Quote:
Originally Posted by paulunderwood
Without checking Feitsma, we now have simple tests for n%8 = {3,5,7}

Transforming the test of x over x^2136*x+16, I have now checked Feitsma's 2^64 PSP2 list against:
(n%8==3n%8==5)&&Mod(Mod(x,n),x^21154*x+1)^((n+1)/2)==1.
Given the test Mod(Mod(x+2),x^2+1)^(n+1)==5 seems to be good enough for n%4==3, then n%8==3 is covered by both tests. Which you choose depends on your needs and FFT arithmetic etc.
I am pleased with this advancement spurred on by the good doctor.
Similar antics for 3 mod 8 and 5 mod 8 can be done with x^220*x+2 with discriminant 2*2^2*3^2. Fietsma's list check out with Euler 2PSPs against (n%8==3n%8==5)&&Mod(Mod(x,n),x^2198*x+1)^((n+1)/2)==kronecker(2,n).