Thread: kronecker(2,n)==-1 test View Single Post 2021-02-18, 02:33 #13 paulunderwood   Sep 2002 Database er0rr F6916 Posts Unique solution test Code: { tst(n,a)=local(A=a^2+a); kronecker(A,n)==-1&& kronecker(a,n)==-1&& Mod(A,n)^((n-1)/2)==-1&& Mod(a,n)^((n-1)/2)==-1&& Mod(Mod(z,n),z^2-(4*a/(a-1)-2)*z+1)^((n+1)/2)==-1 } I am running this test against Richard Pinch's Carmichael number list and with David Broadhurst's CRT Semi-prime Pari/GP script. The latter has only produced unique solutions for a given n, making 2 rounds with different a's sufficient for a primality proof?!?!? Those pesky primes! I just found 2 solutions for 2728624939. However gcd(a1^2-a2^2,n)>1 I have now found n with 3 or 4 solutions too, and gcd(ai^2-aj^2,n)>1 for i != j. I am seeing the same sort of GCDs for A=a+1 and kronecker(A,n)==-1. Hence the corresponding test can be done for suitable {a, a+1} and {a+1,a+2} which reduces the number of sub-tests, and no GCD need be calculated. Last fiddled with by paulunderwood on 2021-02-18 at 05:03  