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\begin{center}
\vskip 1cm{\LARGE\bf 
Product of Some Polynomials \\
\vskip .1in
and Arithmetic Functions
}
\vskip 1cm
\large
Passawan Noppakaew \\
Department of Mathematics \\
Faculty of Science\\
Silpakorn University\\
Nakhon Pathom, 73000 \\
Thailand\\
\href{mailto:noppakaew_p@silpakorn.edu}{\tt noppakaew\_p@silpakorn.edu}\\
\ \\
Prapanpong Pongsriiam\footnote{Prapanpong Pongsriiam is the corresponding author.} \\
Department of Mathematics \\
Faculty of Science\\
Silpakorn University\\
Nakhon Pathom, 73000 \\
Thailand\\
\medskip
and\\
\medskip
Graduate School of Mathematics\\
Nagoya University\\
Nagoya, 464-8602\\
Japan\\
\href{mailto:pongsriiam_p@silpakorn.edu}{\tt pongsriiam\_p@silpakorn.edu}, 
\href{mailto:prapanpong@gmail.com}{\tt prapanpong@gmail.com}  \\
\end{center}

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\begin{abstract}
We study the injectivity and noninjectivity of the function $fg$, where $f$ is a polynomial in a simple form and $g$ is a popular arithmetic function such as the Euler totient function or the sum of divisors function. We also show the connection between our results and Mersenne primes, amicable pairs, and other integer sequences.
\end{abstract}
		
\section{Introduction}
\label{sec1}

		For each $n\in\mathbb{N}$, let $\varphi(n)$ be the number of positive integers that are at most $n$ and relatively prime to $n$, and let $\sigma(n)$ be the sum of positive divisors of $n$. The functions $\varphi$ and $\sigma$ are connected with some popular topics such as Lehmer's problem, Carmichael's conjecture, perfect numbers, Mersenne primes, amicable pairs, and aliquot sequences.		
		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 long-standing open conjecture 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 $\sigma(n) = 2n$ and whether or not there exists an odd integer $n$ with $\sigma(n)=2n$ have been open for a long time. 
		
		Many mathematicians including Ford \cite{For,For1}, Ford, Luca, and Pomerance \cite{FLP}, Ford and Pollack \cite{FPo}, and  Pomerance \cite{Pom}  have contributed to the progress of this area of research.
		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\H{o}s' question on the ranges of $\varphi$ and $\sigma$ by showing that $\varphi(x) = \sigma(y)$ has infinitely many solutions in $x, y\in \mathbb N$.
		We refer the reader to the sequences \seqnum{A000010} and \seqnum{A007617} in the {\it On-Line Encyclopedia of Integer Sequences} (OEIS) \cite{Slo} for more information on the range of $\varphi$, the sequences \seqnum{A000396} and \seqnum{A000668} for perfect numbers and Mersenne primes, and the sequences  \seqnum{A063990}, \seqnum{A063900}, \seqnum{A001065}, \seqnum{A008888}, and \seqnum{A098007} for amicable pairs and aliquot sequences.
		
		It is well known that the function $\varphi$ is not injective but the function $f_0$ defined by $f_0(n)=n\varphi(n)$ is injective. Not every function has this property: both $\sigma$ and the function $A(n)= n\sigma(n)$ are not injective. Nevertheless, we show in Examples \ref{ExA} and \ref{ExB} and  Theorem \ref{thmAsqf} that $A$ is injective on squarefree integers and both $A$ and $\sigma$ are related to Mersenne primes and amicable pairs. Generally speaking, both injectivity and noninjectivity are interesting; if an arithmetic function $f$ is injective, we can conclude that the equation $f(x)=n$ has at most one solution; if $f$ is not injective, then we may like to count the number of solutions to $f(x)=n$, and study the relation between $f$ and each solution.
		
