TITLE: An analogue of Hensel's lifting for Fibonacci numbers
QUESTION [30 upvotes]: Let $F_0, F_1, F_2, \ldots$ be the Fibonacci numbers, defined by $F_0 = 0$, $F_1 = 1$ and $F_n = F_{n-1} + F_{n-2}$ for all $n\geq 2$.
In this question Oleg567 conjectured the following interesting fact about Fibonacci numbers:
$$ k\mid F_n\quad\Longrightarrow\quad k^d\mid F_{k^{d-1}n} $$
where $k \in \mathbb{N}$ and $n \in \mathbb{N}$.
This can be regarded as an analogue of Hensel's lifting lemma.
I would be glad to give a proof (following the share your knowledge, Q&A-style), based on a simple combinatorial identity, and see if others come out with more elegant ones.

REPLY [16 votes]: Here is a different proof, using neither the notion of a prime nor that of an
algebraic integer (which Jack used tacitly in Step 1 of his proof, I believe).
Instead it uses the algebra of $2\times2$ matrices (which, arguably, emulates
that of algebraic integers in $\mathbb{Z}\left[  \sqrt{-5}\right]  $).
We first recall how the usual Hensel lifting lemma is proven:

Lemma 1. Let $A$ be a commutative ring with unity. Let $a$ and $b$ be two
  elements of $A$. Let $p$ be a nonnegative integer. Let $k\in\mathbb{N}$ and $\ell
\in\mathbb{N}$ be such that $k>0$. Assume that $a\equiv b\operatorname{mod}
p^{k}A$. Then, $a^{p^{\ell}}\equiv b^{p^{\ell}}\operatorname{mod}p^{k+\ell}A$.

(Here and in the following, $\mathbb{N}$ means the set $\left\{
0,1,2,\ldots\right\}  $.)
Notice that we do not require $p$ to be a prime; we merely call it $p$ because
it fills the role that is usually reserved for primes in such results.
Proof of Lemma 1. The following proof is mostly copypasted from my witt#3
note (where Lemma 1 appears as Lemma 3):
We will show Lemma 1 by induction over $\ell$. For $\ell=0$, the assertion of
Lemma 1 is trivial. Now, for the induction step, we assume that $a^{p^{\ell}
}\equiv b^{p^{\ell}}\operatorname{mod}p^{k+\ell}A$ for some $\ell\in
\mathbb{N}$, and we want to show that $a^{p^{\ell+1}}\equiv b^{p^{\ell+1}
}\operatorname{mod}p^{k+\ell+1}A$.
We have $a^{p^{\ell}}\equiv b^{p^{\ell}}\operatorname{mod}p^{k+\ell}A$, thus
$a^{p^{\ell}}-b^{p^{\ell}}\in p^{k+\ell}A$.
On the other hand, $a\equiv b\operatorname{mod}p^{k}A$, thus $a-b\in
p^{k}A\subseteq pA$ (since $k>0$) and therefore $a\equiv b\operatorname{mod}
pA$. Hence,
$\sum\limits_{i=0}^{p-1}\left(  \underbrace{a^{p^{\ell}}}_{\substack{\equiv
b^{p^{\ell}}\\\text{(since }a\equiv b\operatorname{mod}pA\text{)}}}\right)
^{i}\left(  b^{p^{\ell}}\right)  ^{p-1-i}\equiv\sum\limits_{i=0}
^{p-1}\underbrace{\left(  b^{p^{\ell}}\right)  ^{i}\left(  b^{p^{\ell}
}\right)  ^{p-1-i}}_{=\left(  b^{p^{\ell}}\right)  ^{p-1}}$
$=\sum\limits_{i=0}^{p-1}\left(  b^{p^{\ell}}\right)  ^{p-1}=p\left(
b^{p^{\ell}}\right)  ^{p-1}\equiv0\operatorname{mod}pA$,
so that $\sum\limits_{i=0}^{p-1}\left(  a^{p^{\ell}}\right)  ^{i}\left(
b^{p^{\ell}}\right)  ^{p-1-i}\in pA$. Hence,
$a^{p^{\ell+1}}-b^{p^{\ell+1}}=\left(  a^{p^{\ell}}\right)  ^{p}-\left(
b^{p^{\ell}}\right)  ^{p}$
$=\underbrace{\left(  a^{p^{\ell}}-b^{p^{\ell}}\right)  }_{\in p^{k+\ell}
A}\cdot\underbrace{\sum\limits_{i=0}^{p-1}\left(  a^{p^{\ell}}\right)
^{i}\left(  b^{p^{\ell}}\right)  ^{p-1-i}}_{\in pA}\in\left(  p^{k+\ell
}A\right)  \cdot\left(  pA\right)  =p^{k+\ell+1}A$,
so that $a^{p^{\ell+1}}\equiv b^{p^{\ell+1}}\operatorname{mod}p^{k+\ell+1}A$,
and the induction step is complete. Thus, Lemma 1 is proven.
(If you are wondering why your favorite version of Lemma 1 is only the $k=1$
case of my Lemma 1, don't worry -- you are not missing much. The general case
follows easily from the $k=1$ case.)
Now, let $\left(  F_{0},F_{1},F_{2},\ldots\right)  $ be the Fibonacci
sequence, defined by $F_{0}=0$, $F_{1}=1$ and $F_{n}=F_{n-1}+F_{n-2}$ for
every $n\geq2$. Let $X$ be the $2\times2$-matrix $\left(
\begin{array}
[c]{cc}
0 & 1\\
1 & 1
\end{array}
\right)  \in\mathbb{Z}^{2\times2}$. A fundamental property of the Fibonacci
numbers is the following fact:

