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This is an exercise from Book of proof from page 197. I tried to use proof by smallest counterexample in the second step. But I don't know if my proof is correct using this approach. It's my first time trying to use this method to proof a proposition.

First I proof if $3\mid n$, then $n$-th fib number is even.

  1. $3\mid 3$, and $F_{3}=2$; $3\mid 6$, and $F_{6}=8$; $3\mid 9$, and $F_{9}=34$
  2. Assume that $n=3k$ for some $k\in\mathbb{N}$, then $F_{3k}=2a$ for some $a\in\mathbb{N}$
  3. Next I want to proof that $F_{3(k+1)}=2b$ for some $b\in\mathbb{N}$ $$ F_{3(k+1)}=F_{3k+3}=F_{3k+1}+F_{3k+2}=2F_{3k+1}+F_{3k}=2F_{3k+1}+2a=2(F_{3k+1}+a) $$ then we got $b=F_{3k+1}+a\in\mathbb{N}$, $F_{3(k+1)}$ is an even number. First part of this proof finished.

Then I proof that if the $n$-th fib number is even, $3\mid n$.

  1. $F_{3}=2$, and $3\mid 3$; $F_{6}=8$, and $3\mid 6$; $F_{9}=34$, and $3\mid 9$
  2. Assume that $F_{n}$ is even for some $n$, thus $F_{n}=2a$ for some $a\in\mathbb{N}$, then $3\mid n$

I don't know where then next even fib is, so I tried the smallest counterexample proof here. But I don't know if it's correct setup such an assumption and if it's enough to only use 2 cases to proof the proposition.

  1. Assume that $F_{m}=2b$ is the next even fib number and $3\nmid m$, then there are 2 situations:
  • $m=n+1$: $F_{n+1}=F_{n}+F_{n-1}=2a+F_{n-1}=2b$, then $F_{n-1}$ is an even number, but it disobeys the rules that if the $n-th$ fib number is even, then $3\mid n$. Because $n-1$ is not a multiple of 3.
  • $m=n+2$: $F_{n+2}=F_{n}+F_{n+1}=2a+F_{n+1}=2b$, then $F_{n+1}$ is an even number, but it disobeys the rules that if the $n-th$ fib number is even, then $3\mid n$. Because $n+1$ is not a multiple of 3.
  1. Therefore if the $n$-th fib number is even, $3\mid n$

QED.

I wonder if my proof is correct. Please tell me if you find any step unreasonable.

Sahaj
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Gwen
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    Looks correct to me. You may want to search the site, for this and similar questions have been looked at earlier. Pisano period is a usefull buzzword (most likely it is studied in some Wikipedia article as well). Also, it may be simpler to do both parts with a SINGLE similar induction, when you declare the claim $P(k)$ to be proved to state that "$F_{3k}$ is even, $F_{3k+1}$ is odd and $F_{3k+2}$ is odd". It seems to me that such a way of organizing it makes it more transparent, no? Check $P(0)$ (=the base case), and use the recurrence to show that $P(k+1)$ is true assuming $P(k)$ is! – Jyrki Lahtonen Sep 20 '24 at 08:12
  • Tag description "solution-verification": "For posts looking for feedback or verification of a proposed solution. "Is my proof correct?" is too broad or missing context. Instead, the question must identify precisely which step in the proof is in doubt, and why so. This should not be the only tag for a question, and should not be used to circumvent site policies regarding duplicate questions. " – Paul Frost Sep 20 '24 at 08:28
  • I edited the problem and tried to change the tag to proof-verification but failed. – Gwen Sep 20 '24 at 08:52
  • "proof-verification" is just a synonym of "solution-verification". See https://math.stackexchange.com/tags/synonyms. Anyway, even you woud delete the tag, your question asks for verification and the above requirement applies. Your edit explained it, and it is fine now. – Kritiker der Elche Sep 20 '24 at 09:02
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    Immediate consequence of the Fib gcd formula $,(f_m,f_n) = f_{(m,n)}$ in the dupe, viz.

    $$\underbrace{\color{#c00}{(2,f_n)} = (f_3,f_n) = f_{\color{#0a0}{(3,n)}}}_{!!\large \neq, 1\iff \color{#c00}{\ 2\ \mid\ f_n} \iff \color{#0a0}{3\ \mid\ n}}$$

     – Bill Dubuque Sep 20 '24 at 17:19
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    If you specialize the linked proof to $,m=3,$ you get a short simple inductive proof of the above using $,f_n = f_3 f_{n-2} + f_2 f_{n-3} = 2 f_{n-2} + f_{n-3},,$ so $,f_n\equiv f_{n-3}\pmod{!2}.\ \ $ – Bill Dubuque Sep 20 '24 at 17:30
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    Genrally the above shows $\ f_m\mid f_n \iff m\mid n.\ $ OP is case $,m = 3,,$ so $,f_m = 2.\ \ $ – Bill Dubuque Sep 20 '24 at 17:37

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