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I was playing around with Fibonacci numbers divisible by their indexes (i.e. the $12$th Fibonacci number, $144$, is divisible by $12$) and found that this works when the index is a power of $5$ or $12$. For example:

$k$ $N = 5^k$ $N$th Fibonacci number
$0$ $1$ $1 = 1 * 1$
$1$ $5$ $5 = 5 * 1$
$2$ $25$ $75025 = 25 * 3001$
$3$ $125$ $59425114757512643212875125 = 125 * 475400918060101145703001$

and...

$k$ $N = 12^k$ $N$th Fibonacci number
$0$ $1$ $1 = 1 * 1$
$1$ $12$ $144 = 12 * 12$
$2$ $144$ $555565404224292694404015791808 = 144* 3859669476841564088624933206$

So far, I found that this holds true up to $5^8 = 390625$ and $12^5 = 248832$, the largest powers of $5$ and $12$ less than $1$ million. Is it provable that this will work for all powers of $5$ and/or $12$, or is there a counter-example?

Ami
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  • This makes me think of p-adic numbers, but I could be wrong as all I've seen on the topic is a wikipedia page and a cool youtube vid about them – gist076923 Mar 27 '23 at 19:36
  • @AmiAshman Welcome to Math SE. FYI, using an Approach0 search, there's Is it true that $5^k \mid f(5^k)$?, which shows your first part is true. For the second part, another Approach0 search came up with a property of self-Fibonacci numbers. This has a link to OEIS A023172, i.e., ... – John Omielan Mar 27 '23 at 20:07
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    "Self-Fibonacci numbers: numbers k that divide Fibonacci(k)". In particular, you're asking if all powers of $12$ are self-Fibonacci numbers. However, the Comments section starts with "Sequence contains all powers of 5, infinitely many multiples of 12 and other numbers (including some factors of Fibonacci(5^j), e.g., 75025)". This implies that not all powers of $12$ are included, but I haven't tried to specifically confirm or prove that myself (e.g., finding a counter-example). – John Omielan Mar 27 '23 at 20:09
  • More generally if $m \mid n$ then $F_m \mid F_n$. It's just a lucky coincidence for $5$ and $12$ that $n \mid F_n$. – JBL Mar 27 '23 at 20:48
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    For the powers of $5$ it may be helpful to write the Fibonacci numbers in the form $$F_n=\frac1{2^{n-1}}\sum_{k=0}^\infty\binom n{2k+1}5^k.$$ – bof Mar 27 '23 at 23:22
  • @JohnOmielan Yes, I saw 75025 when I initially coded this. For 12 as well, there are other multiples of 12 that divide Fibonacci(n). In fact, one could broaden the conjecture to claim that the only numbers N that divide Fibonacci(N), besides 1, are divisible by 5 or 12. I've checked up to ~200000, no counter-example. However, since 12 is divisible by 2 and 3, one could hypothesize that the pattern will continue to all prime numbers. Is there an N that divides Fibonacci(N) whose lowest prime factor is 7? – Ami Mar 28 '23 at 13:32
  • @JBL Maybe, but does this coincidence occur for more numbers? If so, they'll have to be greater than 100000 – Ami Mar 28 '23 at 13:33
  • @AmiAshman Thanks for the extra info. Re: other multiples of $12$, the answer indicates this, as well as proving all powers of $12$ are included. As for why the OEIS comment didn't mention this, it's possible they just didn't check, they mistakenly thought they had a counter-example, they just tried keep it brief as powers of $12$ are included in "infinitely many multiples of 12", etc. Regardless, I think it should be changed to avoid others making the same mistake as me. As for a self-Fibonacci # with a factor of $7$, I don't know. You may wish to ask a new question about this. – John Omielan Mar 28 '23 at 16:16

1 Answers1

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Edit: while the old answer has some interest, it’s very laconic, and, although more easy to generalize (it straightforwardly implies the analog of Claims 1 and 2 for any prime $p \neq 5$), needlessly sophisticated for the problem at hand.

Let $a=\frac{1+\sqrt{5}}{2},b=\frac{1-\sqrt{5}}{2}$. It’s well-known that $F_n=\frac{a^n-b^n}{\sqrt{5}}$, that $a+b=1$, $ab=-1$, and that for all $n \geq 1$, $a^n+b^n$ is an integer.

Claim 1: if $3\mid F_n$, then $3F_n\mid F_{3n}$.

Proof: $\frac{F_{3n}}{F_n}=a^{2n}+b^{2n}+(ab)^n=(a^n-b^n)^2+3(ab)^n=5F_n^2+3(-1)^n$.

Claim 2: if $2\mid F_n$, then $2F_n\mid F_{2n}$.

Proof: $\frac{F_{2n}}{F_n}=a^n+b^n$ is an integer. Note that $(a^n+b^n)^2=(a^n-b^n)^2+4(ab)^n=5F_n^2+4(-1)^n$ is even, so $\frac{F_{2n}}{F_n}$ is even.

Claim 3: let $n\mid m$, then $F_n\mid F_m$.

Proof: as said in the comments, this is well-known, but I include a proof for the sake of completeness. Write $m=pn$.

If $p=2l+1$, then $\frac{F_m}{F_n}=(ab)^{nl}+\sum_{k=0}^{l-1}{(ab)^{nk}(a^{n(p-1-2k)}+b^{n(p-1-2k)})}$ is an integer. On the other hand, if $p=2$, then $\frac{F_m}{F_n}=a^n+b^n$ is an integer. Then use induction on the number of prime factors of $p$.

Now, note that $3\mid F_4$. Thus, by Claims 1 and 3, for all $k,r \geq 1$, $3^{r+1}\mid F_{4k\cdot 3^r}$.

Since $2\mid F_3$, by Claims 2 and 3, for all $k,r \geq 1$, $2^{r+1}\mid F_{3k \cdot 2^r}$.

It follows in particular that for any $k,r \geq 1$, $12^{r+1}\mid F_{k \cdot 12^r}$.

Old answer:

Let $p=2$ or $3$, and $O$ the ring of integers of the unique extension of $\mathbb{Z}_p$ (the $p$-adic integers) unramified of degree two.

Note that the identity $F_n=\frac{\phi^n-\overline{\phi}^n}{\sqrt{5}}$ also holds (meaningfully) in $O$, and, for any $k \geq 0$, $p^k$ divides $F_n$ iff $p^k\mid \phi^n-\overline{\phi}^n$ iff $p^k\mid -(\phi\overline{\phi})^n+\phi^{2n}$ iff $p^k\mid\phi^{2n}-(-1)^n$.

Now take $p=3$, $n=4q \cdot 3^r$, and note that $\phi^4=3\phi+2\equiv -1 \pmod{3}$, and $\phi^8=(3\phi+2)^2 \not\equiv 1\pmod{9}$.

By (a suitable extension with an identical proof of) the LTE lemma, we see that $v_3(F_n)=v_3(\phi^{8q\cdot 3^r}-1)=v_3(\phi^8-1)+v_3(3^rq) \geq {r+1}$.

Take $p=2$, $n=3q\cdot 2^r$ (with an odd $q$ and $r\geq 1$), then $\phi^6=(2\phi+1)^2=8\phi+5$, so by “LTE” again (since $4\mid \phi^6-1$), $v_2(F_n)=v_2((\phi^6)^{2^rq}-1)=v_2(\phi^6-1)+r=2+r$.

In particular, one easily sees that $12^{r+1}\mid F_{k \cdot 12^r}$ for all $r,k\geq 1$.

J. W. Tanner
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Aphelli
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