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\title[Distribution of factorial units]{Joint distribution of factorial units and valuations in translated intervals}
\author{Will Howe}
\email{howew@purdue.edu}
\author{Grant Huffman}
\email{huffma97@purdue.edu}
\author{Jensen Kohlmeyer}
\email{jenwin@purdue.edu}
\author{Liam Kruer}
\email{kruerl@purdue.edu}
\thanks{Prepared with AI assistance from OpenAI Codex in mathematical exploration,
checking, and manuscript preparation.}
\date{November 7, 2026}
\makeatletter
\def\@setaddresses{\par\nobreak\bigskip
  {\footnotesize\noindent\textit{E-mail addresses:}
  \href{mailto:howew@purdue.edu}{\texttt{howew@purdue.edu}},
  \href{mailto:huffma97@purdue.edu}{\texttt{huffma97@purdue.edu}},\par
  \noindent\href{mailto:jenwin@purdue.edu}{\texttt{jenwin@purdue.edu}},
  \href{mailto:kruerl@purdue.edu}{\texttt{kruerl@purdue.edu}}\par}}
\makeatother

\begin{document}
\begin{abstract}
For every fixed finite set of primes, the corresponding prime valuations of
factorials, reduced modulo arbitrary fixed integers, and the prime-free parts,
reduced modulo arbitrary fixed prime powers, have a joint uniform distribution.
We give a proof with a power-saving error on every interval of consecutive
nonnegative integers, uniform in the interval's starting point. A fixed
congruence on the index may be imposed independently. The resulting statement
for the part of a factorial coprime to an arbitrary fixed modulus gives the
densities and relative density asked for in Problems 2 and 3 of Hajdu and Papp.
The proof uses a finite digit-window identity, a strict local character
contraction, and a carry estimate followed by finite Fourier analysis.
\end{abstract}
\maketitle

\section{Statement and scope}

Write $\N_0=\{0,1,2,\ldots\}$ and $\e(x)=\exp(2\pi i x)$.
For a prime $p$ and $n\in\N_0$, put
\[
 E_p(n)=v_p(n!),\qquad U_p(n)=\frac{n!}{p^{E_p(n)}}.
\]
Thus $E_p(0)=0$, $U_p(0)=1$, and $U_p(n)$ is a unit modulo every
power of $p$. All parameters in the following theorems are fixed; only the
length and starting point of the interval vary.

\begin{theorem}\label{thm:local}
Let $\mathcal P$ be a nonempty finite set of distinct primes. For each
$p\in\mathcal P$, fix integers $r_p,b_p\geq1$. Fix also an integer $Q\geq1$.
There are constants $C>0$ and $\delta>0$, depending only on these parameters,
such that the following holds. For all $A\in\N_0$, $N\geq1$, all residues
$a_p\pmod{r_p}$ and $c\pmod Q$, and all $z_p\in\units{p^{b_p}}$,
\begin{align}
 &\#\{A\leq n<A+N:
     E_p(n)\equiv a_p\pmod{r_p},\;
     U_p(n)\equiv z_p\pmod{p^{b_p}}\ (p\in\mathcal P),\;
     n\equiv c\pmod Q\}\notag\\
 &\hspace{22mm}=
 \frac{N}{Q\prod_{p\in\mathcal P}r_p\varphi(p^{b_p})}
   +O(CN^{1-\delta}).\label{eq:localcount}
\end{align}
The implied absolute constant may be taken to be one by enlarging $C$.
In particular, no coprimality assumption on $Q$ is required.
\end{theorem}

For an integer $m>1$, let
\[
 W_m(n)=\frac{n!}{\prod_{p\mid m}p^{E_p(n)}}.
\]
This integer is coprime to $m$.

