the equation $$(a^{n}-2)(b^{n}-2)=x^{2}$$ , $$2\leq a3$$ is odd and $$P_{k},Q_{k}$$ are the Pell and Pell Lucas numbers, respectively. We also...     

Optimal holey packings OHP4(2, 4, n ,3)'s

An optimal holey packing OHP_d(2, k, n, g) is equivalent to a maximal (g + 1)‐ary (n, k, d) constant weight code. In this paper, we provide some recursive constructions for OHP_d(2, k, n, g)'s and use them to investigate the existence of an OHP₄(2, 4, n, 3) for n ≡ 2, 3 (mod 4). Combining this with Wu's result ( 18 ), we prove that the necessary condition for the existence of an OHP₄(2, 4, n, 3), namely, n ≥ 5 is also sufficient, except for n ∈ {6, 7} and except possibly for n = 26. © 2005 Wiley Periodicals, Inc. J Combin Designs 14: 111–123, 2006