		In this article, we study the injectivity and noninjectivity of the product  of polynomials and arithmetic functions. We will replace $n\varphi(n)$ and $n\sigma(n)$ by $g(n)h(n)$ where $g(n)$ is a polynomial and $h(n)$ is an arithmetic function. For simplicity, we focus our attention to polynomials in a simple form such as $g(n)=n^a$ or $g(n)=n+c$ where $a,c$ are any positive integers, while $h(n)$ is a popular arithmetic function such as $\varphi(n)$, $\sigma(n)$, $d(n)$, $s(n)$, $\omega(n)$, $\Omega(n)$, $S_{b}(n)$, $\psi(n)$, and $J_s(n)$, where $d(n)$ is the number of positive divisors of $n$, $s(n)=\sigma(n)-n$ is the sum of proper divisors of $n$, $\omega(n)$ is the number of distinct prime divisors of $n$, $\Omega(n)$ is the number of prime divisors of $n$ counted with multiplicity, $S_{b}(n)$ is the sum of digits of $n$ when $n$ is written in base $b$, $\psi(n)$ is the Dedekind function, and $J_s(n)$ is Jordan's totient function. The functions $\psi$ and $J_s$ are defined by
		\begin{center}
			$\psi(n)=n\prod\limits_{p\mid n}(1+\frac{1}{p})$ and $J_s(n)=n^s\prod\limits_{p\mid n}(1-\frac{1}{p^s})$,
		\end{center} 
	where $s$ is a positive integer. For more information about injectivity or noninjectivity of arithmetic functions, see for example in Guy's book \cite[Section B]{Guy}, Pongsriiam's recent article \cite{Pong}, and the online database OEIS \cite{Slo}.
	
	We organize this article as follows. In Section~\ref{sec2}, we prove some results on injectivity of the function $gh$ where $g$ is a polynomial in a simple form and $h=\varphi, \psi$, and $J_s$. In Section~\ref{sec3}, we show the noninjectivity of $gh$ and a connection to other problems when $h=\sigma, s, d, \omega, \Omega$, and $S_b$. In Section~\ref{sec4}, we study the injectivity of $gh$ where $g$ and $h$ are restricted to squarefree integers. In fact, in Sections \ref{sec1}--\ref{sec4}, the function $g$ is of the form $g(n)=n^a$, but in Section~\ref{sec5}, we set $g(n)=n+c$ where $c$ is a positive integer. We obtain in Section~\ref{sec5} that the function $n\mapsto (n+c)\varphi(n)$ is not injective for infinitely many $c\in\mathbb{N}$. We also provide some related results in Section~\ref{sec6}. Finally, we give a list of open questions in Section~\ref{sec7}.
			
	\section{Results on injectivity}
	\label{sec2}
	In this section, we show that the product of $\varphi$, $\psi$, and $J_s$ with a polynomial in a simple form are injective. Recall that an arithmetic function $f$ is said to be multiplicative if $f(1)=1$ and $f(mn)=f(m)f(n)$ for all $m,n\in\mathbb{N}$ with $(m,n)=1$. It is well known that $\varphi$, $\sigma$, $d$, $\psi$, and $J_s$ are multiplicative. The following formulas are also well known and may be used throughout this article:
	\begin{center}
		$\varphi(n)=n\prod\limits_{p\mid n}(1-\frac{1}{p})$,\ \ 
		$\psi(n)=n\prod\limits_{p\mid n}(1+\frac{1}{p})$,\ \ 
		$J_s(n)=n^s\prod\limits_{p\mid n}(1-\frac{1}{p^s})$,
		
		$d(n)=\prod\limits_{p^\alpha\mid\mid n}(\alpha+1)$,\ \ 
		$\sigma(n)=\prod\limits_{p^\alpha\mid\mid n}(1+p+p^2+\cdots+p^\alpha)=\prod\limits_{p^\alpha\mid\mid n}\left(\dfrac{p^{\alpha+1}-1}{p-1}\right)$.
	\end{center}
	
		
	
	
	We begin our study with $\varphi$. Although it is well known that the function $n\mapsto n\varphi(n)$ is injective, we can extend it to the following form.
		