\begin{theorem}\label{thm:composite}
Fix $m>1$, distinct primes $q_1,\ldots,q_t$, positive integers
$r_1,\ldots,r_t$, and $Q\geq1$, allowing $t=0$.
For every unit $z\pmod m$, residues $a_i\pmod{r_i}$, and $c\pmod Q$,
\begin{align}
 &\#\{A\leq n<A+N:W_m(n)\equiv z\pmod m,
       \ E_{q_i}(n)\equiv a_i\pmod{r_i}\ (1\leq i\leq t),
       \ n\equiv c\pmod Q\}\notag\\
 &\hspace{25mm}=
 \frac{N}{Q\varphi(m)\prod_{i=1}^t r_i}+O(N^{1-\delta}),\label{eq:compositecount}
\end{align}
where $\delta>0$ and the implied constant depend only on the fixed parameters.
The estimate is uniform for $A\in\N_0$, $N\geq1$, and the displayed targets.
\end{theorem}

Taking $Q=1$ and, respectively, $t=0$ or arbitrary $t$ gives the densities
asked for in Problems 2 and 3 of Hajdu and Papp \cite{HP}.
Their Theorems 2.1 and 2.2 establish the single-prime case with valuation
modulus a power of that prime, including an explicit gap bound; their
Theorems 2.3 and 2.4 give finer results in a particular case modulo $8$.
The distribution of the valuation coordinates alone was established by
Berend and Kolesnik \cite{BK}. We include all ingredients needed for the
joint unit and valuation statement here, without asserting priority for
its special cases.

\begin{corollary}\label{cor:gaps}
Every set counted in Theorem~\ref{thm:composite} is infinite and relatively
dense in $\N_0$: some integer $L\geq1$, depending only on the fixed parameters,
has the property that every interval $[A,A+L)$, $A\in\N_0$, contains a
member of the set. Its natural density is
$(Q\varphi(m)\prod_i r_i)^{-1}$.
\end{corollary}
\begin{proof}
Set $d=(Q\varphi(m)\prod_i r_i)^{-1}$. Choose $L$ so large that
$CL^{1-\delta}<dL$. Then the count in every interval of length $L$ is
positive. Dividing \eqref{eq:compositecount} with $A=0$ by $N$ and letting
$N$ tend to infinity gives the density.
\end{proof}

\begin{remark}
The congruence $n\equiv c\pmod Q$ permits any fixed restriction on finitely
many terminal digits, or any other fixed periodic restriction on the index,
by taking a union of residue classes. No independence from unrestricted
digit-sum conditions is asserted. The parameters do not grow with $N$.
The results concern fixed local data and give no conclusion about isolated
solutions of $n!+1=x^2$.
\end{remark}

\section{A digit-window formula}

Fix a prime $p$ and integers $r,b\geq1$ throughout this section. All unit
identities are in $\units{p^b}$. Define
\[
 F(t)=\prod_{\substack{1\leq j\leq t\\p\nmid j}}j,
 \qquad G=F(p^b),\qquad
 w_i(n)=\left\lfloor\frac n{p^i}\right\rfloor\bmod p^b,
\]
where $0\leq w_i(n)<p^b$. Let $s_p(n)$ denote the sum of the base-$p$
digits of $n$.

\begin{lemma}\label{lem:windows}
For $n\in\N_0$,
\begin{equation}\label{eq:windows}
 U_p(n)=G^{E_p(n)-\sum_{j=1}^{b-1}\lfloor n/p^j\rfloor}
       \prod_{i\geq0}F(w_i(n)).
\end{equation}
For every character $\chi:\units{p^b}\longrightarrow\C^\times$ and
$u\in\{0,\ldots,r-1\}$, one may write
\begin{equation}\label{eq:phase}
 \e(uE_p(n)/r)\chi(U_p(n))=\e(\alpha n)f(n),\qquad
 f(n)=B(n)\e(\theta s_p(n))\prod_{i\geq0}\chi(F(w_i(n))),
\end{equation}
where $|f(n)|=1$, $B$ has period $p^{b-1}$, and $\alpha,\theta$ are real.
\end{lemma}
\begin{proof}
Separating the multiples of $p$ from a factorial gives
$U_p(n)=F(n)U_p(\lfloor n/p\rfloor)$ and hence
$U_p(n)=\prod_{i\geq0}F(\lfloor n/p^i\rfloor)$.
The units in each block of length $p^b$ have the same residues, so
$F(t)=G^{\lfloor t/p^b\rfloor}F(t\bmod p^b)$.
The resulting exponent of $G$ is
\[
 \sum_{i\geq0}\left\lfloor\frac n{p^{i+b}}\right\rfloor
 =E_p(n)-\sum_{j=1}^{b-1}\left\lfloor\frac n{p^j}\right\rfloor.
\]
Here $E_p(n)=\sum_{j\geq1}\lfloor n/p^j\rfloor$, since each factor
of $n!$ is counted once for each power of $p$ dividing it. If
$n=\sum_j d_jp^j$, summing this formula over digits also gives
$E_p(n)=(n-s_p(n))/(p-1)$.

