黎曼猜想(Ⅴ)的证明黎曼ζ(s)函数的全部(无限多个)非平凡零点（即复数解）都位于直线σ﹦1／2上而t为一π(n)的函数张熙文(中国)摘要在黎曼猜想（V）中,T和︱s︱必须满足下列六项条件:1.当T＜9146,n﹦N(T)＜T；当T﹦9146,n﹦N(T)﹦T；当T＞9146,n﹦N(T)＞T；2.T∝π(n)·㏒n,T﹦k·π(n)·㏒n，k﹦9π(eπ(n)／n-g),(2≤n＜∞).3.︱s︱2﹦σ2﹢t2﹦(1／2)2﹢t2,t＞0,t2＞0,︱s︱min＞0.5；4.黎曼ζ(s)函数是研究素数分布的规律的,σ既然为一常数,因此,︱s︱和t必为含素数特征π(n)的函数；5．t＜︱s︱＜T,Whenn＜9146,t＜︱s︱＜n＜T；Whenn﹦9146,t＜︱s︱＜T﹦n；Whenn＞9146,t＜︱s︱＜T＜n；6．︱s︱﹦ρ·π(n)·㏒n,ρ﹦eπ(n)／n-0.914720032,(2≤n＜∞).MSC(2010)11-xx.关键词:黎曼猜想,黎曼ζ(s)函数,非平凡零点.定理1.ET﹦9π(eπ(n)／n-0.914720032)·π(n)·㏒n,(2≤n＜∞).(1)证.黎曼猜想(Ⅱ):若以N(T)表示ζ(s)在矩形0≤σ≤1,0＜t＜T中的零点个数,则N(T)﹦(T／2π)(㏒（T／2π）-1)﹢O(㏒T),N(T)个零点包含：1个平凡零点和(n﹣1)个非平凡零点,n﹦N(T).当T＜9146,n﹦N(T)＜T；当T﹦9146,n﹦N(T)﹦T；当T＞9146,n﹦N(T)＞T；我们有T∝π(n)·㏒n,T﹦k·π(n)·㏒n，k﹦9π(eπ(n)／n-g),（2≤n＜∞).当T﹦27.13793542,N(T)﹦n﹦2,π(2)﹦1,27.13793542﹦1·㏒2·k,k﹦39.15176485,39.15176485﹦9π(eπ(n)／n-0.264010847),当T﹦31.23136274,N(T)﹦n﹦3,π(3)﹦2,31.23136274﹦2·㏒3·k,k﹦14.21400573,14.21400573﹦9π(eπ(n)／n-1.445016424),当T﹦35.01237530,N(T)﹦n﹦4,π(4)﹦2,35.01237530﹦2·㏒4·k,k﹦12.62804505,12.62804505﹦9π(eπ(n)／n-1.202095539),当T﹦38.5682878,N(T)﹦n﹦5,π(5)﹦3,38.5682878﹦3·㏒5·k,k﹦7.987941525,7.987941525﹦9π(eπ(n)／n-1.539603161),当T﹦41.95146661,N(T)﹦n﹦6,π(6)﹦3,41.95146661﹦3·㏒6·k,k﹦7.804519771,7.804519771﹦9π(eπ(n)／n-1.372692848),当T﹦45.19632686,N(T)﹦n﹦7,π(7)﹦4,45.19632686﹦4·㏒7·k,k﹦5.806579364,5.806579364﹦9π(eπ(n)／n-1.565429217),当T﹦48.32696318,N(T)﹦n﹦8,π(8)﹦4,48.32696318﹦4·㏒8·k,k﹦5.810089177,5.810089177﹦9π(eπ(n)／n-1.443231401),当T﹦51.36104798,N(T)﹦n﹦9,π(9)﹦4,51.36104798﹦4·㏒9·k,k﹦5.843855074,5.843855074﹦9π(eπ(n)／n-1.352939403),当T﹦54.3120145,N(T)﹦n﹦10,π(10)﹦4,54.3120145﹦4·㏒10·k,k﹦5.89685205,5.89685205﹦9π(eπ(n)／n-1.283266219),当T﹦57.19036654,N(T)﹦n﹦11,π(11)﹦5,57.19036654﹦5·㏒11·k,k﹦4.770047065,4.770047065﹦9π(eπ(n)／n﹣1.406751198),当T﹦60.00450875,N(T)﹦n﹦12,π(12)﹦5,60.00450875﹦5·㏒12·k,k﹦4.829518143,4.829518143﹦9π(eπ(n)／n﹣1.346087532),当T﹦62.7612951,N(T)﹦n﹦13,π(13)﹦6,62.7612951﹦6·㏒13·k,k﹦4.078137379,4.078137379﹦9π(eπ(n)／n﹣1.442278292),