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\begin{center}
\vskip 1cm{\Large\bf 
Quasi-Injectivity of Some Arithmetic Functions
}
\vskip 1cm
\large
Prapanpong Pongsriiam \\
Department of Mathematics \\
Faculty of Science\\
Silpakorn University\\
Nakhon Pathom 73000 \\
Thailand\\
\href{mailto:pongsriiam_p@silpakorn.edu}{\tt pongsriiam\_p@silpakorn.edu} \\
\href{mailto:prapanpong@gmail.com}{\tt prapanpong@gmail.com} \\
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\vskip .2 in
	
\begin{abstract}
 Let $a$ and $b$ be positive integers and $f$ an arithmetic function. In this article, we investigate whether or not a certain condition on the value of $f$ implies $a=b$. For example, if $f$ is the sum of divisors function and $f(an)=f(bn)$ for all positive integers $n$, then $a=b$.
\end{abstract}

	
\section{Introduction}

Let $\varphi$ be the Euler function, which counts the number of positive integers $k\leq n$ with $(k,n) = 1$. For each nonnegative integer $s$ and a positive integer $n$, let $\sigma_s(n)=\sum_{d\mid n}d^s$, where $d$ runs over the positive divisors of $n$, $\sigma(n)=\sigma_1(n)$, and $\tau(n)=\sigma_0(n)$. Problems on the ranges of arithmetic functions have been a popular area of research. For example, it is easy to see that if $n$ is a prime, then $\varphi(n) = n-1$; Lehmer asked whether $\varphi(n)\mid n-1$ implies that $n$ is a prime, but this question is still open. In addition, Carmichael's longstanding open problem on the range of $\varphi$ states that if $\varphi(x) = n$, then there exists $y\in \mathbb N$ distinct from $x$ such that $\varphi(y) = n$ too. Moreover, whether or not there are infinitely many $n\in \mathbb N$ with $\sigma(n) = 2n$ has been an open question for a very long time. 

Many mathematicians have worked on these problems and made some progress. For example, Pomerance \cite{Pom} obtained a result concerning the maximal order of $A(n)$, where $A(n)$ is the number of solutions to $\varphi(x) = n$.  Ford \cite{For} gave a comprehensive study of the range of $\varphi$ including $A(n)$. In particular, Ford \cite{For1} solved Sierpi\'nski's conjecture, and partially solved Carmichael's problem stated above. That is, Ford showed that for each integer $k\geq 2$, there exists a positive integer $n$ for which the equation $\varphi(x)=n$ has exactly $k$ solutions. Furthermore, Ford, Luca, and Pomerance \cite{FLP} completely answered Erd\"os' question on the range of $\varphi$ and $\sigma$ by showing that $\varphi(x) = \sigma(y)$ has infinitely many solutions in $x, y\in \mathbb N$. Ford and Pollack \cite{FPo} also gave a result complementary to that of  Ford, Luca, and Pomerance \cite{FLP}. For more information on the range of $\varphi$ and $\sigma$, we refer the reader to the sequences \seqnum{A000010}, \seqnum{A007617}, and \seqnum{A000396} in OEIS \cite{Slo}. Finally, in a recent Thailand Online Mathematical Olympiad TOMO 2021, an interesting arithmetic problem \cite{problem9} was to show that if $a$, $b\in\mathbb{N}$ and $\tau(\tau(an))=\tau(\tau(bn))$ for all $n\in\mathbb{N}$, then $a=b$. This naturally suggests various generalizations, where $\tau$ may be replaced by $\sigma_s$ or by other arithmetic functions.

\begin{definition}
We call a function $f:\mathbb{N}\to\mathbb{C}$ a quasi-injective function if for all $a$, $b\in\mathbb{N}$, the condition $f(an)=f(bn)$ for all $n\in\mathbb{N}$ implies $a=b$. In addition, if $f:\mathbb{N}\to\mathbb{N}$ and $\ell\in \mathbb N$, then we say that $f$ is quasi-injective of order $\ell$ if $f, f^{(2)}, f^{(3)}, \ldots, f^{(\ell)}$ are quasi-injective, that is, for any $a$, $b$, $k\in\mathbb{N}$ with $1\leq k\leq\ell$, 
	\begin{equation}\label{eq1}
	\text{if $f^{(k)}(an)=f^{(k)}(bn)$ for all $n\in\mathbb{N}$, then $a=b$.}
	\end{equation}
\end{definition}

In the above definition and throughout this article, if $k\in\mathbb{N}$, then $f^{(k)}$ is the $k$-fold composition of $f$. In addition, $\mathbb N$ is the set of positive integers but we may need to replace $\mathbb N$ by the set $\mathbb N_0$ of nonnegative integers. Therefore, the problem in TOMO 2021 mentioned above asks to show that \eqref{eq1} holds when $k=2$ and $f=\tau$. In fact, \eqref{eq1} also holds when $k=1$ and $f=\tau$. So $\tau$ is quasi-injective of order $2$. In general, if $f$ is quasi-injective of order $\ell$, then it is also quasi-injective of order $m\leq\ell$. For the concept of quasi-injectivity in algebra, see, for example, the articles by Yavari \cite{Yav} and Yavari and Ebrahimi \cite{YEb}.

In this article, we study quasi-injectivity of $f$ when $f=\sigma_s$ and other popular arithmetic functions such as the Euler totient function, the Jordan totient function, functions counting prime divisors, the M\"{o}bius function, and the Alladi-Erd\"os function. We also give some open problems at the end of this paper.