Choose $\gamma\in\R$ with $\chi(G)=\e(\gamma)$, choosing $\gamma=0$
when $\chi$ is trivial. Substitution in \eqref{eq:windows} gives
\[
 \theta=-\frac{u/r+\gamma}{p-1},\qquad
 \alpha=-\theta-\gamma\sum_{j=1}^{b-1}p^{-j},\qquad
 B(n)=\e\left(\gamma\sum_{j=1}^{b-1}
                    \frac{n\bmod p^j}{p^j}\right).
\]
These formulas prove \eqref{eq:phase} and the asserted periodicity.
All infinite products and sums above have only finitely many nonidentity
terms. Empty sums, including those when $b=1$, are zero.
\end{proof}

Define the zero-padded window product
\begin{equation}\label{eq:T}
 T(t)=\prod_{i\geq0}F(w_i(p^{b-1}t)).
\end{equation}
It includes every window of $b$ consecutive digits meeting the digits of
$t$, with zero digits on both sides.

\begin{lemma}\label{lem:permutation}
One has $T(pt)=T(t)$ and
\begin{equation}\label{eq:Tpadded}
 T(t)=U_p(p^{b-1}t)G^{-E_p(t)}.
\end{equation}
For $p\nmid t$,
\begin{equation}\label{eq:R}
 \frac{T(t)}{T(t-1)}=R(t),\qquad
 R(t)=t\prod_{h=1}^{b-1}C_h(t),\qquad
 C_h(t)=\prod_{\substack{1\leq a\leq p^h\\p\nmid a}}
                   \bigl(p^h(t-1)+a\bigr).
\end{equation}
The function $R$ permutes $\units{p^b}$.
\end{lemma}
\begin{proof}
Shifting the digits one position in \eqref{eq:T} adds the factor $F(0)=1$,
which proves $T(pt)=T(t)$. In \eqref{eq:windows} with $n=p^{b-1}t$, the
exponent of $G$ is $\sum_{j\geq b}\lfloor p^{b-1}t/p^j\rfloor=E_p(t)$,
proving \eqref{eq:Tpadded}. Further,
\[
 U_p(p^{b-1}t)=U_p(t)\prod_{h=1}^{b-1}F(p^ht).
\]
If $p\nmid t$, then $E_p(t)=E_p(t-1)$ and
$U_p(t)/U_p(t-1)=t$. The quotient of the $h$th factors in the last product
is $C_h(t)$, proving \eqref{eq:R}.

Each $C_h(t)$ is a unit and is constant modulo $p$. Thus $R$ is a
permutation modulo $p$. If $k\geq1$, comparison of individual factors gives
\[
 C_h(t+ap^k)\equiv C_h(t)\pmod{p^{k+1}},
\]
because $h\geq1$. Consequently, for $0\leq a<p$,
\begin{equation}\label{eq:lifting}
 R(t+ap^k)\equiv R(t)+ap^k\prod_{h=1}^{b-1}C_h(t)
                         \pmod{p^{k+1}}.
\end{equation}
The coefficient of $ap^k$ is a unit. Therefore the $p$ lifts of a unit
class modulo $p^k$ map bijectively onto the $p$ lifts of its image.
Induction from $k=1$ proves the assertion modulo $p^b$. For $b=1$ this
is simply the map $R(t)=t$.
\end{proof}

