20200118, 20:51  #1 
"Sam"
Nov 2016
280_{10} Posts 
Counterexample to test ?
Is anyone able to come up with a counterexample to the following test? I'm sure there are some, I just can't construct any. Please let me know.
Define a(n) (case where b=1 see below) to be https://oeis.org/A116201 Let n = 2 mod 3 and Jacobi(13  n) = 1. If a(n^2+1) = 0 mod n, then n is prime. In general, there should be an integer b such that Jacobi([4*b7]  n) = 1 and Jacobi([b^2+4*b+8]  n) = 1 and a similar test would apply. This stems from the discriminant of the characteristic polynomial: x^4 + x^3  b*x^2  x + 1. This polynomial has discriminant (4*a  7)^2*(a^2 + 4*a + 8). The reasoning for the above test is similar to that of Fermat's Little Theorem. If n is prime and: n = 1 mod 3, then a(n1) divides n. n = 2 mod 3 and J(13  n)=1, then a(n+1) divides n. n = 2 mod 3 and J(13  n)=1, then a(n^2+1) divides n. The first two conditions, counterexamples are easily construable. But the third is a little difficult. Here's how to construct one (if they exists): We all know the properties of Carmichael numbers: p1  N1 for all primes p  N. And then there is something similar for other numbers N, which is that p+1  N+1 for all primes p  N. And finally we look at integers N such that p^2+1  N^2+1 for all primes p  N. While I have been able to verify that if N > 10^9 if such an N exists, I have not been able to prove that they exist. I imagine they would be very difficult to construct, but still possible. So if someone is able to prove that they do NOT exist at all, then the test is proved. Otherwise, it will remain a theoretical PRP test. 
20200119, 16:21  #2  
Feb 2017
Nowhere
3,251 Posts 
Quote:
See also this paper. Looking at the OEIS entry, I asked myself, "Why is this a divisibility sequence?" The characteristic polynomial f = x^4  x^3  x^2  x + 1 for the recurrence is a quartic with Galois group D4 (the dihedral group with 8 elements, AKA the "group of the square"). I knew that the sequence could be expressed in the form a(k) = trace(y*x^k) where y = Mod(p, f), with p a nonzero polynomial in x of degree less than 4. Luckily, I know how to compute y. And what the computation showed was that for the given sequence, y^2  13 = 0. I'm sure that this is the key to the sequence being a divisibility sequence, but I'm too lazy to work out the details. The field k_{1} = Q(sqrt(13)) is a subfield of the splitting field K of f; K/Q has degree 8. Also, f splits into two quadratic factors in k_{1}[x], so K is a fourgroup extension of k_{1}. The splitting field K has two other quadratic subfields, k_{2} = Q(sqrt(39)) and k_{3} = Q(sqrt(3)). The polynomial f remains irreducible in k_{2}[x] and k_{3}[x], so K is a cyclic quartic extension of k_{2} and k_{3}. The field k_{3} is also the field of cube roots of unity. The extension K/k_{2} is interesting because the class group of k_{2} is cyclic of order 4, and K is the Hilbert Class Field of k_{2}. 

Thread Tools  
Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Modifying the Lucas Lehmer Primality Test into a fast test of nothing  Trilo  Miscellaneous Math  25  20180311 23:20 
Double check LL test faster than first run test  lidocorc  Software  3  20081203 15:12 
Will the torture test, test ALL available memory?  swinster  Software  2  20071201 17:54 
A counterexampleanyoneII  devarajkandadai  Miscellaneous Math  3  20051212 16:04 
A primality test for Fermat numbers faster than Pépin's test ?  T.Rex  Math  0  20041026 21:37 