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The dumb conjecture of the day

  1. May 23, 2008 #1
    Let [tex]Q(\sqrt{k})[/tex], for some positive integer k, be the extension of the field of rationals with basis [tex](1, \sqrt{k})[/tex]. For example, in [tex]Q(\sqrt{5})[/tex] the element [tex]({1 \over 2}, {1 \over 2})[/tex] is the golden ratio = [tex]{1 \over 2} + {1 \over 2}\sqrt{5}[/tex].

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

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

    Conj. 2: When [tex]k \equiv 1 \ ("mod" \ 4)[/tex], [tex]B(n)[/tex] is the closed-form formula for the recurrence sequence
    [tex]x_n = x_{n-1} + A \; x_{n-2}[/tex], n = 2, 3, 4, ...​
    [tex]x_0 = 0, \ \ x_1 = 1[/tex]​
    with [tex]A = {{k - 1} \over 4}[/tex].
     
  2. jcsd
  3. May 23, 2008 #2

    Hurkyl

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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 [itex]p^n[/itex] and [itex](1-p)^n[/itex] would be a basis for the solution space to the unconstrained version of your difference equation.
     
  4. May 23, 2008 #3
    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.
     
  5. May 23, 2008 #4
    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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