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 2016-08-01, 22:56 #12 paulunderwood     Sep 2002 Database er0rr 7×491 Posts I have extended the nub of this thread a little further, but it still seems to be computationally useless. For example b = 3 and p = 61 so that mp = 2^61 - 1, I ran this: Code: ? p=61;mp=2^p-1;D=mp-1;V=factor(D,1000000);forbigdiv(D,d -> r=lift(Mod(3,mp)^(D/d));if(2^logint(r,2)==r,print(">>>"d))) >>>1 >>>61 >>>3 >>>183 >>>9 >>>549 Where "forbigdiv" is given here This means 3^((mp-1)/549) == 2^n for some n OR 3^4200078341008550 == 2^n for some n, thus implying M61 is prime. I will experiment with other bases b, for example p ... Last fiddled with by paulunderwood on 2016-08-01 at 23:21
2016-08-01, 23:38   #13
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

100000101100012 Posts

Quote:
 Originally Posted by paulunderwood Code: ? p=61;mp=2^p-1;D=mp-1;V=factor(D,1000000);forbigdiv(D,d -> r=lift(Mod(3,mp)^(D/d));if(2^logint(r,2)==r,print(">>>"d))) >>>1 >>>61 >>>3 >>>183 >>>9 >>>549
is your intent for D/d to get the largest one's out of the way first as D/2 will be the biggest exponent and will also be a divisor of D.

Last fiddled with by science_man_88 on 2016-08-01 at 23:46

2016-08-02, 00:05   #14
paulunderwood

Sep 2002
Database er0rr

7×491 Posts

Quote:
 Originally Posted by science_man_88 is your intent for D/d to get the largest one's out of the way first as D/2 will be the biggest exponent and will also be a divisor of D.
Some order is possible, I guess.

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