	\begin{theorem} \label{thm1}
		For each $a, b \in\mathbb{N}$, the arithmetic function $F$ defined by $F(n)=n^a\varphi(n)^b$ for all $n\in\mathbb{N}$ is an injective function. In particular, the function $f_0$ is injective.
	\end{theorem}
	\begin{proof}
		To show that $F$ is injective, let $m,n\in\mathbb{N}$ and $F(m)=F(n)$. If $m=1$, then $n^a\varphi(n)^b=F(n)=F(1)=1$, which implies $n=1$. Similarly, if $n=1$, then $m=1$. Therefore $m=1$ if and only if $n=1$. So we assume that $m,n\geq2$. Let
		\begin{center}
			$m=p_1^{m_1}p_2^{m_2}\cdots p_k^{m_k}$ and $n=q_1^{n_1}q_2^{n_2}\cdots q_\ell^{n_{\ell}}$,
		\end{center}
		where $p_1&lt;p_2&lt;\cdots&lt;p_k$, $q_1&lt;q_2&lt;\cdots&lt;q_{\ell}$ are primes and $m_i, n_j$ are positive integers for all $i,j$. By the well known formula for $\varphi$ and the fact that $F(m)=F(n)$, we obtain
		\begin{equation}\label{eq1}
			\prod_{i=1}^{k}p_i^{am_i+b(m_i-1)} \prod_{i=1}^{k}(p_i-1)^b = \prod_{i=1}^{\ell}q_i^{an_i+b(n_i-1)}\prod_{i=1}^{\ell}(q_i-1)^b. 
		\end{equation}
		For convenience, we write LHS and RHS to denote the left-hand side and the right-hand side of \eqref{eq1}, respectively. Suppose that $p_k\geq q_{\ell}$. Since the exponent of $p_k$ in LHS is at least $2m_i-1\geq 1$, we see that $p_k$ divides LHS. So $p_k$ divides RHS too. Since $p_k$ does not divide $q_i-1$ and $q_j$ for any $i=1,2,\ldots,\ell$ and $j=1,2,\ldots,\ell-1$, we see that $p_k$ divides $q_\ell$. So $p_k=q_{\ell}$. Similarly, if $p_k\leq q_{\ell}$, then we start with the fact that $q_{\ell}$ divides RHS, and so $q_{\ell}$ divides LHS too, which leads to $q_{\ell}=p_k$. In any case $p_k=q_{\ell}$. Furthermore, by the unique factorization, the exponent of $p_k$ and $q_{\ell}$ are the same. Therefore $am_k+b(m_k-1)=an_{\ell}+b(n_{\ell}-1)$, which implies $m_k=n_{\ell}$. Thus \eqref{eq1} reduces to 
		\begin{equation}\label{eq2}
			\prod_{i=1}^{k-1}p_i^{am_i+b(m_i-1)} \prod_{i=1}^{k-1}(p_i-1)^b = \prod_{i=1}^{\ell-1}q_i^{an_i+b(n_i-1)}\prod_{i=1}^{\ell-1}(q_i-1)^b. 
		\end{equation}
		We observe that \eqref{eq2} is obtained from \eqref{eq1} by the change of $k$ to $k-1$ and $\ell$ to $\ell-1$. So we can use the same argument to conclude that $p_{k-1}=q_{\ell-1}$ and $m_{k-1}=n_{\ell-1}$. Doing this process repeatedly, it will eventually stop. If $k&lt;\ell$, then it leads to the equation $1=R$ where $R$ is divisible by $q_1$, which is a contradiction. Similarly, the inequality $k&gt;\ell$ is not possible. Therefore $k=\ell$. So when the process stops, we obtain $k=\ell$, $p_i=q_i$, and $m_i=n_i$ for all $i$. Therefore $m=n$, as required.
	\end{proof}

	Since $\psi(n)$ and $\varphi(n)$ are similar, we expect that the function $n\mapsto n\psi(n)$ should also be injective. Nevertheless, there is a little problem with the primes $2$ and $3$. So we first prove the following lemma.
	
