Proving Conjectures About the Binet Formula for Q(\sqrt{k})

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SUMMARY

The discussion centers on the Binet formula for the field extension Q(√k), specifically addressing two conjectures regarding integer outputs and recurrence sequences. Conjecture 1 states that B(n) yields only integers if and only if k ≡ 1 (mod 4). Conjecture 2 posits that for k ≡ 1 (mod 4), B(n) serves as the closed-form solution for the recurrence sequence defined by x_n = x_{n-1} + A x_{n-2}, with A = (k - 1)/4. The participants explore the implications of these conjectures and their interdependencies, emphasizing the significance of the congruence condition on k.

PREREQUISITES
  • Understanding of field extensions, specifically Q(√k)
  • Familiarity with the Binet formula and its applications
  • Knowledge of recurrence relations and sequences
  • Basic modular arithmetic, particularly congruences
NEXT STEPS
  • Study the generalized Binet formula for recurrence relations, particularly G(n) = aG(n-1) + bG(n-2)
  • Research integer sequences generated by Binet-like formulas
  • Explore the implications of modular arithmetic in number theory
  • Review the reference: Vella, A. and Vella, D. (2006) on cycle lengths in generalized Fibonacci sequences
USEFUL FOR

Mathematicians, number theorists, and students interested in field theory, recurrence relations, and integer sequences will benefit from this discussion.

dodo
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Let Q(\sqrt{k}), for some positive integer k, be the extension of the field of rationals with basis (1, \sqrt{k}). For example, in Q(\sqrt{5}) the element ({1 \over 2}, {1 \over 2}) is the golden ratio = {1 \over 2} + {1 \over 2}\sqrt{5}.

Given an extension Q(\sqrt{k}), let B(n) denote the 'Binet formula',
B(n) = {{p^n - (1-p)^n} \over \sqrt k}, n = 0, 1, 2, ...​
where p = ({1 \over 2}, {1 \over 2}).

Conj. 1: B(n) produces only integers, iif k \equiv 1 \ ("mod" \ 4).

Conj. 2: When k \equiv 1 \ ("mod" \ 4), B(n) is the closed-form formula for the recurrence sequence
x_n = x_{n-1} + A \; x_{n-2}, n = 2, 3, 4, ...​
x_0 = 0, \ \ x_1 = 1​
with A = {{k - 1} \over 4}.
 
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Isn't the second conjecture a straightforward exercise? And you don't even have to bother solving the ordinary linear difference equation: you already know the (putative) solution! (Why do you have the congruence condition on k for conjecture 2?)

And isn't conjecture 1 a trivial consequence of conjecture 2?


I suppose one thing might help to see conjecture 2: if it's right, then p^n and (1-p)^n would be a basis for the solution space to the unconstrained version of your difference equation.
 
I'm not sure I follow. (And in any case it's not trivial to me.) :D

As I see it, the first conjecture is a necessary previous step, for the second to be applicable. Possibly because I was interested in recurrence sequences returning only integers and with A = 1, 2, 3, ..., and frankly did not pay attention to the fact that B(n) could in fact be also the closed form for rational values of A and rational-producing sequences.

I'll try to digest what you're saying.
 
There is a generalised Binet formula for the recurrence relation G(n) = aG(n-1) + bG(n-2), which explains much more than the problem raised.
The system does not allow me to post URLs to other sites as I have made less than 15 posts(why ?), but googling "generalised fibonacci" helps, and I also give a "traditional" reference:
Vella, A. and Vella, D. (2006) Calculating exact cycle lengths in the generalised Fibonacci sequence modulo p. Math. Gaz. 90 (available on the internet).
 

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