20190205, 06:46  #1 
Sep 2002
Database er0rr
4811_{10} 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
3·7·311 Posts 
Did you mean kronecker(b*(a+2),n)?

20190205, 15:32  #3 
Sep 2002
Database er0rr
17×283 Posts 

20190205, 23:21  #4 
Sep 2002
Database er0rr
4811_{10} 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^{2}·3·499 Posts 
Would you post your script?

20190206, 03:45  #6 
Sep 2002
Database er0rr
17×283 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
3·7·311 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
4811_{10} Posts 
Quote:


20190212, 06:32  #10 
Sep 2002
Database er0rr
4811_{10} Posts 

20190215, 05:22  #11 
Sep 2002
Database er0rr
17·283 Posts 
Testing has reached 2*10^11 with Pari/GP, having now switched over to the much quicker attached GMP script.

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