\section{Preliminaries and lemmas}

In this section, we recall some basic terminologies and give some useful results for the reader's convenience. From this point on, $p$ is always a prime, $s$ and $n$ are positive integers, $\mu$ is the M\"{o}bius function, $\varphi$ is the Euler totient function, $\omega(n)$ is the number of distinct prime factors of $n$, $\Omega(n)$ is the number of prime divisors of $n$ counted with multiplicity, and $J_s$ is the Jordan totient function. So $J_s(n)$ is the number of $s$-tuples $(a_1, a_2, \ldots, a_s)$ such that $1\leq a_i\leq n$ for all $i=1$, $2$, $\ldots$, $s$ and $(a_1, a_2, \ldots, a_s, n)=1$. Therefore $\varphi = J_1$. Furthermore, it is well known that
\[
\varphi(n)=n\displaystyle\prod_{p\mid n}\left(1-\frac{1}{p}\right) \quad\text{and}\quad J_s(n)=n^s\displaystyle\prod_{p\mid n}\left(1-\frac{1}{p^s}\right).
\]
An arithmetic function $f$ is called multiplicative if $f$ is not the zero function and $f(mn)=f(m)f(n)$ for all $m$, $n\in\mathbb{N}$ with $(m,n)=1$; and $f$ is called additive if $f(mn)=f(m)+f(n)$ for all $m$, $n\in\mathbb{N}$ with $(m,n)=1$. In addition, $\mu(1)=1$, $\mu(n)=(-1)^{\omega(n)}$ if $n$ is squarefree, and $\mu(n)=0$ if $n$ is not squarefree. It is well known that the functions $\tau$, $\sigma_s$, $\mu$, and $J_s$ are multiplicative, while $\omega$ and $\Omega$ are additive. Furthermore, 
\[
\text{if $c\in\mathbb{N}$, then $\displaystyle\sigma_s(p^c)=\frac{p^{(c+1)s}-1}{p^s-1}$.}
\]
For more details about this, see for instance, the books by Hardy and Wright \cite{HWr}
and McCarthy \cite{Mc}.

Recall that the $p$-adic valuation of $n$, denoted by $v_p(n)$, is the
exponent of $p$ in the prime factorization of $n$. A useful formula for
$v_p(x^n-y^n)$, sometimes called the ``Lifting the Exponent Lemma'',
is well known and popular among students taking mathematical Olympiad exams.
For the proof, see, for example, \cite[pp.~14--15]{ANT}.

\begin{lemma}[``Lifting the Exponent Lemma'']\label{lemma2}
	Let $x$, $y\in\mathbb{Z}$, $n\in\mathbb{N}$, and let $p$ be a prime such that $p\nmid x$ and $p\nmid y$. Then the following statements hold:
\begin{enumerate}
	\item[{\rm (i)}] If $p$ is odd and $p\mid x-y$, then $v_p(x^n-y^n)=v_p(x-y)+v_p(n)$.
	\item[{\rm (ii)}] If $p$ is odd, $n$ is odd, and $p\mid x+y$, then $v_p(x^n+y^n)=v_p(x+y)+v_p(n)$.
\end{enumerate}
When $p=2$ (so $x$ and $y$ are odd integers), we have
\begin{enumerate}
	\item[{\rm (iii)}] if $n$ is odd, then $v_2(x^n-y^n)=v_2(x-y)$ and $v_2(x^n+y^n)=v_2(x+y)$;
	\item[{\rm (iv)}] if $n$ is even, then $v_2(x^n-y^n)=v_2(x^2-y^2)+v_2(n)-1$.
\end{enumerate}
\end{lemma}

The conditions (A), (B), (C) in the following lemma are used throughout this article.
\begin{lemma}\label{lemma4}
	Let $a$, $b\in\mathbb{N}$ and let $f:\mathbb{N}\to\mathbb{N}$ be a multiplicative function satisfying the following conditions:
	\begin{enumerate}
		\item[{\rm (A)}] $f(p^k)&gt;f(p^r)$ for all primes $p$ and nonnegative integers $k&gt;r$;
		\item[{\rm (B)}] for any prime $p$ and nonnegative integers $x$, $y$, $c_1$, $c_2$, if $c_1\neq c_2$ and $f(p^{x+c_2})f(p^{y+c_1})=f(p^{y+c_2})f(p^{x+c_1})$, then $x=y$;
		\item[{\rm (C)}] for each prime $p$ dividing $ab$, there are $c_1$, $c_2\in\mathbb{N}\cup\{0\}$ such that $c_1\neq c_2$ and $f(ap^c)=f(bp^c)$ for $c\in\{c_1, c_2\}$.
	\end{enumerate}
	Then $a=b$.
\end{lemma}
\begin{proof}
	Since $f:\mathbb{N}\to\mathbb{N}$, $f(n)\geq 1$ for all $n$. If $a=1$ but $b\geq 2$, then there are a prime $p$ and $k\in\mathbb{N}$ such that $b=p^kb_1$, $p\nmid b_1$, and so by the condition (C), there exists $c\in\mathbb{N}$ such that
	\[
	f(p^c)=f(ap^c)=f(bp^c)=f(p^{c+k}b_1)=f(p^{c+k})f(b_1)\geq f(p^{c+k})&gt;f(p^c),
	\]
which is a contradiction. So if $a=1$, then $b=1=a$. Similarly, if $b=1$, then $a=1=b$. So assume throughout that $a$, $b\geq 2$. Let
	\[
	a=p_1^{a_1}p_2^{a_2}\ldots p_k^{a_k} \quad\text{and}\quad b=p_1^{b_1}p_2^{b_2}\ldots p_k^{b_k},
	\]
where $p_1$, $p_2$, $\ldots$, $p_k$ are distinct primes and $a_1$, $a_2$, $\ldots$, $a_k$, $b_1$, $b_2$, $\ldots$, $b_k$ are nonnegative integers with $a_j^2+b_j^2\neq 0$ for any $j=1$, $2$, $\ldots$, $k$. To show that $a=b$, it suffices to show that $a_j=b_j$ for all $j$. So let  $j\in\{1, 2, \ldots, k\}$ and $p=p_j$. Then $p\mid ab$. By the condition (C), there are $c_1$, $c_2\in\mathbb{N}\cup\{0\}$ such that $c_1\neq c_2$, $f(ap^{c_1})=f(bp^{c_1})$, and $f(ap^{c_2})=f(bp^{c_2})$. Then
	\begin{align}
	f(p_j^{a_j+c_1})\displaystyle\prod_{\substack{1\leq i\leq k \\ i\neq j}}f\left(p_i^{a_i}\right)&amp;=f(p_j^{b_j+c_1})\prod_{\substack{1\leq i\leq k \\ i\neq j}}f\left(p_i^{b_i}\right) \label{eq2}\\
	f(p_j^{a_j+c_2})\displaystyle\prod_{\substack{1\leq i\leq k \\ i\neq j}}f\left(p_i^{a_i}\right)&amp;=f(p_j^{b_j+c_2})\prod_{\substack{1\leq i\leq k \\ i\neq j}}f\left(p_i^{b_i}\right)\label{eq3}
	\end{align}
	Dividing \eqref{eq3} by \eqref{eq2} gives
	\begin{equation}\label{eq4}
	\frac{f(p_j^{a_j+c_2})}{f(p_j^{a_j+c_1})} = \frac{f(p_j^{b_j+c_2})}{f(p_j^{b_j+c_1})}
	\end{equation}
	By \eqref{eq4} and the condition (B), we obtain $a_j=b_j$, as required.
\end{proof}

