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carpetpool 2020-01-18 20:51

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 [url]https://oeis.org/A116201[/url]

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*b-7] | 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(n-1) 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: p-1 | N-1 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.

Dr Sardonicus 2020-01-19 16:21

[QUOTE=carpetpool;535466]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 [url]https://oeis.org/A116201[/url]
<snip>[/QUOTE][url=https://arxiv.org/pdf/1903.06825.pdf]This paper[/url] may be pertinent regarding the existence of counterexamples. In particular, it proves that there are Carmichael numbers composed entirely of primes p for which the given polynomial (mod p) splits into linear factors.

See also [url=https://arxiv.org/pdf/math/9812089.pdf]this paper[/url].

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[sub]1[/sub] = 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[sub]1[/sub][x], so K is a four-group extension of k[sub]1[/sub].

The splitting field K has two other quadratic subfields, k[sub]2[/sub] = Q(sqrt(-39)) and k[sub]3[/sub] = Q(sqrt(-3)). The polynomial f remains irreducible in k[sub]2[/sub][x] and k[sub]3[/sub][x], so K is a cyclic quartic extension of k[sub]2[/sub] and k[sub]3[/sub].

The field k[sub]3[/sub] is also the field of cube roots of unity.

The extension K/k[sub]2[/sub] is interesting because the class group of k[sub]2[/sub] is cyclic of order 4, and K is the Hilbert Class Field of k[sub]2[/sub].


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