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How to prove that every third Fibonacci number is even?

  1. Mar 21, 2017 #1
    1. The problem statement, all variables and given/known data
    "Consider the sequence ##F_1##, ##F_2##, ##F_3##, . . . , where
    ##F_1 = 1##, ##F_2 = 1##, ##F_3 = 2##, ##F_4 = 3##, ##F_5 = 5## and ##F_6 = 8##.
    The terms of this sequence are called Fibonacci numbers.
    (a) Define the sequence of Fibonacci numbers by means of a recurrence relation.
    (b) Prove that ##2 | F_n## if and only if ##3 | n##.

    2. Relevant equations
    (a) ##F_n=F_{n-2}+F_{n-1}##

    3. The attempt at a solution
    (b)
    Basically, I'm going to express every Fibonacci number in terms of ##mod2## and express ##n## as ##n=3x## for some ##x∈ℕ##.

    For ##x=1##, ##F_3=F_1+F_2=1mod2+1mod2=0mod2##.
    Then assuming that ##F_{3k}=0mod2## for some ##k>1##, I need to prove that ##F_{3(k+1)}=F_{3k+1}+F_{3k+2}##.

    So I have ##F_{3k+1}=0mod2+1mod2## and ##F_{3k+2}=0mod2+1mod2##.
    Adding them up gives ##F_{3(k+1)}=F_{3k+1}+F_{3k+2}=0mod2+1mod2+0mod2+1mod2=0mod2##.

    And I'm pretty sure this isn't sufficient to complete the inductive proof. Can anyone check my work? Thanks.
     
  2. jcsd
  3. Mar 21, 2017 #2

    PeroK

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    Science Advisor
    Homework Helper
    Gold Member

    You we're supposed to show iff.

    The proof would be simpler if you considered the sequence in sets of 3 and explicitly showed that, modulo 2, the sequence is:

    ##1, 1, 0,1,1,0 \dots##
     
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