The next lemma is used in the calculation of the $p$-adic valuation of $\sigma_s(n)$.
\begin{lemma}\label{lemma5}
	Assume that $p$ and $q$ are primes, $q\equiv 1 \pmod{p}$, and $c\in\mathbb{N}\cup\{0\}$. Then the following statements hold.
	\begin{enumerate}
		\item[{\rm (i)}] If $p$ is odd, then $v_p(\sigma_s(q^c))=v_p(c+1)$.
		\item[{\rm (ii)}] If $p=2$ and $s$ is even, then $v_p(\sigma_s(q^c))=v_p(c+1)$.
		\item[{\rm (iii)}] If $p=2$, $s$ is odd, and $c$ is odd, then $v_p(\sigma_s(q^c))=v_p(c+1)+v_p(q+1)-1$.
		\item[{\rm (iv)}] If $p=2$, $s$ is odd, and $c$ is even, then $v_p(\sigma_s(q^c))=0$.
	\end{enumerate}
\end{lemma}
\begin{proof}
	Since $p\nmid q$, $p\nmid 1$, and $p\mid q-1$, we obtain by Lemma \ref{lemma2} that if $p$ is odd, then
	\begin{align*}
	v_p(\sigma_s(q^c)) &amp;= v_p(q^{(c+1)s}-1)-v_p(q^s-1)\\
	&amp;= (v_p(q-1)+v_p((c+1)s))-(v_p(q-1)+v_p(s))=v_p(c+1).
	\end{align*}
	Similarly, if $p=2$ and $s$ is even, then
	\[
	v_p(\sigma_s(q^c))=(v_2(q^2-1)+v_2((c+1)s)-1)-(v_2(q^2-1)+v_2(s)-1)=v_2(c+1);
	\]
	if $p=2$, $s$ is odd, and $c$ is odd, then
	\begin{align*}
	v_p(\sigma_s(q^c))&amp;=v_2(q^2-1)+v_2((c+1)s)-1-v_2(q-1)\\
	&amp;=v_2(c+1)+v_2(q+1)-1;
	\end{align*}
	and if $p=2$, $s$ is odd, and $c$ is even, then $v_p(\sigma_s(q^c))=0$. This completes the proof.
\end{proof}

\section{Main results}

\begin{theorem}\label{thm1}
Let $a$, $b$, $s\in\mathbb{N}$. Then the functions $\tau$ and $\sigma_s$ satisfy the conditions {\rm (A)} and {\rm (B)} in Lemma \ref{lemma4}.
\end{theorem}
\begin{proof}
If $k&gt;r$, then every divisor of $p^r$ is also a divisor of $p^k$ while $p^k$ is not a divisor of $p^r$. From this, it is easy to see that $\tau$ and $\sigma_s$ satisfy the condition (A). Let $p$ be a prime, $x$, $y$, $c_1$, $c_2\in\mathbb{N}\cup\{0\}$, and $c_1\neq c_2$. Suppose that 
	\[
	\tau(p^{x+c_2})\tau(p^{y+c_1})=\tau(p^{y+c_2})\tau(p^{x+c_1}).
	\]
	Then $(x+c_2+1)(y+c_1+1)=(y+c_2+1)(x+c_1+1)$. Dividing both sides by $(y+c_1+1)(x+c_1+1)$, subtracting both sides by $1$, and then dividing both sides by $c_2-c_1$ leads to $x=y$, as required. Next, suppose that
	\begin{equation}\label{eq5}
	\sigma_s(p^{x+c_2})\sigma_s(p^{y+c_1})=\sigma_s(p^{y+c_2})\sigma_s(p^{x+c_1}).
	\end{equation}
	Let $x_1=x+c_1+1$, $x_2=x+c_2+1$, $y_1=y+c_1+1$, and $y_2=y+c_2+1$. Then \eqref{eq5} implies that
	\begin{equation}\label{eq6}
	\frac{p^{x_2s}-1}{p^{x_1s}-1}=\frac{p^{y_2s}-1}{p^{y_1s}-1}.
	\end{equation}
	Observing that $y_2-y_1=c_2-c_1=x_2-x_1$ and subtracting both sides of \eqref{eq6} by 1, we obtain
	\begin{equation}\label{eq7}
	\frac{p^{x_1s}(p^{(c_2-c_1)s}-1)}{p^{x_1s}-1}=\frac{p^{y_1s}(p^{(c_2-c_1)s}-1)}{p^{y_1s}-1}.
	\end{equation}
	Dividing both sides of \eqref{eq7} by $p^{(c_2-c_1)s}-1$, and then subtracting both sides by $1$ leads to $p^{x_1s}=p^{y_1s}$. Therefore $x_1=y_1$, which implies $x=y$, as required. This shows that $\tau$ and $\sigma_s$ satisfy the condition (B).
\end{proof}

