20160801, 22:56  #12 
Sep 2002
Database er0rr
3·11·107 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^p1;D=mp1;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 This means 3^((mp1)/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 20160801 at 23:21 
20160801, 23:38  #13 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
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 20160801 at 23:46 
20160802, 00:05  #14 
Sep 2002
Database er0rr
DCB_{16} Posts 

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