20190205, 06:46  #1 
Sep 2002
Database er0rr
D76_{16} Posts 
Amazing 6
Given nonsquare odd candidate prime n and b=6 and any r so that a=b^r, where kronecker(a^24,n)==1 and gcd(a^3a,n)==1. then the test
(b*x)^((n+1)/2) == b*kronecker(b*(a+2)) (mod n, x^2a*x+1) implies n is prime! Proof? Counterexample? A strong test? 
20190205, 13:41  #2 
Feb 2017
Nowhere
2·1,787 Posts 
Did you mean kronecker(b*(a+2),n)?

20190205, 15:32  #3 
Sep 2002
Database er0rr
3446_{10} Posts 

20190205, 23:21  #4 
Sep 2002
Database er0rr
110101110110_{2} Posts 
Since I stipulate that gcd(a^3a,n)==1, I may as well say gcd(210,n)==1, since 6 divides a, 5 divides 6^r1 for all r and 7 divides 6^r+1 for all r.
Testing has now reached 5*10^10 
20190206, 03:19  #5 
Aug 2006
2×5×593 Posts 
Would you post your script?

20190206, 03:45  #6 
Sep 2002
Database er0rr
6566_{8} Posts 
Sure. Here you go. It needs some enhancements and formatting :
Code:
b=6;forstep(n=1,10000000000000,2,if(n%10000000==1,print(">>"n));if(!ispseudoprime(n)&&!issquare(n)&&Mod(b,n)^(n1)==1,z=znorder(Mod(b,n));for(r=0,z,a=lift(Mod(b,n)^r);if(gcd(a^3a,n)==1&&kronecker(a^24,n)==1&&Mod(Mod(1,n)*b*x,x^2a*x+1)^((n+1)/2)==b*kronecker(b*(a+2),n),print([n,r,a]))))) Last fiddled with by paulunderwood on 20190206 at 03:48 
20190206, 13:34  #7 
Feb 2017
Nowhere
2·1,787 Posts 
Fascinating. Assuming this is a property possessed by all primes p satisfying the conditions, it implies that
if kronecker(a^2  4, p) = 1, then lift(Mod(x,x^2  Mod(a,p)*x + 1)^((p+1)/2))) == kronecker(a+2, p). I'm convinced that this is true, but I don't see why it's true. What is clear to me is, it's either kronecker(a+2,p) or kronecker(a2, p). I'm probably overlooking something obvious... 
20190206, 14:02  #8  
Sep 2002
Database er0rr
2·1,723 Posts 
Quote:


20190212, 06:32  #10 
Sep 2002
Database er0rr
110101110110_{2} Posts 

20190215, 05:22  #11 
Sep 2002
Database er0rr
2·1,723 Posts 
Testing has reached 2*10^11 with Pari/GP, having now switched over to the much quicker attached GMP script.

Thread Tools  
Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Don't miss this amazing trick that the moon is going to do!  Uncwilly  Astronomy  6  20180201 05:40 
Amazing result in P1  Miszka  Information & Answers  2  20140704 17:11 
Amazing academic resource  Xyzzy  Lounge  6  20120325 22:57 
Amazing!!!  R.D. Silverman  Factoring  5  20060126 09:14 
Two Amazing Things  clowns789  Hardware  1  20031227 16:57 