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 Showing results 1 to 11 of 11 Search took 0.01 seconds. Search: Posts Made By: wsc812
 Forum: Math 2013-03-21, 23:18 Replies: 3 Views: 1,430 Posted By wsc812 GraphPlot[Table[i->Mod[2*i^2-1,Mq],{i,Mq}]] GraphPlot[Table[i->Mod[2*i^2-1,Mq],{i,Mq}]]
 2012-12-20, 03:33 Replies: 36 Views: 3,317 Posted By wsc812 see my complete proof on mathoverflow ... see my complete proof on mathoverflow http://mathoverflow.net/questions/115149/is-there-a-composite-number-that-satisfies-these-conditions/115262#115262
 2012-12-13, 10:18 Replies: 36 Views: 3,317 Posted By wsc812 I have found the Complete proof for N=4k+3!!!!! I have found the Complete proof for N=4k+3!!!!!
 2012-12-10, 02:45 Replies: 36 Views: 3,317 Posted By wsc812 how to prove... how to prove q_1t^3+(k_2-1)t^2-k_2((q_1^2-1)k_1+1)^2=0 has no Positive integer root, t is variable ,q_1 is constant and k_1,k_2 are parameter q_1>0,k_1>0,k_2>0 ,and all character represents...
 2012-12-07, 02:10 Replies: 36 Views: 3,317 Posted By wsc812 123 123
 2012-12-06, 21:09 Replies: 36 Views: 3,317 Posted By wsc812 I found all these psp(3+2i) are not spsp(13), so... I found all these psp(3+2i) are not spsp(13), so we can find its factors easily. now whether we can give a certanty primality test no matter what it's 4k+1 or 4k+3
 2012-12-06, 19:49 Replies: 36 Views: 3,317 Posted By wsc812 these numbers are all psp(13) and have no 4k+3... these numbers are all psp(13) and have no 4k+3 factors ,we may decompose it by using extraction squre root constantly until it become a complex. e.g (3+2i)^20300=80852+1631i (mod 192401 )...
 2012-12-06, 18:31 Replies: 36 Views: 3,317 Posted By wsc812 2465=5*17*29 10585=5*29*73 162401=17*41*233 2465=5*17*29 10585=5*29*73 162401=17*41*233
 2012-12-06, 11:29 Replies: 36 Views: 3,317 Posted By wsc812 sorry! R. Gerbicz .I'm not good at English ,I... sorry! R. Gerbicz .I'm not good at English ,I don't understand your meaning .now my proof is in progress,and at first we can conclude there is no 4k+1 factors for N=4k+3 if a counter-examples...
 2012-12-06, 08:48 Replies: 36 Views: 3,317 Posted By wsc812 is algorithm complexity O(log(N)) For N=4k+3? we... is algorithm complexity O(log(N)) For N=4k+3? we may not select base a+bi, a= b for effectively testing, and calculate (a+bi)^{N+1}=a^2+b^2 (mod N) instead of (a+bi)^N=a-bi(mod N)
 2012-12-05, 13:21 Replies: 36 Views: 3,317 Posted By wsc812 a new Deterministic primality testing we know that if $q=4k+3$ ($q$ is a prime), then $(a+bI)^q=a^q+b^q(I)^{4k+3}(mod q) =a -bI$ for every gaussian integer $(a+bi)$ ,Now consider a composite $N=4k+3$ satisfies this condistion for...
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