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 Showing results 1 to 4 of 4 Search took 0.01 seconds. Search: Posts Made By: wsc811
 Forum: Miscellaneous Math 2013-08-29, 10:36 Replies: 6 why continued fractions gives one factor for N=(4m+3)(4n+3) Views: 1,580 Posted By wsc811 you can observe the following list N=4181 as complex {0,1,0,65} {1,-44,65,-3} {2,7,67,19} {3,-25,66,-5} {4,-28,59,-4} {5,-49,53,-2} {6,-44,45,-2} {7,-53,43,-2} {8,-4,63,-32}
 Forum: Miscellaneous Math 2013-08-28, 01:33 Replies: 6 why continued fractions gives one factor for N=(4m+3)(4n+3) Views: 1,580 Posted By wsc811 Or you can use complex number d = 11 - 4 I; pell = -1; P = 0; Q = 1; x = (P + Sqrt[d])/Q; a = Round[x]; i = 0; While[(x[i] != 1/(x - a) && P[i] != pell) || i == 1, P[i + 1] = Q[i] a[i] - P[i]; ...
 Forum: Miscellaneous Math 2013-08-28, 01:27 Replies: 6 why continued fractions gives one factor for N=(4m+3)(4n+3) Views: 1,580 Posted By wsc811 my mathematica code d = 23*43; pell = -1; P = 0; Q = 1; x = (P + Sqrt[d])/Q; a = IntegerPart[x]; i = 0; While[(x[i] != 1/(x - a) && P[i] != pell) || i == 1, P[i + 1] = Q[i] a[i] - P[i];...
 Forum: Miscellaneous Math 2013-08-25, 07:56 Replies: 6 why continued fractions gives one factor for N=(4m+3)(4n+3) Views: 1,580 Posted By wsc811 why continued fractions gives one factor for N=(4m+3)(4n+3) for example N=989=23*43 ,Sqrt[N]={a0;a1,a2,...,2a0} {n,Q,P,a} {0,1,0,31} {1,28,31,2} {2,13,25,4} {3,20,27,2} {4,41,13,1} {5,5,28,11} {6,52,27,1} {7,7,25,8}
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