Lemma 2. We have $X^{m}=F_{m-1}I_{2}+F_{m}X$ for every positive integer
  $m$.

(Here, $I_{2}$ denotes the $2\times2$ identity matrix.)
Proof of Lemma 2. We will prove Lemma 2 by induction over $m$.
Induction base: We have $\underbrace{F_{1-1}}_{=F_{0}=0}I_{2}
+\underbrace{F_{1}}_{=1}X=0I_{2}+1X=X=X^{1}$, so that $X^{1}=F_{1-1}
I_{2}+F_{1}X$. In other words, Lemma 2 holds for $m=1$. This completes the
induction base.
Induction step: Let $M$ be a positive integer. Assume that Lemma 2 holds for
$m=M$. We need to prove that Lemma 2 holds for $m=M+1$.
We have $X^{M}=F_{M-1}I_{2}+F_{M}X$ (since Lemma 2 holds for $X=M$). It is
straightforward to check that $X^{2}=X+I_{2}$. The recursive definition of the
Fibonacci number $F_{M+1}$ shows that $F_{M+1}=F_{M-1}+F_{M}$. Now,
$X^{M+1}=X\underbrace{X^{M}}_{=F_{M-1}I_{2}+F_{M}X}=X\left(  F_{M-1}
I_{2}+F_{M}X\right)  =F_{M-1}X+F_{M}\underbrace{X^{2}}_{=X+I_{2}}$
$=F_{M-1}X+F_{M}\left(  X+I_{2}\right)  =F_{M-1}X+F_{M}X+F_{M}I_{2}$
$=\underbrace{\left(  F_{M-1}+F_{M}\right)  }_{=F_{M+1}}X+\underbrace{F_{M}
}_{=F_{M+1-1}}I_{2}=F_{M+1}X+F_{M+1-1}I_{2}$
$=F_{M+1-1}I_{2}+F_{M+1}X$.
In other words, Lemma 2 holds for $m=M+1$. This completes the induction step,
and thus Lemma 2 is proven by induction.
As a simple application of Lemma 2, let us prove Jack's Claim 1 and a bit more:

Proposition 3. Let $m\in\mathbb{N}$ and $n\in\mathbb{N}$. Then, $F_{m}\mid
F_{mn}$. Also, if $m$ and $n$ are positive, we have $F_{mn-1}\equiv\left(  F_{m-1}\right)  ^{n}\operatorname{mod} F_{m}$.

