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Math Help - A Tricky Fibonacci Sequence Identity

  1. #1
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    A Tricky Fibonacci Sequence Identity

    The Fibonacci Sequence is define as Fn=Fn-2 + Fn-1 where F0=0 and F1=1;

    Suppose that a Fibonacci number Fn is divisible by some positive integer d, and Fn+1=k (2).

    Show that Fn (congruent) kF0 (mod d) and Fn+1 (congruent) kF1 (mod d).

    I tried to use the given information i.e d|Fn --> Fn=dL for some L (1)

    and I used the definition of Fn and the given Fn+1=k and added both equation (1) and (2) to arrive at an identity similar but different to what is required.

    Can you kindly shed some light on this matter!

    Thanks
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  2. #2
    MHF Contributor chiph588@'s Avatar
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     d\mid F_n\implies F_n\equiv 0 \bmod{d} and  F_0=0 ...

     F_1=1 and  F_{n+1}=k \implies F_{n+1}\equiv k\bmod{d} . Can you see why?
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  3. #3
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    Hmm.. I'm a little concused as to why Fn+1- k / d ??

    Cna you explain why please?

    Thanks
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  4. #4
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    Is it because Fn cong 0 mod can be modified to Fn+1 cong k*F0 mod d ?
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