一个素数和两个素数的平方和问题

本文证明了最多有O(N13/30+ε)个例外之外,所有的正的奇整数n≤N,n≡0或1(mod 3)能表示成一个素数和两个素数的平方和.     

算术级数中的五素数平方和定理

设N是充分大的正整数满足N≡5mod 24,l和d是满足(l,d)=1的整数.A0,A>1是满足A0=600A+2000的正常数.本文证明对所有的整数0相似文献   

常差分方程奇异摄动问题的渐近方法

在本文中,我们讨论如下差分方程问题(P_ε):(L.y)_k≡εy(k+1)+a(k,ε)y(k)+b(k,ε)y(k-1)=f(k,ε)(1≤k≤N-1)B₁y≡-y(0)+c₁y(1)=a,B₂y≡-c₂y(N-1)+y(N)=β这里ε是一个小参数,c₁,c₂,a,β为常数,a(k,ε),b(k,ε),f(k,ε)(1≤k≤N)是k和ε的函数.首先,我们讨论了常系数的情形;接着引进伸长变换对变系数的情形进行了讨论,给出了解的一致渐近展开式;最后给出了一个数值例子.     

ON THE DIOPHANTINE EQUATION X4-Dy2=1(II)

For the Diophantine equation x^4 — Dy^2 = 1 (1) where D>0 and is not a perfect square, we prove the following theorems in this paper. Theorem 1. If D\[{\not \equiv }\]7 (mod 8)，D=p1p2...ps，s≥2，where pi（i = 1，…，s) are distincyt primes，p1≡1(mod 4) such that either 2p1=a^2+b^2，а≡\[ \pm \]3(mod 8)，b三\[ \pm \]3(mod 8) or there is a j(2≤j≤s), for which Legendre symbal \[\left( {\frac{{{p_j}}}{{{p_1}}}} \right) = - 1\],and pi≡7(mod8) (i=2,..., s) or pi≡3(mod 8) (i=2,..., s), then (1) has no solutions in positive integer x,y. Theorem 2. If D=p1...ps,s≥2, where pi（i = 1，…，s) are distinct primes, and pi≡3(mod 4)（i = 1，…，s), then (1) has no solutions in positive integer x, y. Theorem 3. The equation (1) with D=2p1...ps has no solutions in positive integer x, y, if (1) p1≡(mod 4), pi≡7(mod 8) (i = 2, ???, s), snch that either 2p1 = a^2+b^2 a≡\[ \pm \]3(mod 8)，b≡\[ \pm \]3(mod 8)or there is a j (2≤j≤s)，for which \[\left( {\frac{{{p_j}}}{{{p_1}}}} \right) = - 1\]; or (2) p1≡5(mod8),pi≡3(mod8) (i = 2,..., s)； or ⑶p1≡5(mod8)，pi≡7(mod 8) (i=2，…，s). Corollary of theorem 3. If D = 2pq, p≡5(mod 8), q≡3(mod 4), where p, q are distinct primes, then (1) has no solutions in positive integer x, y. Theorem 4. If D=2p1...ps, pi≡3(mod 4)(0 = 1,...,s), then (1) has no solutions In positive integer x, y.     

The Exceptional Set in the Two Prime Squares and a Prime Problem

In this paper we prove that, with at most O（N^5/12＋ε） exceptions, all positive odd integers n ≤ N with n ≡ 0 or 1（mod 3） can be written as a sum of a prime and two squares of primes.     

数学问题解答

20 0 0年 6月号问题解答(解答由问题提供人给出 )1 2 56　求 77 7　 (n个 7,n≥ 3)的末四位数 .解　∵ 74≡ 1 (mod1 0 0 )∴　 74 x ≡ 1 ((mod1 0 0 ) ,x∈ N又　 7≡ - 1 (mod4) ,故 77≡ (- 1 ) 7≡- 1 (mod4) .因而 77 7　 (n - 1个 7,n - 1≥ 2 )≡- 1 (mod4) .所以可设77 7　 (n - 1个 7,n - 1≥ 2 ) =4x 3,x∈N∴　 77 7≡ 74 x 3≡ 73≡ 43(mod1 0 0 )于是可设 77 7　 (n个 7,n≥ 3) =710 0 m 4 3,m∈ N (1 )而　 74 ≡ 2 4 0 1 (mod1 0 0 0 0 )∴　 78≡ 480 1 (mod1 0 0 0 0 )716≡ 960 1 (mod1 0 0 0 0 )732 ≡ 92 0 1 (mod1…     

代数数域中的均值定理

设 K 是 n 次代数数域.令Ψ(x,u,η)=(?)∧(b),其中 u～b mod η(?)α、β∈Z_k,α≡β(modη),α(?)0,β(?)0,(α,η)=(β,η)=1,(α)u=(β)b、h(η)表等价类 modη的类数,T(η)=(U∶U＇),其中 U 表示域 K 中全体单位所成的群,U＇={ε|ε∈U,ε(?)0,ε≡1(modη}.我们证明了下述定理:对于任一正常数 A,存在一正常数 B=B(A)>0,当 Q=x~(1/(n+1))(log x)~(-B),x≥1时有sum from Nη≤Q(?)1/(T(η))|ψ(z,u,η)-z/(h(η))|(?)x/(log~Ax).     

混合幂的除数和（英文）

令S_k(x)=∑d(n_1~2+n_2~2+n_3~k),3≤k∈N.1≤n_1,n_2≤x~(1/2)1≤n_3≤x~(1/k)本文得到了渐近公式S_k(x)=A(k)x~(1+1/k)logx+B(k)x~(1+1/k)+O(x~(1+1/k-δ(k)+ε)),这里A(k),B(k)是只与k有关的常数,δ(3)=5/(42),δ(4)=1/(16),δ(5)=1/(40),并且当6≤k≤7时δ(k)=1/(k2~(k-1)),当k≥8时δ(k)=1/(2k~2(k-1)).     

有向圈的笛卡尔积有向图的双控制数

Let γ*(D) denote the twin domination number of digraph D and let Cm Cn denote the Cartesian product of C_m and C_n, the directed cycles of length m, n ≥ 2. In this paper, we determine the exact values: γ*(C_2?C_n) = n; γ*(C_3 ?C_n) = n if n ≡ 0(mod 3),otherwise, γ*(C_3?C_n) = n + 1; γ*(C_4?C_n) = n + n/2 if n ≡ 0, 3, 5(mod 8), otherwise,γ*(C_4?C_n) = n + n/2 + 1; γ*(C_5?C_n) = 2n; γ*(C_6?C_n) = 2n if n ≡ 0(mod 3), otherwise,γ*(C_6?C_n) = 2n + 2.     

On the diophantine equation x 2 − Dy 2 = n

We propose a method to determine the solvability of the diophantine equation x2-Dy2=n for the following two cases:(1) D = pq,where p,q ≡ 1 mod 4 are distinct primes with(q/p)=1 and(p/q)4(q/p)4=-1.(2) D=2p1p2 ··· pm,where pi ≡ 1 mod 8,1≤i≤m are distinct primes and D=r2+s2 with r,s ≡±3 mod 8.