Subfield: Number Theory


Question: For a prime $p$ and a positive integer $n$, denote by $\nu_{p}(n)$ the exponent of $p$ in the prime factorization of $n$ !. Given a positive integer $d$ and a finite set $\left\{p_{1}, \ldots, p_{k}\right\}$ of primes. Show that there are infinitely many positive integers $n$ such that $d \mid \nu_{p_{i}}(n)$ for all $1 \leq i \leq k$.

Solution:
For arbitrary prime $p$ and positive integer $n$, denote by $\operatorname{ord}_{p}(n)$ the exponent of $p$ in $n$. Thus,
$$
\nu_{p}(n)=\operatorname{ord}_{p}(n !)=\sum_{i=1}^{n} \operatorname{ord}_{p}(i)
$$
Lemma. Let $p$ be a prime number, $q$ be a positive integer, $k$ and $r$ be positive integers such that $p^{k}>r$. Then $\nu_{p}\left(q p^{k}+r\right)=\nu_{p}\left(q p^{k}\right)+\nu_{p}(r)$.

Proof. We claim that $\operatorname{ord}_{p}\left(q p^{k}+i\right)=\operatorname{ord}_{p}(i)$ for all $0<i<p^{k}$. Actually, if $d=\operatorname{ord}_{p}(i)$ then $d<k$, so $q p^{k}+i$ is divisible by $p^{d}$, but only the first term is divisible by $p^{d+1}$; hence the sum is not.

Using this claim, we obtain
$$
\nu_{p}\left(q p^{k}+r\right)=\sum_{i=1}^{q p^{k}} \operatorname{ord}_{p}(i)+\sum_{i=q p^{k}+1}^{q p^{k}+r} \operatorname{ord}_{p}(i)=\sum_{i=1}^{q p^{k}} \operatorname{ord}_{p}(i)+\sum_{i=1}^{r} \operatorname{ord}_{p}(i)=\nu_{p}\left(q p^{k}\right)+\nu_{p}(r)
$$
For any integer $a$, denote by $\bar{a}$ its residue modulo $d$. The addition of residues will also be performed modulo $d$, i. e. $\bar{a}+\bar{b}=\overline{a+b}$. For any positive integer $n$, let $f(n)=\left(f_{1}(n), \ldots, f_{k}(n)\right)$, where $f_{i}(n)=\overline{\nu_{p_{i}}(n)}$.

Define the sequence $n_{1}=1, n_{\ell+1}=\left(p_{1} p_{2} \ldots p_{k}\right)^{n_{\ell}}$. We claim that
$$
f\left(n_{\ell_{1}}+n_{\ell_{2}}+\ldots+n_{\ell_{m}}\right)=f\left(n_{\ell_{1}}\right)+f\left(n_{\ell_{2}}\right)+\ldots+f\left(n_{\ell_{m}}\right)
$$
for any $\ell_{1}<\ell_{2}<\ldots<\ell_{m}$. (The addition of $k$-tuples is componentwise.) The base case $m=1$ is trivial.

Suppose that $m>1$. By the construction of the sequence, $p_{i}^{n_{\ell_{1}}}$ divides $n_{\ell_{2}}+\ldots+n_{\ell_{m}}$; clearly, $p_{i}^{n_{\ell_{1}}}>n_{\ell_{1}}$ for all $1 \leq i \leq k$. Therefore the Lemma can be applied for $p=p_{i}, k=r=n_{\ell_{1}}$ and $q p^{k}=n_{\ell_{2}}+\ldots+n_{\ell_{m}}$ to obtain
$$
f_{i}\left(n_{\ell_{1}}+n_{\ell_{2}}+\ldots+n_{\ell_{m}}\right)=f_{i}\left(n_{\ell_{1}}\right)+f_{i}\left(n_{\ell_{2}}+\ldots+n_{\ell_{m}}\right) \quad \text { for all } 1 \leq i \leq k
$$
and hence
$$
f\left(n_{\ell_{1}}+n_{\ell_{2}}+\ldots+n_{\ell_{m}}\right)=f\left(n_{\ell_{1}}\right)+f\left(n_{\ell_{2}}+\ldots+n_{\ell_{m}}\right)=f\left(n_{\ell_{1}}\right)+f\left(n_{\ell_{2}}\right)+\ldots+f\left(n_{\ell_{m}}\right)
$$
by the induction hypothesis.

Now consider the values $f\left(n_{1}\right), f\left(n_{2}\right), \ldots$ There exist finitely many possible values of $f$. Hence, there exists an infinite sequence of indices $\ell_{1}<\ell_{2}<\ldots$ such that $f\left(n_{\ell_{1}}\right)=f\left(n_{\ell_{2}}\right)=\ldots$ and thus
$$
f\left(n_{\ell_{m+1}}+n_{\ell_{m+2}}+\ldots+n_{\ell_{m+d}}\right)=f\left(n_{\ell_{m+1}}\right)+\ldots+f\left(n_{\ell_{m+d}}\right)=d \cdot f\left(n_{\ell_{1}}\right)=(\overline{0}, \ldots, \overline{0})
$$
for all $m$. We have found infinitely many suitable numbers.
We use the same Lemma and definition of the function $f$.

Let $S=\{f(n): n \in \mathbb{N}\}$. Obviously, set $S$ is finite. For every $s \in S$ choose the minimal $n_{s}$ such that $f\left(n_{s}\right)=s$. Denote $N=\max _{s \in S} n_{s}$. Moreover, let $g$ be an integer such that $p_{i}^{g}>N$ for each $i=1,2, \ldots, k$. Let $P=\left(p_{1} p_{2} \ldots p_{k}\right)^{g}$.

We claim that
$$
\{f(n) \mid n \in[m P, m P+N]\}=S \tag{1}
$$
for every positive integer $m$. In particular, since $(\overline{0}, \ldots, \overline{0})=f(1) \in S$, it follows that for an arbitrary $m$ there exists $n \in[m P, m P+N]$ such that $f(n)=(\overline{0}, \ldots, \overline{0})$. So there are infinitely many suitable numbers.

To prove (1), let $a_{i}=f_{i}(m P)$. Consider all numbers of the form $n_{m, s}=m P+n_{s}$ with $s=\left(s_{1}, \ldots, s_{k}\right) \in S$ (clearly, all $n_{m, s}$ belong to $[m P, m P+N]$ ). Since $n_{s} \leq N<p_{i}^{g}$ and $p_{i}^{g} \mid m P$, we can apply the Lemma for the values $p=p_{i}, r=n_{s}, k=g, q p^{k}=m P$ to obtain
$$
f_{i}\left(n_{m, s}\right)=f_{i}(m P)+f_{i}\left(n_{s}\right)=a_{i}+s_{i}
$$

hence for distinct $s, t \in S$ we have $f\left(n_{m, s}\right) \neq f\left(n_{m, t}\right)$.

Thus, the function $f$ attains at least $|S|$ distinct values in $[m P, m P+N]$. Since all these values belong to $S, f$ should attain all possible values in $[m P, m P+N]$.