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Math Help - Another fibonacci proposition

  1. #1
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    Another fibonacci proposition

    Hey all I would really love some help with this fibonacci proposition:

    For all k,m in the Natural Numbers, fib(mk) is divisible by fib(m)

    Thanks a bunch!
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by CaptainBlack View Post
    Now:

    F_{m(k+1)}=F_{mk}+F_{m(k-1)}

    CB
    I don't think this is true. For example, F_{2(k+1)}=F_{2k+2}=F_{2k+1}+F_{2k}, not F_{2k}+F_{2k-2}.
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by Bruno J. View Post
    I don't think this is true. For example, F_{2(k+1)}=F_{2k+2}=F_{2k+1}+F_{2k}, not F_{2k}+F_{2k-2}.
    Opps, will probably have to go back to the other method that I discarded as too complicated (if I can recall it now..).

    CB
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  4. #4
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    I think the original problem can be solved using the identity

    F_{m+n} = F_{m-1}F_n + F_mF_{n+1}

    which holds for all positive integers m,n. This identity can be proved by fixing m, say, and inducting on n. Then, prove that

    F_m divides F_{mk}

    with another inductive argument, using the above identity.
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