While Proposition 3 will not be used in the following, I believe it is a
worthwhile demonstration of how useful Lemma 2 can be even without Lemma 1.
Proof of Proposition 3. If $m=0$, then $F_{mn}=F_{0n}=F_{0}=0$, whence
$F_{m}\mid F_{mn}$ in this case. Hence, we WLOG assume that we don't have
$m=0$. Thus, $m$ is a positive integer.
If $n=0$, then $F_{mn}=F_{m0}=F_{0}=0$, whence $F_{m}\mid F_{mn}$ in this
case. Hence, we WLOG assume that we don't have $n=0$. Thus, $n$ is a positive integer.
The matrix ring $\mathbb{Z}^{2\times2}$ is not commutative. However, its
subring $\mathbb{Z}\left[  X\right]  $ generated by the matrix $X$ is
commutative (because its elements are polynomials in $X$, and clearly any two
such polynomials commute). Let $A$ be this commutative subring $\mathbb{Z}
\left[  X\right]  $. Thus, clearly, $X\in A$ and $I_{2}\in A$, and any
polynomial in $X$ (with integer coefficients) lies in $A$.
Lemma 2 shows that $X^{m}=F_{m-1}I_{2}+F_{m}X$, so that $X^{m}-F_{m-1}
I_{2}=F_{m}X\in F_{m}A$ (since $X\in A$). In other words, $X^{m}\equiv
F_{m-1}I_{2}\operatorname{mod}F_{m}A$. Taking this congruence to the $n$-th
power, we obtain $\left(  X^{m}\right)  ^{n}\equiv\left(  F_{m-1}I_{2}\right)
^{n}\operatorname{mod}F_{m}A$. Thus,
$X^{mn}=\left(  X^{m}\right)  ^{n}\equiv\left(  F_{m-1}I_{2}\right)
^{n}=\left(  F_{m-1}\right)  ^{n}I_{2}\operatorname{mod}F_{m}A$.
But applying Lemma 2 to $mn$ instead of $m$, we find that $X^{mn}
=F_{mn-1}I_{2}+F_{mn}X$. Thus,
$F_{mn-1}I_{2}+F_{mn}X=X^{mn}\equiv\left(  F_{m-1}\right)  ^{n}I_{2}
\operatorname{mod}F_{m}A$,
so that $\left(  F_{mn-1}I_{2}+F_{mn}X\right)  -\left(  F_{m-1}\right)
^{n}I_{2}\in F_{m}\underbrace{A}_{\subseteq\mathbb{Z}^{2\times2}}\subseteq
F_{m}\mathbb{Z}^{2\times2}$. Since
$\left(  F_{mn-1}I_{2}+F_{mn}X\right)  -\left(  F_{m-1}\right)  ^{n}
I_{2}=F_{mn-1}I_{2}-\left(  F_{m-1}\right)  ^{n}I_{2}+F_{mn}X$
$=\left(  F_{mn-1}-\left(  F_{m-1}\right)  ^{n}\right)  \underbrace{I_{2}
}_{=\left(
\begin{array}
[c]{cc}
1 & 0\\
0 & 1
\end{array}
\right)  }+F_{mn}\underbrace{X}_{=\left(
\begin{array}
[c]{cc}
0 & 1\\
1 & 1
\end{array}
\right)  }$
$=\left(  F_{mn-1}-\left(  F_{m-1}\right)  ^{n}\right)  \left(
\begin{array}
[c]{cc}
1 & 0\\
0 & 1
\end{array}
\right)  +F_{mn}\left(
\begin{array}
[c]{cc}
0 & 1\\
1 & 1
\end{array}
\right)  $
$=\left(
\begin{array}
[c]{cc}
F_{mn-1}-\left(  F_{m-1}\right)  ^{n} & F_{mn}\\
F_{mn} & F_{mn-1}-\left(  F_{m-1}\right)  ^{n}+F_{mn}
\end{array}
\right)  $,
this rewrites as $\left(
\begin{array}
[c]{cc}
F_{mn-1}-\left(  F_{m-1}\right)  ^{n} & F_{mn}\\
F_{mn} & F_{mn-1}-\left(  F_{m-1}\right)  ^{n}+F_{mn}
\end{array}
\right)  \in F_{m}\mathbb{Z}^{2\times2}$. In other words, all entries of the
matrix $\left(
\begin{array}
[c]{cc}
F_{mn-1}-\left(  F_{m-1}\right)  ^{n} & F_{mn}\\
F_{mn} & F_{mn-1}-\left(  F_{m-1}\right)  ^{n}+F_{mn}
\end{array}
\right)  $ are divisible by $F_{m}$. In other words, the integers
$F_{mn-1}-\left(  F_{m-1}\right)  ^{n}$, $F_{mn}$, $F_{mn}$ and $F_{mn-1}
-\left(  F_{m-1}\right)  ^{n}+F_{mn}$ are divisible by $F_{m}$.
Now, $F_{m}\mid F_{mn}$ (since $F_{mn}$ is divisible by $F_{m}$) and
$F_{mn-1}\equiv\left(  F_{m-1}\right)  ^{n}\operatorname{mod}F_{m}$ (since
$F_{mn-1}-\left(  F_{m-1}\right)  ^{n}$ is divisible by $F_{m}$). Proposition
3 is proven.
Now, here is Jack's main result:

