关于Diophantine方程(20n)x+(21n)y=(29n)z

设a,b,c是满足条件a2+ b2=c2的两两互素的正整数.Jesmanowicz于1956年猜想对于任意给定的正整数n,方程(an)x+(bn)y=(cn)z仅有解(x,y,z)=(2,2,2).本文证明了方程(20n)x+(21n)y=(29n)z有唯一解(x,y,z)=(2,2,2).     

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).     

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).     

指数Diophantine方程(bn)~x+(2n)~y=((b+2)n)~z的例外解

设b是大于3的正奇数.运用初等方法讨论了方程(bn)^x+(2n)x+(2n)^y=((b+2)n)y=((b+2)n)^z适合(x,y,z)≠(1,1,1)的正整数解(x,y,z,n).证明了:i)对于任何给定的正整数N,存在无穷多个b可使该方程有满足min{x,y,z}≥N的正整数解(x,y,z,n);ii)对于任何给定的b,该方程仅有有限多组正整数解(x,y,z,n)满足y>z=x.     

关于丢番图方程(12n)~x+(35n)~y=(37n)~z

运用同余及元素阶的性质,证明对任意正整数n,丢番图方程(12n)x+(35n)y=(37n)z仅有正整数解(x,y,z)=(2,2,2).     

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(m2+1)x+(cm2-1)y=(am)z

设m,a,c均是大于1的正整数.当am≡1(mod 4)或am≡3(mod 8),3■m或2■a,2|m,3■m时,得到了丢番图方程(m²+1)^x+(cm²-1)^y=(am)^z,1+c=a²,m≥2只有正整数解(x,y,z)=(1,1,2).特别地,当a≡1,3,5 (mod 8),a≠3或a≡7 (mod 8),a≡2(mod 3)时,方程2^x+(a²-2)^y=(a)^z只有正整数解(x,y,z)=(1,1,2).     

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).     

关于丢番图方程（195n）x+（28n）y=（197n）z

运用同余及元素阶的性质,证明了对任意的正整数n,丢番图方程(195n)x+(28n)y=(197n)z仅有正整数解(x, y, z)=(2,2,2)。     

An algorithm for linear programming which requires O(((m+n)n 2+(m+n)1.5 n)L) arithmetic operations

We present an algorithm for linear programming which requires O(((m+n)n ²+(m+n)^1.5 n)L) arithmetic operations wherem is the number of constraints, andn is the number of variables. Each operation is performed to a precision of O(L) bits.L is bounded by the number of bits in the input. The worst-case running time of the algorithm is better than that of Karmarkar's algorithm by a factor of .     

On the functional equationS(m,n)F[n/(m,n)] =F(n)h[n/(m,n)]

In this paper, we discuss the pairs (f, h) of arithmetical functions satisfying the functional equation in the title, whereF is the product off andh under the Dirichlet convolution; that is,F(n) = Σ _d|n ?(d)h(n/d) andS(m n) = Σd|(m, n) ?(d)h(n/d). The well-known Hölder's identity is a special case of this functional equation (?(n) =n, h(n) = μ(n)). We also generalize the functional equation in the title to any arbitrary regular arithmetical convolution and discuss the pairs of solutions (f, h) of the generalized functional equation and pose some problems relating to the characterization of all pairs of solutions.     

Triangular embeddings of K((i-2) n,n,…,n)

For complete i-partite graphs of the form K(n₁, n, n, …, n) the largest value of n₁ that allows the graph to be triangularly-embedded into a surface is (i-2)n. In this paper the author constructs triangular embeddings into surfaces of some complete partite graphs of the form K((i-2)n, n, …, n). The embeddings are exhibited using embedding schemes but the surfaces into which K((i-2)n, n, …, n) are triangularly embedded can be seen to be particularly nice branched covers of a surface into which K(i-2, 1, 1,…,1) is triangularly embedded.