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Old 2020-01-19, 16:21   #2
Dr Sardonicus
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Feb 2017

59·71 Posts

Originally Posted by carpetpool View Post
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
This paper 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 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 k1 = 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 k1[x], so K is a four-group extension of k1.

The splitting field K has two other quadratic subfields, k2 = Q(sqrt(-39)) and k3 = Q(sqrt(-3)). The polynomial f remains irreducible in k2[x] and k3[x], so K is a cyclic quartic extension of k2 and k3.

The field k3 is also the field of cube roots of unity.

The extension K/k2 is interesting because the class group of k2 is cyclic of order 4, and K is the Hilbert Class Field of k2.
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