While studying the Fibonacci sequence I encountered this problem in the handout, and I can not understand how to do it.
Show that if the Fibonacci sequence has a term divisible by a natural number $m$, then there are infinitely many such terms.
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lumpy
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benjamin_ee
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See the solution of puzzle 20 here https://mathigon.org/puzzles#2021 – Hemant Agarwal Aug 21 '24 at 13:47
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The duplication identity for the Fibonacci numbers states: $$ F_{2m} = F_m \left( F_m + 2 F_{m-1} \right) $$
Hence, for some $k \in \mathbb{N}$, if $F_m$ is divisible by $k$, every element of the sequence $F_{2m}, F_{4m}, \ldots, F_{2^r m}, \ldots$ is also divisible by $k$.
Sasha
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Sorry but the way that I do not understand a way of answering this question – benjamin_ee Aug 15 '13 at 19:16
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@marcelolpjunior Since Fibonacci numbers are integers, $F_{2m}$ is an integer multiple of $F_{m}$. Hence if $F_m$ is divisible by som $k$, so is the $F_{2k}$. I hope this helps. – Sasha Aug 15 '13 at 19:51
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Hint: Consider the sequence modulo $m$. This sequence is (immediately) periodic.
I'm assuming you know what modular arithmetic is.
Claim: The sequence modulo $m$ is immediately periodic.
Proof:
Every term of the sequence is uniquely defined by the 2 terms preceding it. Hence, given any pair of consecutive terms, we know what the rest of the terms are.
Consider pair of consecutive terms taken modulo $m$. There are $m \times m$ possibilities for the pairs. Since this is finite, and we have infinitely many pairs, there must be (at least) 1 pair that repeats (at least once). From the first observation, this shows that the sequence is (eventually) periodic. [The period is a factor of the difference of the indices of the terms.]
In fact, if we know what 2 consecutive terms are, then we know what the previous term is, and hence every term that comes before it. Thus, the sequence is immediately periodic.
Calvin Lin
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I began my studies in the theory of numbers this year, still do not know modular arithmetic. – benjamin_ee Aug 15 '13 at 18:41
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I'm really sorry, but I honestly do not understand how to do it, how to solve the problem. – benjamin_ee Aug 15 '13 at 19:14
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@marcelo: I am puzzled. You seemed to be doing modular arithmetic just fine in this answer. What's the deal here? Are you in high school? I find it hard to believe that a college level elementary course in numbre theory wouldn't get to congruences, i.e. modular arithmetic very quickly. – Jyrki Lahtonen Aug 15 '13 at 19:51
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@marcelolpjunior I added more details for the periodic part, assuming that you know modular arithmetic. Is this clear now? – Calvin Lin Aug 15 '13 at 20:06
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@JyrkiLahtonen I'm sorry. Studied modular arithmetic just one day, so this issue but have not learned to do demonstrations using the. – benjamin_ee Aug 15 '13 at 20:42
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I thought so:$$\fbox{Lemma 1: $a\mid b\Longleftrightarrow u_a\mid u_b$, in Fibonacci sequence}$$We have $m\mid u_n$ and $\exists a\in\mathbb{N}$ with $a\geq n$ such that $n\mid a$ therefore, by Lemma 1: $u_n\mid u_a$; and if $n\mid a\Longrightarrow n\mid ak\;\;\;\forall k\in \mathbb{N}$ soon $$u_n\mid u_{ak}\;\;\forall k\in\mathbb{N}$$ as $m\mid u_n$ and $u_n\mid u_{ak}\;\;\forall k\in\mathbb{N}$ then $m\mid u_{ak}\;\;\forall k\in\mathbb{N}$ Since $k$ can take any natural $u_{ak}$ has infinite values.$\;\;\;\;\Box$$$$$Correct?
benjamin_ee
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I don't understand how the lemma could be true? The second entry in the Fibonacci sequence is $=1$ and thus divides all the others, even those with an odd index? – Jyrki Lahtonen Aug 15 '13 at 21:07
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1@JyrkiLahtonen: I think the $\implies$ is true (I show that in my answer), but the other direction is not (as you have explained). – robjohn Aug 15 '13 at 22:07
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Since $$ F_n=\frac{\phi^n-(-1/\phi)^n}{\sqrt5} $$ we also have $$ F_{mn}=\frac{\phi^{mn}-(-1/\phi)^{mn}}{\sqrt5} $$ Therefore, $$ \begin{align} \frac{F_{mn}}{F_n} &=\frac{\phi^{mn}-(-1/\phi)^{mn}}{\phi^n-(-1/\phi)^n}\\ &=\sum_{k=1}^m\phi^{(m-k)n}(-1/\phi)^{(k-1)n}\\ &=\sum_{k=1}^{\lfloor m/2\rfloor}\left(\phi^{(m-k)n}(-1/\phi)^{(k-1)n}+\phi^{(k-1)n}(-1/\phi)^{(m-k)n}\right)\\ &\hphantom{=}+(-1)^{(m-1)n/2}[m\text{ is odd}]\\ &=\sum_{k=1}^{\lfloor m/2\rfloor}(-1)^{(k-1)n}\left(\phi^{(m-2k+1)n}+(-1/\phi)^{(m-2k+1)n}\right)\\ &\hphantom{=}+(-1)^{(m-1)n/2}[m\text{ is odd}]\\ &=\sum_{k=1}^{\lfloor m/2\rfloor}(-1)^{(k-1)n}L_{(m-2k+1)n}\\ &\hphantom{=}+(-1)^{(m-1)n/2}[m\text{ is odd}]\\ \end{align} $$ where $L_n$ are Lucas Numbers and $[\dots]$ are Iverson Brackets.
The important thing is that $\dfrac{F_{mn}}{F_n}\in\mathbb{Z}$.
robjohn
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