Lucas number cubic and quadratic reciprocity
If q is a Sophie Germain prime (p = 2*q+1 is also prime), then it is well known that a^q = 1 modulo p if and only if a is a quadratic residue modulo p (leave a = 0, 1, 1 aside as special cases).
In the same sense, if q is a prime, and p = q*k+1 is also prime, then a^q = 1 modulo p if and only if a is a kth power residue modulo p. Equivalently stated, the former is true if and only if x^k = a modulo p is solvable. This is a nice and easy "reciprocity" law to determine weather or not p divides a^q1.
Is it possible to apply some kind of "reciprocity" law to the Lucas numbers and their generalizations such as the CompanionPell numbers and Lucas Polynomial sequences?
Let L(n) be the nth Lucas Number. If q is a Sophie Germain prime and p = 2*q+1, then p divides L(q) if and only if p = q = 4 modulo 5.
What if q and p = q*k+1 are primes where k > 2? What is the "reciprocity" law to determine weather or not p divides L(q) for given primes p,q with the conditions above?
For instance, if p = 6*q+1 where p and q are primes, already given that p = 4 modulo 5, what other rule is there to determine weather or not p divides L(q) or it divides L(3*q)?