\begin{corollary}\label{corollary1}
	Let $a$, $b\in\mathbb{N}$ and $f=\tau$ or $f=\sigma_s$. Suppose that for each prime $p\mid ab$, we can find distinct nonnegative integers $c_1$, $c_2$ such that $f(ap^c)=f(bp^c)$ for $c\in\{c_1, c_2\}$. Then $a=b$. In particular, if $f(an)=f(bn)$ for all $n\in\mathbb{N}$, then $a=b$. In other words, $\tau$ and $\sigma_s$ are quasi-injective.
\end{corollary}
\begin{proof}
	By Theorem \ref{thm1}, $f$ satisfies the conditions (A) and (B), and the above supposition is, in fact, the condition (C). Therefore $a=b$, as required.
\end{proof}

Modifying the proof of Theorem \ref{thm1}, we see that $\tau$ is in fact quasi-injective of order $\ell$ for any $\ell\in\mathbb{N}$, as shown in the next theorem.

\begin{theorem}\label{thm2}
	If $m$, $a$, $b$ are positive integers and $\tau^{(m)}(an)=\tau^{(m)}(bn)$ for all $n\in\mathbb{N}$, then $a=b$.
\end{theorem}
\begin{proof}
	We prove by induction on $m$. If $m=1$, the result follows from Corollary \ref{corollary1}. So let $m\geq 2$ and assume that the result holds for $m-1$. Let $a$, $b\in\mathbb{N}$ and 
	\begin{equation}\label{eq8}
	\tau^{(m)}(an)=\tau^{(m)}(bn) \text{ for all } n\in\mathbb{N}.
	\end{equation}
	Let $k\in\mathbb{N}$ and $d\in\mathbb{N}\cup\{0\}$. Choosing a prime $p\nmid abk$ and substituting $n=p^dk$ in \eqref{eq8}, we obtain
	\begin{equation}\label{eq9}
	\tau^{(m-1)}((d+1)\tau(ak))=\tau^{(m-1)}(\tau(an))=\tau^{(m-1)}(\tau(bn))=\tau^{(m-1)}((d+1)\tau(bk)).
	\end{equation}
	Since \eqref{eq9} holds for all $d\in\mathbb{N}\cup\{0\}$, we obtain by the induction hypothesis that $\tau(ak)=\tau(bk)$. Since $k$ is arbitrary, we obtain by Corollary \ref{corollary1} that $a=b$.
\end{proof}

We now know that if $\tau^{(m)}(an)=\tau^{(m)}(bn)$ for all $n$, then $a=b$; and if $\sigma_s(an)=\sigma_s(bn)$ for all $n$, then $a=b$. We would like to extend the result for $\sigma_s$ to $\sigma_s^{(m)}$ for any $m$ but it seems much more complicated than that of $\tau^{(m)}$, so we do it only for $\sigma_s^{(2)}$. We conjecture that the result holds for $m\geq 3$ as well but we currently do not have a proof.

\begin{theorem}\label{thm3}
	For each positive integer $x\geq 2$ and for each prime $p$ dividing $x$, there are positive integers $m$, $n$, $c_1$, $c_2$, $A$, $B$ such that $c_2&gt;c_1$, $(ABmn,x)=1$, $\sigma_s(m)=p^{c_1}A$, and $\sigma_s(n)=p^{c_2}B$.
\end{theorem}
\begin{proof}
	Let $x=p_1^{a_1}p_2^{a_2}\ldots p_k^{a_k}$, where $p_1$, $p_2$, $\ldots$, $p_k$ are distinct primes and $a_1$, $a_2$, $\ldots$, $a_k$ are positive integers. Let $p$ be a prime dividing $x$, say $p=p_j$ for some $j\in\{1, 2, \ldots, k\}$. Let 
	\[
	M=\prod_{1\leq i\leq k}p_i \quad\text{and}\quad M_j=\frac{M}{p}.
	\]
	By Dirichlet's theorem for primes in arithmetic progressions, there exists a prime $q\equiv 1\pmod{M}$. To find the integers $m$, $n$, $c_1$, $c_2$, $A$, $B$ as above, we divide the calculation into two cases:

