Subfield: Number Theory


Question: Let $a>1$ be a positive integer, and let $d>1$ be a positive integer coprime to $a$. Let $x_{1}=1$ and, for $k \geqslant 1$, define

$$
x_{k+1}= \begin{cases}x_{k}+d & \text { if } a \text { doesn't divide } x_{k} \\ x_{k} / a & \text { if } a \text { divides } x_{k}\end{cases}
$$

Find the greatest positive integer $n$ for which there exists an index $k$ such that $x_{k}$ is divisible by $a^{n}$.

Solution:
By trivial induction, $x_{k}$ is coprime to $d$.

By induction and the fact that there can be at most $a-1$ consecutive increasing terms in the sequence, it also holds that $x_{k}<d a$ if $x_{k}=x_{k-1}+d$ and that $x_{k}<d$ if $x_{k}=\frac{x_{k-1}}{a}$ or $k=1$. This gives the upper bound on the exponent.

This implies that the sequence is (eventually) periodic, and that both increasing and decreasing steps happen infinitely many times. Let $a^{-k}$ be the multiplicative inverse of $a^{k}$ modulo $d$. The sequence contains elements congruent to $1, a^{-1}, a^{-2}, \ldots$ modulo $d$.

Let $x_{k_{0}}$ the first element such that $x_{k_{0}} \equiv a^{-n}(\bmod d)$. We have either $k_{0}=1$ or $x_{k_{0}}=$ $x_{k_{0}-1} / a$; in both cases $x_{k_{0}}<d<a^{n}<d a$ and therefore

$$
x_{k_{0}} \in\left\{a^{n}-d, a^{n}-2 d, \ldots, a^{n}-(a-1) d\right\}
$$

In this set no element is divisible by $a$, so therefore the sequence will visit the value $a^{n}$ in the next $a-1$ steps.
By trivial induction, $x_{k}$ is coprime to $d$.

By induction and the fact that there can be at most $a-1$ consecutive increasing terms in the sequence, it also holds that $x_{k}<d a$ if $x_{k}=x_{k-1}+d$ and that $x_{k}<d$ if $x_{k}=\frac{x_{k-1}}{a}$ or $k=1$. This gives the upper bound on the exponent.

Like above, $x_{k}$ is relatively prime to $d$ and $x_{k}<a d$ for all $k$.

Let

$$
S=\left\{x \in \mathbb{Z}_{>0}: 0<x<a d, \operatorname{gcd}(x, d)=1\right\} \quad \text { and } \quad f: S \rightarrow S, f(x)= \begin{cases}x+d & \text { if } a \nmid x \\ x / a & \text { if } a \mid x\end{cases}
$$

So, we have $x_{1}=1$ and $x_{k+1}=f\left(x_{k}\right)$.

Notice that the recurrence formula is invertible. Suppose that $f(x)=y$ for some $x, y \in S$. If $y>d$ then $y-d \in S$ but $a \cdot y \notin S$; otherwise, if $y<d$ then $a \cdot y \in S$ but $y-d \notin S$. So, the function $f$ is a permutation on $S$, and its inverse is

$$
f^{-1}(y)=\left\{\begin{array}{ll}
y-d & \text { if } y>d, \\
a \cdot y & \text { if } y<d
\end{array} \quad \text { for all } y \in S\right.
$$

It follows that the sequence is periodic, and therefore attains the initial value 1 infinitely many times. Let $k_{1}>1$ be an index such that $x_{k_{1}}=1$. Then

$$
x_{k_{1}}=1, \quad x_{k_{1}-1}=f^{-1}(1)=a, x_{k_{1}-2}=f^{-1}(a)=a^{2}, \ldots, \quad x_{k_{1}-n}=a^{n}
$$
By trivial induction, $x_{k}$ is coprime to $d$.

By induction and the fact that there can be at most $a-1$ consecutive increasing terms in the sequence, it also holds that $x_{k}<d a$ if $x_{k}=x_{k-1}+d$ and that $x_{k}<d$ if $x_{k}=\frac{x_{k-1}}{a}$ or $k=1$. This gives the upper bound on the exponent.

Like above, $x_{k}$ is relatively prime to $d$ and $x_{k}<a d$ for all $k$.

We wish to prove that there are $n$ consecutive decreasing indices. Let $m_{0}=0$ and $m_{k}$ be the $k$-th smallest decreasing index (an index $k$ is decreasing if $x_{k}=x_{k-1} / a$ ) and define a sequence $\left(y_{k}\right)_{k}=\left(x_{m_{k}}\right)_{k}$, i.e. the subsequence consisting only of elements with decreasing indices. For each $k \geqslant 1$ we can write

$$
y_{k+1}=\frac{y_{k}+z_{k} d}{a}
$$

for some $z_{k} \in\{0,1, \ldots, a-1\}$, where $z_{k}=0$ if and only if $y_{k}$ and $y_{k+1}$ are consecutive elements in the original sequence. Then

$$
y_{k}=\frac{1+\left(z_{0}+z_{1} a+\cdots+z_{k-1} a^{k-1}\right) d}{a^{k}}
$$

because $y_{k}$ is bounded the sequence is (eventually) periodic. Consider some $y_{u}=y_{u+v}$ where $v \geqslant 1$. We can write

$$
y_{u}=\frac{1+\left(z_{0}+z_{1} a+\cdots+z_{u-1} a^{u-1}\right) d}{a^{u}}
$$

and

$$
y_{u+v}=\frac{1+\left(z_{0}+z_{1} a+\cdots+z_{u+v-1} a^{u+v-1}\right) d}{a^{u+v}}
$$

and so

$$
\frac{1+\left(z_{0}+z_{1} a+\cdots+z_{u+v-1} a^{u+v-1}\right) d}{a^{u+v}}=\frac{1+\left(z_{0}+z_{1} a+\cdots+z_{u-1} a^{u-1}\right) d}{a^{u}}
$$

Rearranging gives

$$
\frac{a^{v}-1}{d}=z_{0}+z_{1} a+\cdots+z_{v-1} a^{v-1}+\left(z_{v}-z_{0}\right) a^{v}+\cdots+\left(z_{u+v-1}-z_{u-1}\right) a^{u+v-1}
$$

Since $d>a^{n} / a=a^{n-1}$ the LHS is strictly less than $a^{v-n}$. This implies that on the RHS, the coefficients of $a^{v-n}, a^{v-n+1}, \ldots$ must all be zero, i.e. $z_{v-n}=z_{v-n+1}=\cdots=z_{v-1}=0$. This implies that there are $n$ consecutive decreasing indices in the original sequence.

Final answer: $n$ is the exponent with $d<a^{n}<a d$