Theorem 4. Let $k\in\mathbb{N}$ and $n\in\mathbb{N}$ be such that $k\mid
F_{n}$. Let $d$ be a positive integer. Then, $k^{d}\mid F_{k^{d-1}n}$.

Proof of Theorem 4. First, let us assume that $d=1$. Thus, $k^{d}
=k^{1}=k\mid F_{n}$. Also, $k^{d-1}n=\underbrace{k^{1-1}}_{=k^{0}=1}n=n$.
Hence, $k^{d}\mid F_{n}$ rewrites as $k^{d}\mid F_{k^{d-1}n}$. Thus, Theorem 4
holds under the assumption that $d=1$.
Now, let us forget that we assumed that $d=1$. Having proven Theorem 4 in the case when $d = 1$, we can now WLOG assume that
$d\neq1$. Assume this. Hence, $d\geq2$, and thus $0^{d-1}=0$. Consequently, $F_{0^{d-1}
n}=F_{0n}=F_{0}=0$ is divisible by every integer. In particular, $0^{d}\mid
F_{0^{d-1}n}$. Hence, Theorem 4 holds if $k=0$. We thus WLOG assume that
$k\neq0$. Thus, $k$ is a positive integer.
Also, $k^{d}\mid F_{k^{d-1}0}$ (since $F_{k^{d-1}0}=F_{0}=0$). Hence, Theorem
4 holds if $n=0$. We thus WLOG assume that $n\neq0$. Thus, $n$ is a positive integer.
The matrix ring $\mathbb{Z}^{2\times2}$ is not commutative. However, its
subring $\mathbb{Z}\left[  X\right]  $ generated by the matrix $X$ is
commutative (because its elements are polynomials in $X$, and clearly any two
such polynomials commute). Let $A$ be this commutative subring $\mathbb{Z}
\left[  X\right]  $. Thus, clearly, $X\in A$ and $I_{2}\in A$, and any
polynomial in $X$ (with integer coefficients) lies in $A$. In particular, $\mathbb{Z} X \subseteq A$.
We have $F_{n}\in k\mathbb{Z}$ (since $k\mid F_{n}$). Lemma 2 (applied to
$m=n$) shows that $X^{n}=F_{n-1}I_{2}+F_{n}X$, so that $X^{n}-F_{n-1}
I_{2}=\underbrace{F_{n}}_{\in k\mathbb{Z}}X\in k\underbrace{\mathbb{Z}
X}_{\subseteq A}\subseteq kA=k^{1}A$. In other words, $X^{n}\equiv
F_{n-1}I_{2}\operatorname{mod}k^{1}A$. Thus, Lemma 1 (applied to $X^{n}$,
$F_{n-1}I_{2}$, $k$, $1$ and $d-1$ instead of $a$, $b$, $p$, $k$ and $\ell$)
shows that $\left(  X^{n}\right)  ^{k^{d-1}}\equiv\left(  F_{n-1}I_{2}\right)