\bigskip
	
\noindent \textbf{Case 1:} $p_i$ is odd for all $i\neq j$. By the Chinese remainder theorem, there exists a positive integer $w$ such that
	\begin{equation}\label{eq10}
	w\equiv -1\pmod{p} \quad\text{and}\quad w\equiv 0\pmod{M_j}.
	\end{equation}
	Let $L$ be a positive integer larger than $v_p(w+1)+v_p(q+1)$. Applying the Chinese remainder theorem again to obtain $y\in\mathbb{N}$ such that
	\begin{equation}\label{eq11}
	y\equiv -1\pmod{p^L} \quad\text{and}\quad y\equiv 0\pmod{M_j}.
	\end{equation}
	Let $m=q^w$ and $n=q^y$. By Lemma \ref{lemma5}, we obtain
	\[
	v_{p_i}(\sigma_s(m))=v_{p_i}(\sigma_s(q^w))=v_{p_i}(w+1)=0 \text{ for all } i\neq j.
	\]
	In addition, if $p$ is odd or if $p=2$ and $s$ is even, then $v_p(\sigma_s(m))=v_{p}(\sigma_s(q^w))=v_p(w+1)&gt;0$; if $p=2$ and $s$ is odd, then $w$ is odd, $q$ is odd, $v_p(q+1)\geq 1$, and
	\[
	v_p(\sigma_s(m))=v_p(\sigma_s(q^w))=v_p(w+1)+v_p(q+1)-1&gt;0.
	\]
	This shows that $v_p(\sigma_s(m))&gt;0$ and $v_{p_i}(\sigma_s(m))=0$ for all $i\neq j$. Let $c_1=v_p(\sigma_s(m))$. Then $c_1\in\mathbb{N}$ and $\sigma_s(m)=p^{c_1}A$, where $A\in\mathbb{N}$ and $(A,x)=1$. Similarly, we obtain $v_{p_i}(\sigma_s(n))=0$ for all $i\neq j$ and
	\[
v_p(\sigma_s(n))\geq v_p(y+1)\geq L&gt;v_p(w+1)+v_p(q+1)\geq v_p(\sigma_s(m)).
	\]
	Let $c_2=v_p(\sigma_s(n))$. Then $c_2\in\mathbb{N}$, $c_2&gt;c_1$, and $\sigma_s(n)=p^{c_2}B$, where $B\in\mathbb{N}$ and $(B,x)=1$. Since $(m,x)=(n,x)=(A,x)=(B,x)=1$, we obtain $(mnAB,x)=1$.

\bigskip
	
\noindent\textbf{Case 2:} $p_i=2$ for some $i\neq j$. Without loss of generality, assume that $p_1=2$. We still choose $w$, $L$, $y$, $m=q^w$ and $n=q^y$ as in Case 1 and apply Lemma \ref{lemma5} to calculate the $p$-adic valuation of $\sigma_s(m)$ and $\sigma_s(n)$. If $i\neq 1$, then $p_i$ is odd and so
	\[
	v_{p_i}(\sigma_s(m))=v_{p_i}(w+1)=0, v_{p_i}(\sigma_s(n))=v_{p_i}(y+1)=0.
	\]
	In addition, $p$ is odd, and so
	\[
	v_p(\sigma_s(m))=v_p(w+1)&gt;0, v_p(\sigma_s(n))=v_p(y+1)\geq L&gt;v_p(\sigma_s(m)).
	\]
	Since $p_1=2$ and $w$, $y\equiv 0\pmod{M_j}$, $w$ and $y$ are even. If $s$ is even, then $v_{p_1}(\sigma_s(m))=v_2(w+1)=0=v_2(y+1)=v_{p_1}(\sigma_s(n))$. If $s$ is odd, then $v_{p_1}(\sigma_s(m))$ and $v_{p_1}(\sigma_s(n))$ are also zero. This shows that $v_p(\sigma_s(n))&gt;v_p(\sigma_s(m))&gt;0$ and $v_{p_i}(\sigma_s(m))=v_{p_i}(\sigma_s(n))=0$ for all $i\neq j$. Therefore we can choose $c_1$, $c_2$, $A$, $B$ as in the previous case. This completes the proof.
\end{proof}