	\begin{lemma}\label{lemofPsi}
		For each $a, b \in\mathbb{N}$, let $F$ be the arithmetic function defined by $F(n)=n^a\psi(n)^b$ for all $n\in\mathbb{N}$. Let $m\in\mathbb{N}$ and $r, s\in\mathbb{N}\cup\{0\}$. If $F(m)=F(2^r3^s)$, then $m=2^r3^s$.
	\end{lemma}
	\begin{proof}
		If $r=s=0$, then $F(m)=F(1)=1$, which implies $m=1=2^r3^s$. So assume that $r\neq 0$ or $s\neq 0$. By the definition of $F$ and the formula for $\psi$, we observe that each $x,y\in\mathbb{N}$, we have
		\begin{equation}\label{lempsieq}
			F(2^x)=2^{xa+xb-b}3^b, F(3^y)=2^{2b}3^{ya+yb-b}, \text{ and } F(2^x3^y)=2^{xa+xb+b}3^{ya+yb}.
		\end{equation}
		Therefore if $p$ is a prime factor of $m$, then $p$ also divides $F(m)=F(2^r3^s)$, and so $p\leq 3$. Thus $m=2^u3^v$ for some nonnegative integers $u$ and $v$.

\bigskip
		
\noindent \textbf{Case 1:} $r=0$. Then $s\neq 0$ and $F(m)=F(2^r3^s)=F(3^s)=2^{2b}3^{sa+sb-b}$. Suppose, by way of contradiction, that $v=0$. Then $2^{ua+ub-b}3^b=2^{2b}3^{sa+sb-b}$, which implies that $ua+ub=3b$ and $sa+sb=2b$. Since $sa+sb=2b$, we have $s=1$ and $a=b$. Then $2ub=ua+ub=3b$ which is not possible. Thus $v\neq 0$. If $u\neq 0$, then we have that $2^{ua+ub+b}3^{va+vb}=2^{2b}3^{sa+sb-b}$, which implies $b=ua+ub&gt;b$, a contradiction. So $u=0$ and $2^{2b}3^{va+vb-b}=2^{2b}3^{sa+sb-b}$. This implies that $v=s$ and $m=3^s=2^r3^s$.

\bigskip
		
\noindent \textbf{Case 2:} $s=0$. Then $r\neq 0$. By using an argument similar to Case 1, one can show that $m=2^r=2^r3^s$.

\bigskip
		
\noindent \textbf{Case 3:} $r\neq 0$ and $s\neq 0$. Then $F(m)=2^{ra+rb+b}3^{sa+sb}$. By considering \eqref{lempsieq} and the exponents of $2$ and $3$ in $F(m)$, we see that $u\neq 0$ and $v\neq 0$. This implies that $2^{ua+ub+b}3^{va+vb}=2^{ra+rb+b}3^{sa+sb}$. Then $u=r$, $v=s$, and so $m=2^r3^s$. This completes the proof.
	\end{proof}
	
	\begin{theorem} \label{thmPsi}
		For each $a, b \in\mathbb{N}$, the arithmetic function $F$ defined by $F(n)=n^a\psi(n)^b$ for all $n\in\mathbb{N}$ is an injective function.
	\end{theorem}
	\begin{proof}
		To show that $F$ is injective, let $m,n\in\mathbb{N}$ and $F(m)=F(n)$. It is easy to see that $m=1$ if and only if $n=1$. So we assume that $m,n\geq2$. Let
		\begin{center}
			$m=p_1^{m_1}p_2^{m_2}\cdots p_k^{m_k}$ and $n=q_1^{n_1}q_2^{n_2}\cdots q_\ell^{n_{\ell}}$,
		\end{center}
		where $p_1&lt;p_2&lt;\cdots&lt;p_k$, $q_1&lt;q_2&lt;\cdots&lt;q_{\ell}$ are primes and $m_i, n_j$ are positive integers for all $i,j$. Since $F(m)=F(n)$, we obtain
		\begin{equation}\label{Deq1}
			\prod_{i=1}^{k}p_i^{am_i+b(m_i-1)} \prod_{i=1}^{k}(p_i+1)^b = \prod_{i=1}^{\ell}q_i^{an_i+b(n_i-1)}\prod_{i=1}^{\ell}(q_i+1)^b. 
		\end{equation}
		For simplicity, we write LHS and RHS to denote the left-hand side and the right-hand side of \eqref{Deq1}, respectively. If $p_{k}\leq 3$, then Lemma \ref{lemofPsi} implies that $m=n$. So assume that $p_{k}&gt;3$. Suppose that $p_k\geq q_{\ell}$. Since the exponent of $p_k$ in LHS is at least $2m_k-1\geq 1$, we see that $p_k$ divides LHS. So $p_k$ divides RHS too. Since $p_k&gt;3$, we see that $p_k$ does not divide $q_i+1$ and $q_j$ for any $i=1,2,\ldots,\ell$ and $j=1,2,\ldots,\ell-1$. Then $p_k$ divides $q_\ell$, and so $p_k=q_{\ell}$. Similarly, if $p_k\leq q_{\ell}$, then we start with the fact that $q_{\ell}$ divides RHS, and so $q_{\ell}$ divides LHS too, which leads to $q_{\ell}=p_k$. In any case $p_k=q_{\ell}$. Furthermore, by the unique factorization, the exponent of $p_k$ and $q_{\ell}$ are the same. Therefore $am_k+b(m_k-1)=an_{\ell}+b(n_{\ell}-1)$, which implies $m_k=n_{\ell}$. Thus \eqref{Deq1} reduces to 
		\begin{equation}\label{Deq2}
			\prod_{i=1}^{k-1}p_i^{am_i+b(m_i-1)} \prod_{i=1}^{k-1}(p_i+1)^b = \prod_{i=1}^{\ell-1}q_i^{an_i+b(n_i-1)}\prod_{i=1}^{\ell-1}(q_i+1)^b. 
		\end{equation}
		