\section{Contraction and Fourier decay in one base}

\begin{lemma}\label{lem:contraction}
Suppose $(u,\chi)$ is nontrivial, meaning $u\not\equiv0\pmod r$ or
$\chi\ne1$. With $\theta$ as in Lemma~\ref{lem:windows}, let
\[
 h_0=b+1,\qquad W(t)=\e(\theta s_p(t))\chi(T(t)).
\]
There exists $\rho<1$ such that, for every real $\lambda$,
\begin{equation}\label{eq:contraction}
 \left|p^{-h_0}\sum_{0\leq t<p^{h_0}}W(t)\e(\lambda t)\right|\leq\rho.
\end{equation}
\end{lemma}
\begin{proof}
The left side is a continuous function of $\lambda\pmod1$, with values
at most one, and therefore has a maximum. Suppose that maximum were one.
Equality in the triangle inequality for numbers of modulus one implies
that all the summands agree. The summand at $t=0$ is one, so
\begin{equation}\label{eq:resonance}
 W(t)\e(\lambda t)=1\qquad(0\leq t<p^{h_0}).
\end{equation}
Both $s_p(pt)=s_p(t)$ and $T(pt)=T(t)$. Comparing $t=1$ and $t=p$
in \eqref{eq:resonance} gives $(p-1)\lambda\in\Z$.
Because $t\equiv s_p(t)\pmod{p-1}$,
\[
 \chi(T(t))=\e(-(\lambda+\theta)s_p(t))
                   \qquad(0\leq t<p^{h_0}).
\]
For every $1\leq t<p^b$ with $p\nmid t$, subtracting one changes just
the last nonzero digit, and $s_p(t)-s_p(t-1)=1$.
Lemma~\ref{lem:permutation} now gives
\[
 \chi(R(t))=\e(-(\lambda+\theta)).
\]
Since $R(t)$ runs through every unit modulo $p^b$, the character $\chi$
is constant on its entire group. It takes the value one at the identity,
so $\chi=1$. Hence $\gamma=0$, and the equation at $t=1$ gives
$\lambda+\theta\in\Z$. Together with $(p-1)\lambda\in\Z$ this implies
$u/r=-(p-1)\theta\in\Z$, contrary to nontriviality. Compactness gives
the strict bound \eqref{eq:contraction}.
\end{proof}

\begin{proposition}\label{prop:fourier}
For a nontrivial pair $(u,\chi)$, the function $f$ in \eqref{eq:phase}
satisfies, for some $K>0$, $\eta>0$,
\begin{equation}\label{eq:fourier}
 \sup_{\xi\in\R}\left|p^{-k}\sum_{0\leq n<p^k}f(n)\e(\xi n)\right|
       \leq Kp^{-\eta k}\qquad(k\geq0).
\end{equation}
\end{proposition}
\begin{proof}
Regard the $k$ digits of $n$ as independent uniform digits. Fix the bottom
$b-1$ digits, so that $B(n)$ is fixed. In the remaining positions choose
\[
 J=\max\left(0,\left\lfloor\frac{k-b+1}{h_0+2b-2}\right\rfloor\right)
\]
disjoint consecutive blocks, each consisting of $b-1$ lower buffer digits,
$h_0$ free digits, and $b-1$ upper buffer digits. Fix all buffers and
any unused digits. No window of length $b$ meets two free blocks.
Consequently the summand in \eqref{eq:fourier}, as a function of the free
blocks, factors into one function for each block times a constant of
modulus one. The digit-sum phase and the linear phase factor in the same way.

