20130919, 23:57  #23 
Sep 2002
Database er0rr
2^{2}×3×17×19 Posts 
I've got 'em covered. One more day to go on a 4770k at 4GHz.
Last fiddled with by paulunderwood on 20130920 at 00:08 
20130920, 22:10  #24 
Sep 2002
Database er0rr
2^{2}×3×17×19 Posts 
Code:
Running N+1 test using discriminant 2, base 1+sqrt(2) (2^13347311+1)/3 is Lucas PRP! (442900.4907s+0.0204s) Code:
Running N+1 test using discriminant 2, base 1+sqrt(2) (2^13372531+1)/3 is Lucas PRP! (443577.1559s+0.0284s) Last fiddled with by paulunderwood on 20130920 at 22:14 
20131017, 21:27  #25  
Sep 2002
Database er0rr
2^{2}·3·17·19 Posts 
Quote:
I have emailed Henri Lifchitz requesting that the annotations for the PRPs in his database lists all our PRP tests. Last fiddled with by paulunderwood on 20131017 at 21:37 

20131018, 00:58  #26  
Einyen
Dec 2003
Denmark
2×7×227 Posts 
Quote:
(and yes I did spend a bit of time looking for an answer myself, before I get told to study myself :) But not too much since it is not that important) Last fiddled with by ATH on 20131018 at 01:02 

20131018, 01:33  #27 
Sep 2002
Database er0rr
2^{2}·3·17·19 Posts 
The modular reduction is done over "n" and "L^2+1". Here is a worked example for n=11 where L^2==1 (mod 11, L^2+1)
Code:
(L+2)^2==L^2+4*L+4==4*L+3 (L+2)^3==(4*L+3)*(L+2)==4*L^2+11*L+6==0*L+64==2 (L+2)^6==2^2==4 (L+2)^12==4^2==16==5 To cover all n, I take minimal x>=0 such that jacobiSymbol(x^24,n)==1, and check: Code:
(L+2)^(n+1)==5+2*x (mod n, L^2x*L+1) Last fiddled with by paulunderwood on 20131018 at 01:47 
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