New Wagstaff PRP exponents - Page 3

paulunderwood · 2013-09-19, 23:57

Quote:

Originally Posted by ATH

I lost the other test on (2^13347311+1)/3 on the way, 10 days is a long time without any intermediate savefiles. Several other people seems to have these test running on faster computers than mine though, so I didn't restart it.

I've got 'em covered. One more day to go on a 4770k at 4GHz.

paulunderwood · 2013-09-20, 22:10

Code:

Running N+1 test using discriminant 2, base 1+sqrt(2)                                    
(2^13347311+1)/3 is Lucas PRP! (442900.4907s+0.0204s)

and confirming ATH's result:

Code:

Running N+1 test using discriminant 2, base 1+sqrt(2)                                    
(2^13372531+1)/3 is Lucas PRP! (443577.1559s+0.0284s)

The tests for (L+2)^(n+1)==5 (mod n, L^2+1) will take another 3 to 4 weeks

paulunderwood · 2013-10-17, 21:27

Quote:

Originally Posted by paulunderwood

The tests for (L+2)^(n+1)==5 (mod n, L^2+1) will take another 3 to 4 weeks

The tests passed for both PRPs.

I have emailed Henri Lifchitz requesting that the annotations for the PRPs in his database lists all our PRP tests.

ATH · 2013-10-18, 00:58

Quote:

Originally Posted by paulunderwood

The tests for (L+2)^(n+1)==5 (mod n, L^2+1) will take another 3 to 4 weeks

Sorry for my ignorance but n is the Wagstaff PRP right, but what is L ? and what is meant by "n, L^2 +1" in the modular expression? My guess is that (L+2)^(n+1)==5 for both (mod n) and (mod L^2+1) ?

(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)

paulunderwood · 2013-10-18, 01:33

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+6-4==2
(L+2)^6==2^2==4
(L+2)^12==4^2==16==5

This will work for all prime n==3 (mod 4), but is not proven that all composites will fail. Wagstaff are 3 (mod 4)

To cover all n, I take minimal x>=0 such that jacobiSymbol(x^2-4,n)==-1, and check:

Code:

(L+2)^(n+1)==5+2*x (mod n, L^2-x*L+1)

This is the basis of my JavaScript program, which has the running time of 2 Miller-Rabin rounds. It is explained in my pre-print "Quadratic Composite Tests" where I used $\lambda$ instead of "L". The general "minimal x" method is good for n < 10^13 -- I am currently testing n < 7*10^13.

2013-09-20, 22:10	#24
paulunderwood Sep 2002 Database er0rr 11·383 Posts	Code: Running N+1 test using discriminant 2, base 1+sqrt(2) (2^13347311+1)/3 is Lucas PRP! (442900.4907s+0.0204s) and confirming ATH's result: Code: Running N+1 test using discriminant 2, base 1+sqrt(2) (2^13372531+1)/3 is Lucas PRP! (443577.1559s+0.0284s) The tests for (L+2)^(n+1)==5 (mod n, L^2+1) will take another 3 to 4 weeks Last fiddled with by paulunderwood on 2013-09-20 at 22:14

2013-10-18, 01:33	#27
paulunderwood Sep 2002 Database er0rr 11×383 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+4L+4==4L+3 (L+2)^3==(4L+3)(L+2)==4L^2+11L+6==0L+6-4==2 (L+2)^6==2^2==4 (L+2)^12==4^2==16==5 This will work for all prime n==3 (mod 4), but is not proven that all composites will fail. Wagstaff are 3 (mod 4) To cover all n, I take minimal x>=0 such that jacobiSymbol(x^2-4,n)==-1, and check: Code: (L+2)^(n+1)==5+2x (mod n, L^2-xL+1) This is the basis of my JavaScript program, which has the running time of 2 Miller-Rabin rounds. It is explained in my pre-print "Quadratic Composite Tests" where I used $\lambda$ instead of "L". The general "minimal x" method is good for n < 10^13 -- I am currently testing n < 710^13. Last fiddled with by paulunderwood on 2013-10-18 at 01:47

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