If both buffers adjoining one free block are zero, its window factors are
exactly $\chi(T(t))$, where $0\leq t<p^{h_0}$ is that block's digit value.
Its digit-sum phase is $\e(\theta s_p(t))$, and its linear phase is
$\e(\lambda t)$ for some real $\lambda$. Its normalized sum is therefore
at most $\rho$ by Lemma~\ref{lem:contraction}. For other buffer values the
triangle inequality gives the bound one. Windows lying entirely in fixed
digits contribute only constants and do not affect these bounds.

Average now over the fixed digits. The $2(b-1)$ buffer digits for distinct
blocks are disjoint, so the events that both buffers vanish are independent,
each with probability $p^{-2(b-1)}$. The absolute normalized sum is at most
\[
 \left(1-p^{-2(b-1)}(1-\rho)\right)^J.
\]
This decays exponentially in $k$, which is \eqref{eq:fourier} after an
adjustment of $K$ for small $k$. When $b=1$ the buffers and initial fixed
digits are empty, and every block gives the factor $\rho$; the same proof
applies.
\end{proof}

\section{Carries and distinct bases}

For $L=p^k$ let $f_{[k]}$ be the periodic function of period $L$ whose
values on $0,\ldots,L-1$ equal those of $f$.

\begin{lemma}\label{lem:carry}
Suppose $k\geq b-1$ and $1\leq h<p^{k-b+1}$. Except for at most
\[
 h\left(\frac{N}{p^{k-b+1}}+2\right)
\]
integers $n\in[A,A+N)$, one has
\begin{equation}\label{eq:carry}
 f(n+h)\overline{f(n)}
       =f_{[k]}(n+h)\overline{f_{[k]}(n)}.
\end{equation}
The bound is independent of $A\in\N_0$.
\end{lemma}
\begin{proof}
Put $D=p^{k-b+1}$. Exclude the $h$ integers immediately preceding each
multiple of $D$. There are at most $N/D+2$ relevant multiples for the
given interval. At all other integers, $\lfloor(n+h)/D\rfloor
=\lfloor n/D\rfloor$. Thus the digits in positions $k-b+1$ and above
are unchanged in passing from $n$ to $n+h$.

Truncation modulo $p^k$ preserves the lower $k$ digits. The omitted digits
make equal contributions to the two digit sums, so cancel in the ratio.
Every window that crosses or lies above the truncation position has its
lowest position at least $k-b+1$ and is unchanged, so its factors also
cancel. All remaining windows agree with those of the truncated integers.
Finally $B$ has period $p^{b-1}$, which divides $p^k$, so it agrees with
its truncated values. These observations prove \eqref{eq:carry}.
\end{proof}

We record the elementary analytic facts used to combine primes. If $v$ is
a function of period $L$ with $|v|=1$, put
\[
 \widehat v(\ell)=\frac1L\sum_{a=0}^{L-1}v(a)\e(-\ell a/L),\qquad
 c_v(h)=\frac1L\sum_{a=0}^{L-1}v(a+h)\overline{v(a)}.
\]
Orthogonality of the $L$th roots of unity gives
\begin{equation}\label{eq:parseval}
 \sum_{\ell=0}^{L-1}|\widehat v(\ell)|^2=1,\qquad
 \sum_{h=0}^{L-1}|c_v(h)|^2
       =L\sum_{\ell=0}^{L-1}|\widehat v(\ell)|^4.
\end{equation}
Indeed, expansion and summation of the geometric series gives the first
identity, while
$c_v(h)=\sum_\ell|\widehat v(\ell)|^2\e(\ell h/L)$ gives the second.
In particular, if $\max_\ell|\widehat v(\ell)|\leq KL^{-\eta}$ and
$H<L$, Cauchy--Schwarz gives
\begin{equation}\label{eq:mean-corr}
 \frac1H\sum_{h=1}^H|c_v(h)|
    \leq K(L/H)^{1/2}L^{-\eta}.
\end{equation}