^{k^{d-1}}\operatorname{mod}k^{1+\left(  d-1\right)  }A$. Since
$\left(  X^{n}\right)  ^{k^{d-1}}=X^{nk^{d-1}}=X^{k^{d-1}n}=F_{k^{d-1}
n-1}I_{2}+F_{k^{d-1}n}X$ (by Lemma 2, applied to $m=k^{d-1}n$)
and $1+\left(  d-1\right)  =d$, this rewrites as follows:
$F_{k^{d-1}n-1}I_{2}+F_{k^{d-1}n}X\equiv\left(  F_{n-1}I_{2}\right)
^{k^{d-1}}\operatorname{mod}k^{d}A$.
Thus, $F_{k^{d-1}n-1}I_{2}+F_{k^{d-1}n}X-\left(  F_{n-1}I_{2}\right)
^{k^{d-1}}\in k^{d}\underbrace{A}_{\subseteq\mathbb{Z}^{2\times2}}\subseteq
k^{d}\mathbb{Z}^{2\times2}$.
Since
$F_{k^{d-1}n-1}I_{2}+F_{k^{d-1}n}X-\underbrace{\left(  F_{n-1}I_{2}\right)
^{k^{d-1}}}_{=\left(  F_{n-1}\right)  ^{k^{d-1}}I_{2}}$
$=F_{k^{d-1}n-1}I_{2}+F_{k^{d-1}n}X-\left(  F_{n-1}\right)  ^{k^{d-1}}I_{2}$
$=F_{k^{d-1}n-1}I_{2}-\left(  F_{n-1}\right)  ^{k^{d-1}}I_{2}+F_{k^{d-1}n}X$
$=\left(  F_{k^{d-1}n-1}-\left(  F_{n-1}\right)  ^{k^{d-1}}\right)
\underbrace{I_{2}}_{=\left(
\begin{array}
[c]{cc}
1 & 0\\
0 & 1
\end{array}
\right)  }+F_{k^{d-1}n}\underbrace{X}_{=\left(
\begin{array}
[c]{cc}
0 & 1\\
1 & 1
\end{array}
\right)  }$
$=\left(  F_{k^{d-1}n-1}-\left(  F_{n-1}\right)  ^{k^{d-1}}\right)  \left(
\begin{array}
[c]{cc}
1 & 0\\
0 & 1
\end{array}
\right)  +F_{k^{d-1}n}\left(
\begin{array}
[c]{cc}
0 & 1\\
1 & 1
\end{array}
\right)  $
$=\left(
\begin{array}
[c]{cc}
F_{k^{d-1}n-1}-\left(  F_{n-1}\right)  ^{k^{d-1}} & F_{k^{d-1}n}\\
F_{k^{d-1}n} & F_{k^{d-1}n-1}-\left(  F_{n-1}\right)  ^{k^{d-1}}+F_{k^{d-1}n}
\end{array}
\right)  $,
this rewrites as
$\left(
\begin{array}
[c]{cc}
F_{k^{d-1}n-1}-\left(  F_{n-1}\right)  ^{k^{d-1}} & F_{k^{d-1}n}\\
F_{k^{d-1}n} & F_{k^{d-1}n-1}-\left(  F_{n-1}\right)  ^{k^{d-1}}+F_{k^{d-1}n}
\end{array}
\right)  \in k^{d}\mathbb{Z}^{2\times2}$.
Thus, the integers $F_{k^{d-1}n-1}-\left(  F_{n-1}\right)  ^{k^{d-1}}$,
$F_{k^{d-1}n}$, $F_{k^{d-1}n}$ and $F_{k^{d-1}n-1}-\left(  F_{n-1}\right)
^{k^{d-1}}+F_{k^{d-1}n}$ are all divisible by $k^{d}$. In particular,
$F_{k^{d-1}n}$ is divisible by $k^{d}$. This proves Theorem 4.