\begin{theorem}\label{thm4}
	Let $a$ and $b$ be positive integers and
	\begin{equation}\label{eq12}
	\sigma_s(\sigma_s(an))=\sigma_s(\sigma_s(bn)) \text{ for all } n\in\mathbb{N}.
	\end{equation}
	Then $a=b$.
\end{theorem}
\begin{proof}
	Let $p$ be a prime dividing $ab$ and $d\in\mathbb{N}\cup\{0\}$. Let $n_1$ be a positive integer such that $(n_1,ab)=1$. Substituting $n=p^dn_1$ in \eqref{eq12}, we obtain
	\begin{equation}\label{eq13}
	\sigma_s(\sigma_s(ap^d)\sigma_s(n_1))=\sigma_s(\sigma_s(bp^d)\sigma_s(n_1)),
	\end{equation}
which holds for any $n_1\in\mathbb{N}$ with $(n_1,ab)=1$. Next, let $a_1=\sigma_s(ap^{d})$, $b_1=\sigma_s(bp^{d})$, $x=aba_1b_1$ and let $q$ be a prime dividing $a_1b_1$. Then $q\mid x$. By Theorem \ref{thm3}, there are positive integers $m$, $n$, $c_1$, $c_2$, $A$, $B$ such that $c_2&gt;c_1$, $(ABmn,x)=1$, $\sigma_s(m)=q^{c_1}A$, and $\sigma_s(n)=q^{c_2}B$. Since \eqref{eq13} holds for any $n_1\in\mathbb{N}$ with $(n_1, ab)=1$ and $(m,ab)=(n,ab)=1$, we can substitute $n_1=m$ and $n_1=n$ in \eqref{eq13} to obtain
	\begin{equation}\label{eq14}
	\sigma_s(a_1q^{c_1}A)=\sigma_s(b_1q^{c_1}A) \quad\text{and}\quad \sigma_s(a_1q^{c_2}B)=\sigma_s(b_1q^{c_2}B).
	\end{equation}
	Since $(AB,x)=1$, we see that $(A,qa_1b_1)=(B,qa_1b_1)=1$. Therefore \eqref{eq14} reduces to
	\[
	\sigma_s(a_1q^{c_1})\sigma_s(A)=\sigma_s(b_1q^{c_1})\sigma_s(A) \quad\text{and}\quad \sigma_s(a_1q^{c_2})\sigma_s(B)=\sigma_s(b_1q^{c_2})\sigma_s(B),
	\]
which imply
	\begin{equation}\label{eq16}
	\sigma_s(a_1q^{c_1})=\sigma_s(b_1q^{c_1}) \quad\text{and}\quad \sigma_s(a_1q^{c_2})=\sigma_s(b_1q^{c_2}).
	\end{equation}
	This shows that for any prime $q\mid a_1b_1$, we can find distinct $c_1$, $c_2\in\mathbb{N}$ such that \eqref{eq16} holds. By Corollary \ref{corollary1}, we obtain $a_1=b_1$. Therefore
	\begin{equation}\label{eq18}
	\sigma_s(ap^d)=\sigma_s(bp^d).
	\end{equation}
	Since \eqref{eq18} holds for each prime $p\mid ab$ and each $d\in\mathbb{N}\cup\{0\}$, we apply Corollary \ref{corollary1} again to obtain $a=b$. This completes the proof.
\end{proof}

Before proceeding to the case of $J_s$, we give an example to show that $\omega$, $\Omega$, and $\mu$ are not quasi-injective.

\begin{example}\label{ex1}
	Let $k\geq 2$, $p_1$, $p_2$, $\ldots$, $p_k$ distinct primes, $c=p_1p_2\ldots p_k$, $a=p_1c$, and $b=p_2c$. Then $\omega(a)=\omega(b)$, $\Omega(a)=\Omega(b)$, and $\mu(a)=\mu(b)$. Since $\Omega$ is completely additive, that is, $\Omega(mn)=\Omega(m)+\Omega(n)$ for all $m$, $n\in\mathbb{N}$, we see that $\Omega(an)=\Omega(bn)$ for all $n\in\mathbb{N}$ but $a\neq b$. Therefore $\Omega$ is not quasi-injective. Similarly, it is not difficult to see that $\omega(an)=\omega(bn)$ and $\mu(an)=\mu(bn)$ for all $n\in\mathbb{N}$ but $a\neq b$, and so $\omega$ and $\mu$ are not quasi-injective.
\end{example}

We can generalize the idea in Example \ref{ex1} as follows.

\begin{theorem}\label{thm11}
	Let $f:\mathbb{N}\to\mathbb{C}$ be completely additive. Then $f$ is quasi-injective if and only if $f$ is injective.
\end{theorem}
\begin{proof}
	Suppose $f$ is quasi-injective, $a$, $b\in\mathbb{N}$, and $f(a)=f(b)$. Then for each $n\in\mathbb{N}$, we have
	\begin{equation}\label{eq17}
	f(an)=f(a)+f(n)=f(b)+f(n)=f(bn).
	\end{equation}
	Since \eqref{eq17} holds for all $n\in\mathbb{N}$ and $f$ is quasi-injective, we obtain $a=b$, as required. The converse is obvious.
\end{proof}

\begin{example}\label{ex2}
	Let $A$ be the Alladi-Erd\"{o}s function defined by $A(n)=\sum_{p^\alpha\parallel n}\alpha p$. It is easy to verify that $A$ is completely additive, that is, $A(mn)=A(m)+A(n)$ for all $m$, $n\in\mathbb{N}$. In addition, $A(2^3)=6=A(3^2)$, so $A$ is not injective. By Theorem \ref{thm11}, $A$ is not quasi-injective. 
	
A variation of Alladi-Erd\"{o}s function can defined by $A_0(n)=\sum_{p\mid n}p$. Then $A_0(6)=2+3=5=A_0(12)$, and it is not difficult to see that $A_0(6n)=A_0(12n)$ for all $n\in\mathbb{N}$. Therefore $A_0$ is not quasi-injective. 
\end{example}
\begin{theorem}
Suppose that $f:\mathbb{N}\to\mathbb{C}$ is strongly additive, that is, $f$ is additive and $f(p^k)=f(p)$ for all primes $p$ and positive integers $k$. If there are distinct positive integers $a$, $b\geq 2$ such that $f(a)=f(b)$ and $a$, $b$ have the same prime factors, then $f$ is not quasi-injective.
\end{theorem}
\begin{proof}
Since $a$ and $b$ have the same prime factors, we write
	\[
	a=p_1^{a_1}p_2^{a_2}\ldots p_k^{a_k} \quad\text{and}\quad b=p_1^{b_1}p_2^{b_2}\ldots p_k^{b_k},
	\]
where $p_1$, $p_2$, $\ldots$, $p_k$ are distinct primes and $a_1$, $a_2$, $\ldots$, $a_k$, $b_1$, $b_2$, $\ldots$, $b_k$ are positive integers. We show that $f(an)=f(bn)$ for all $n\in\mathbb{N}$. For each $n\geq 2$, we write $n=p_1^{n_1}p_2^{n_2}\ldots p_k^{n_k}m$, where $m\in\mathbb{N}$, $(m,ab)=1$, and $n_1$, $n_2$, $\ldots$, $n_k$ are nonnegative integers, and so 
	\[
	f(an)=f(p_1^{a_1+n_1}p_2^{a_2+n_2}\ldots p_k^{a_k+n_k}m)=f(a)f(m)=f(b)f(m)=f(bn).
	\]
	Therefore $f(an)=f(bn)$ for all $n\in\mathbb{N}$ but $a\neq b$. This completes the proof.
\end{proof}