We shall also use the following elementary form of van der Corput's
inequality. For $z_n$ supported on an interval of $N$ integers, with
$|z_n|\leq1$, and $1\leq H\leq N$,
\begin{equation}\label{eq:vdc}
 \left|\frac1N\sum_n z_n\right|^2
 \leq \left(1+\frac HN\right)
 \left(\frac1H+\frac2H\sum_{h=1}^{H-1}
            \left|\frac1N\sum_n z_{n+h}\overline{z_n}\right|\right).
\end{equation}
To verify it, write $H\sum_nz_n=\sum_j\sum_{h=0}^{H-1}z_{j+h}$.
At most $N+H-1$ inner sums are nonzero. Apply Cauchy--Schwarz, expand
their squared moduli, and bound each off-diagonal correlation in absolute
value. This gives \eqref{eq:vdc}.

\begin{proposition}\label{prop:joint}
For each $p\in\mathcal P$ choose a character pair $(u_p,\chi_p)$ as in
Lemma~\ref{lem:windows}. If at least one pair is nontrivial, then for
some $\delta>0$,
\begin{equation}\label{eq:jointchars}
 \sum_{A\leq n<A+N}\e(\beta n)
 \prod_{p\in\mathcal P}\e(u_pE_p(n)/r_p)\chi_p(U_p(n))
       \ll N^{1-\delta}
\end{equation}
uniformly for $A\in\N_0$, $N\geq1$, and $\beta\in\R$.
\end{proposition}
\begin{proof}
By \eqref{eq:phase}, the summand has the form
$g(n)=\e(\beta' n)\prod_pf_p(n)$ for a real $\beta'$.
Choose a prime $p_*$ with a nontrivial character pair, and take an exponent
$0<\eta\leq1/4$ for which Proposition~\ref{prop:fourier} holds in this base.
For a positive integer $H$, choose powers
\[
 L_p=p^{k_p}\asymp_p H^{1+\eta},\qquad L=\prod_{p\in\mathcal P}L_p.
\]
For sufficiently large $H$, we have $k_p\geq b_p-1$ and
$H<L_p/p^{b_p-1}$ for every $p$.

For $1\leq h\leq H$, apply Lemma~\ref{lem:carry} to each base.
Replacing the ratios $f_p(n+h)\overline{f_p(n)}$ by their periodic versions
changes their normalized average by
\[
 O\left(\sum_p h/L_p+h/N\right),
\]
with constants allowed to depend on the fixed prime powers. The restriction
that both $n,n+h$ lie in the original interval changes a full interval
average by at most $h/N$. The product of the periodic ratios has period $L$.
Its average over any interval of $N$ integers differs from its complete
period average by $O(L/N)$, independently of the starting point.
The moduli $L_p$ are pairwise coprime. The Chinese remainder theorem
therefore makes that complete average equal to $\prod_pc_p(h)$, where
\[
 c_p(h)=\frac1{L_p}\sum_{a=0}^{L_p-1}
                   f_{p,[k_p]}(a+h)\overline{f_{p,[k_p]}(a)}.
\]
The linear twist contributes only the factor $\e(\beta'h)$.
Since $|c_p(h)|\leq1$, the normalized correlation of $g$, restricted to
the interval, has absolute value at most
\begin{equation}\label{eq:corrbound}
 |c_{p_*}(h)|+O\left(\sum_p h/L_p+h/N+L/N\right).
\end{equation}

Proposition~\ref{prop:fourier} bounds every discrete Fourier coefficient
of $f_{p_*,[k_{p_*}]}$ by $K L_{p_*}^{-\eta}$. Equations
\eqref{eq:parseval}--\eqref{eq:mean-corr} thus imply
\begin{align*}
 \frac1H\sum_{h=1}^H|c_{p_*}(h)|
 &\ll (L_{p_*}/H)^{1/2}L_{p_*}^{-\eta}\\
 &\ll H^{-\eta/2-\eta^2}.
\end{align*}
Write $s=|\mathcal P|$ and choose
$H=\lfloor N^{1/(2s(1+\eta))}\rfloor$.
Then $L\ll N^{1/2}$ and $H/L_p\ll H^{-\eta}$.
Averaging \eqref{eq:corrbound} in \eqref{eq:vdc} yields
\[
 \left|\frac1N\sum_{A\leq n<A+N}g(n)\right|^2
 \ll H^{-1}+H^{-\eta/2-\eta^2}+H^{-\eta}+H/N+N^{-1/2}
 \ll N^{-\eta/(4s(1+\eta))}.
\]
Taking square roots proves \eqref{eq:jointchars}, for example with
$\delta=\eta/(8s(1+\eta))$. Values of $N$ for which the choices above
are too small are absorbed by increasing the implied constant. No bound
depends on $A$ or $\beta$.
\end{proof}

