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carpetpool 2018-06-27 03:12

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 k-th 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^q-1.

Is it possible to apply some kind of "reciprocity" law to the Lucas numbers and their generalizations such as the Companion-Pell 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)?


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