\begin{theorem}\label{thm5}
	Let $a$, $b$, $c$, $k$, $m$, $s$ be positive integers. Then the following statements hold.
	\begin{enumerate}
		\item[{\rm (i)}] If $a\mid b$, then $J_s(ab)=a^sJ_s(b)$.
		\item[{\rm (ii)}] If $a\mid c$ and $b\mid c$, then $b^sJ_s(ac)=a^sJ_s(bc)$.
		\item[{\rm (iii)}] If $a\mid b$, then $J_s^{(m)}(a)\mid J_s^{(m)}(b)$.
		\item[{\rm (iv)}] $J_s(a^kb)=a^{(k-1)s}J_s(ab)$.
		\item[{\rm (v)}] If $k\geq m+1$, then $J_s^{(m)}(a^kb)=a^{s^m}J_s^{(m)}(a^{k-1}b)$.
		\item[{\rm (vi)}] If $k\geq m+1$, then $a^{s^m}\mid J_s^{(m)}(a^k)$.
		\item[{\rm (vii)}] If $k\geq m+1$ and $c\geq m+1$, then $a^{s^m}\mid J_s^{(m)}(a^kb^c)$ and $b^{s^m}\mid J_s^{(m)}(a^kb^c)$.
	\end{enumerate}
\end{theorem}
\begin{proof}
	For (i) and (iii), assume that $a\mid b$. Then $p\mid ab$ if and only if $p\mid b$. Therefore
	\[
	J_s(ab)=(ab)^s\displaystyle\prod_{p\mid ab}\left(1-\frac{1}{p^s}\right)=a^sb^s\displaystyle\prod_{p\mid b}\left(1-\frac{1}{p^s}\right)=a^sJ_s(b),
	\]
which proves (i). If $a=1$, then (iii) is obvious. So assume that $a\geq 2$. Let $a=p_1^{a_1}p_2^{a_2}\ldots p_k^{a_k}$ and $b=p_1^{b_1}p_2^{b_2}\ldots p_k^{b_k}b_0$, where $p_1$, $p_2$, $\ldots$, $p_k$ are distinct primes, $a_i$, $b_i$, $b_0\in\mathbb{N}$, $b_i\geq a_i$ for all $i=1$, $2$, $\ldots$, $k$, and $(b_0, p_1p_2\ldots p_k)=1$. Then 
	\begin{align*}
	J_s(a)&amp;=\prod_{i=1}^{k}\left(p_i^{(a_i-1)s}\right)\prod_{i=1}^{k}\left(p_i^{s}-1\right),\\
	J_s(b)&amp;=\prod_{i=1}^{k}\left(p_i^{\left(b_i-1\right)s}\right)\prod_{i=1}^{k}\left(p_i^{s}-1\right)J_s(b_0).
	\end{align*}
	From this, it is easy to see that $J_s(a)\mid J_s(b)$. So (iii) is proved for $m=1$. If $m\geq 2$, we can apply the result when $m=1$ repeatedly and obtain the chain of implication as follows:
	\[
	a\mid b \Rightarrow J_s(a)\mid J_s(b) \Rightarrow J_s^{(2)}(a)\mid J_s^{(2)}(b) \Rightarrow \cdots \Rightarrow J_s^{(m)}\mid J_s^{(m)}(b).
	\]
	Therefore (iii) is proved for every $m\geq 1$. For (ii), if $a\mid c$ and $b\mid c$, then we obtain by (i) that 
	\[
	b^sJ_s(ac)=b^sa^sJ_s(c)=a^sb^sJ_s(c)=a^sJ_s(bc).
	\]
	Next, we prove (iv) by induction on $k$. If $k=1$, then (iv) is obvious. So let $k\geq 1$ and assume that (iv) holds for $k$. Since $a\mid a^kb$, we obtain by (i) that
	\[
	J_s(a^{k+1}b)=J_s(a(a^kb))=a^sJ_s(a^kb)=a^{ks}J_s(ab),
	\]
where the last equality is obtained from the induction hypothesis. Next, we prove (v) by induction on $m$. If $m=1$ and $k\geq 2$, then we obtain by (i) that $J_s^{(m)}(a^kb)=J_s(a\cdot a^{k-1}b)=a^sJ_s(a^{k-1}b)$. So let $m\geq 1$ and assume that the result holds for $m$. Suppose $k\geq m+2$. By the induction hypothesis, we obtain
	\begin{equation}\label{eq19}
	J_s^{(m+1)}(a^kb)=J_s(J_s^{(m)}(a^kb))=J_s(a^{s^m}J_s^{(m)}(a^{k-1}b)).
	\end{equation}
	Since $k-1\geq m+1$, we apply the induction hypothesis again to conclude that $a^{s^m}\mid J_s^{(m)}(a^{k-1}b)$. Then (i) implies that the right-hand side of \eqref{eq19} is equal to $a^{s^{m+1}}J_s^{(m+1)}(a^{k-1}b)$. Therefore \eqref{eq19} becomes
	\[
	J_s^{(m+1)}(a^kb)=a^{s^{m+1}}J_s^{(m+1)}(a^{k-1}b),
	\]
which proves (v). Then (vi) is a special case of (v) when $b=1$. For (vii), if $k\geq m+1$ and $c\geq m+1$, then we use (vi) and (iii) to obtain
	\[
	a^{s^m}\mid J_s^{(m)}(a^k)\mid J_s^{(m)}(a^kb^c) \quad\text{and}\quad b^{s^m}\mid J_s^{(m)}(b^c)\mid J_s^{(m)}(a^kb^c).
	\]
	Here $x\mid y\mid z$ means that $x\mid y$ and $y\mid z$. Hence the proof is complete.
\end{proof}