\section{Fourier inversion and composite moduli}

\begin{proof}[Proof of Theorem~\ref{thm:local}]
For a finite abelian group $H$, its characters satisfy
\[
 \mathbf 1_{x=y}=\frac1{|H|}\sum_{\psi\in\widehat H}
                          \psi(x)\overline{\psi(y)}.
\]
For completeness, decompose $H$ into cyclic groups. On each cyclic factor
this is the geometric-series identity; multiplication gives the displayed
formula. Apply it to the group
\[
 (\Z/Q\Z)\times\prod_{p\in\mathcal P}
       (\Z/r_p\Z)\times\units{p^{b_p}}.
\]
The trivial character gives exactly the main term in
\eqref{eq:localcount}. Every character with at least one nontrivial prime
pair is bounded by Proposition~\ref{prop:joint}, with $\beta=v/Q$ for some
integer $v$. If all prime pairs are trivial but the index character is
not, its sum is a geometric series of ratio $\e(v/Q)\ne1$, bounded by a
constant depending only on $Q$. The finite number of characters permits
one to take the minimum of their positive exponents and the maximum of
their constants. If there are no nontrivial characters, the formula is exact.
This proves the theorem uniformly in every target.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:composite}]
Let $\mathcal P$ contain the prime divisors of $m$ and the primes $q_i$.
For $p\mid m$ write $p^{b_p}\parallel m$; take $b_p=1$ for the other
primes. Put $r_p=r_i$ if $p=q_i$ and $r_p=1$ otherwise. Enlarge each
valuation modulus to
\[
 R_q=\lcm\bigl(r_q,\ \{\varphi(p^{b_p}):p\mid m,\ p\ne q\}\bigr),
\]
interpreting the least common multiple of the empty additional list as
$r_q$. For $p\mid m$ the identity of integers defining $W_m$ gives,
in $\units{p^{b_p}}$,
\begin{equation}\label{eq:correction}
 W_m(n)=U_p(n)\prod_{\substack{q\mid m\\q\ne p}}q^{-E_q(n)}.
\end{equation}
The powers on the right are determined by $E_q(n)\pmod{R_q}$, since the
order of a unit modulo $p^{b_p}$ divides $\varphi(p^{b_p})$.

Fix an enlarged valuation vector satisfying the original congruences.
Then multiplication by the known correction factor in
\eqref{eq:correction} is a bijection of $\units{p^{b_p}}$.
For each $p\mid m$ it prescribes exactly one unit value corresponding
to $z\pmod{p^{b_p}}$. The Chinese remainder theorem identifies these
prescriptions with the one target $z\pmod m$.
All other unit coordinates are unrestricted.

Theorem~\ref{thm:local}, with moduli $R_q$, assigns uniform weight to
each enlarged valuation vector and unit vector, independently of
$n\pmod Q$. There are $\prod_q(R_q/r_q)$ allowed enlarged valuation
vectors. Summing over these and the unrestricted unit coordinates leaves
the proportion
\[
 \frac1{Q\prod_{p\mid m}\varphi(p^{b_p})\prod_q r_q}
       =\frac1{Q\varphi(m)\prod_{i=1}^t r_i}.
\]
Only a fixed finite number of error terms is summed, so the power saving
and its uniformity in $A$ remain valid.
\end{proof}

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\end{document}