\begin{theorem}\label{thm6}
	Let $a$, $b$, $m$, $s\in\mathbb{N}$ and $J_s^{(m)}(an)=J_s^{(m)}(bn)$ for all $n\in\mathbb{N}$. Then $a=b$. In other words, $J_s$ is quasi-injective of any order.
\end{theorem}
\begin{proof}
	Substituting $n=a^mb^m$ in the above condition, we have $J_s^{(m)}(a^{m+1}b^m)=J_s^{(m)}(b^{m+1}a^m)$. By (v) of Theorem \ref{thm5}, we obtain
	\[
	J_s^{(m)}(a^{m+1}b^m)=a^{s^m}J_s^{(m)}(a^mb^m) \quad\text{and}\quad J_s^{(m)}(b^{m+1}a^m)=b^{s^m}J_s^{(m)}(b^ma^m).
	\]
	Therefore
	\[
	a^{s^m}J_s^{(m)}(a^mb^m)=b^{s^m}J_s^{(m)}(b^ma^m),
	\]
which implies $a=b$. This completes the proof.
\end{proof}


\section{Conclusion and some open problems}

We have proved that $\tau$ and $J_s$ are quasi-injective of order $m$ for any $m\in\mathbb{N}$ while we only show that $\sigma_s$ is quasi-injective of order $2$. We believe that it can be extended to any order. In addition, Example \ref{ex1} shows that $\mu$, $\omega$, and $\Omega$ are not quasi-injective. This leads us to the following problems.

\begin{question}
If $a$, $b$, $m$, $s\in\mathbb{N}$, $m\geq 3$, and $\sigma_s^{(m)}(an)=\sigma_s^{(m)}(bn)$ for all $n\in\mathbb{N}$, can we conclude that $a=b$?
\end{question}

\begin{question}
For each $m\geq 2$, is there a function $f:\mathbb{N}\to\mathbb{N}$ such that $f$ is quasi-injective of order $m-1$ but not of order $m$?
\end{question}

We may also consider the mix between $\tau$, $\sigma_s$, and $J_s$.

\begin{question}
Let $a$, $b$, $k$, $m$, $s\in\mathbb{N}$. Are the functions $\tau^{(m)}\circ \sigma_s^{(k)}$, $\sigma_s^{(k)}\circ \tau^{(m)}$, $\tau^{(m)}\circ J_s^{(k)}$, $J_s^{(k)}\circ \tau^{(m)}$, $\sigma_s^{(m)}\circ J_s^{(k)}$, and $J_s^{(k)}\circ \sigma_s^{(m)}$ quasi-injective? That is, if $f$ is one of the above functions  and $f(an)=f(bn)$ for all $n$, can we show that $a=b$? This may be easy when $k$ or $m$ is less than $3$. Can we say something when both $k$ and $m$ are larger than $2$?
\end{question}

\begin{question}
Suppose $f$ and $g$ are quasi-injective. Is the composition $f\circ g$ quasi-injective? Can we categorize those functions $f$ and $g$ for which $f\circ g$ must be quasi-injective? An obvious sufficient condition for $f\circ g$ to be quasi-injective is that $g$ is both surjective and completely multiplicative, but there may be a weaker condition. 
\end{question}

For each $n\in\mathbb{N}$ and $b\geq 2$, let $S_b(n)$ be the sum of digits of $n$ in base $b$, and let $S(n)=S_{10}(n)$ be the sum of the decimal digits of $n$.

\begin{question}
If $a$, $c\in\mathbb{N}$ and $S(an)=S(cn)$ for all $n\in\mathbb{N}$, is it true that $a=c$? More generally, if $a$, $c$, $m\in\mathbb{N}$ and $S^{(m)}(an)=S^{(m)}(cn)$ for all $n\in\mathbb{N}$, can we prove that $a=c$? Can we replace $S$ by $S_b$ for any $b\geq 2$?
\end{question}

\section{Acknowledgments}

The author's research project was funded jointly by the Faculty of Science
Silpakorn University, and the National Research Council of Thailand
(NRCT), Grant Number NRCT5-RSA63021-02. The author would also like to
show support for the group of students who organized Thailand Online
Mathematical Olympiad TOMO 2021. This article, in particular, the author
hopes, is evidence that they put an interesting arithmetic problem
in the exam and that they organized the contest very well. Finally,
the author is grateful to the referee for his or her suggestions that
improved the quality and exposition of this paper.


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\noindent 2010 {\it Mathematics Subject Classification}: Primary 11A25;
Secondary 11A05.
	
\noindent \emph{Keywords: } divisor function, Euler's function, Jordan's function, multiplicative function, additive function.
	

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\vspace*{+.1in}
\noindent
Received June 24 2021;
revised version received  November 8 2021.
Published in {\it Journal of Integer Sequences}, November 8 2021.
	
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\noindent (Concerned with sequences 
\seqnum{A000010},
\seqnum{A000396}, and
\seqnum{A007617}.)
	
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\noindent
Return to
\htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}.
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