		We observe that \eqref{Deq2} is obtained from \eqref{Deq1} by the change of $k$ to $k-1$ and $\ell$ to $\ell-1$. If $p_{k-1}\leq 3$, then we apply Lemma \ref{lemofPsi} to obtain $m=n$. If $p_{k-1}&gt;3$, then we repeat the above process and reduce \eqref{Deq2} by the change of $k-1$ to $k-2$ and $\ell-1$ to $\ell-2$. By repeating this process, we eventually obtain $m=n$. This completes the proof.
	\end{proof}
	Similar result also holds when the function $\varphi$ is replaced by $J_{2}$ as shown below.
	
	\begin{lemma}\label{lemofJ2}
		For each $a, b \in\mathbb{N}$, let $F$ be the arithmetic function defined by $F(n)=n^aJ_2(n)^b$ for all $n\in\mathbb{N}$. Let $m\in\mathbb{N}$ and $r, s\in\mathbb{N}\cup\{0\}$. If $F(m)=F(2^r3^s)$, then $m=2^r3^s$.
	\end{lemma}
	\begin{proof}
		Since this lemma can be proved in the same way as Lemma \ref{lemofPsi}, we skip some details. If $r=s=0$, then $m=2^r3^s$. So assume that $r\neq 0$ or $s\neq 0$. If $p$ is a prime factor of $m$, then $p$ also divides $F(m)=F(2^r3^s)$, and so $p\leq 3$. Thus $m=2^u3^v$ for some nonnegative integers $u$ and $v$.
		
		First, assume that $r=0$. Then $s\neq 0$ and $F(m)=2^{3b}3^{sa+2sb-2b}$. If $v=0$, then $2^{ua+2ub-2b}3^b=F(m)=2^{3b}3^{sa+2sb-2b}$, which implies that $ua+2ub=5b$ and $sa+2sb=3b$. Since $sa+2sb=3b$, we have that $s=1$ and $a=b$. Then $3ub=ua+2ub=5b$, a contradiction. Thus $v\neq 0$. From this point, we can still consider the exponents of $2$ and $3$ like the proof of Lemma \ref{lemofPsi} to obtain $m=n$. For the cases $s=0$ or ($r\neq 0$ and $s\neq 0$), we can also compare the exponents of $2$ and $3$ to obtain the desired result. So the proof is completed.
	\end{proof}
	
	Next, we show that for certain $a,b\in\mathbb{N}$, the function $n\mapsto n^aJ_s(n)^b$ is injective. When $s=2$, we can use any positive integers $a,b$ as follows.
	
	\begin{theorem} \label{thmJ2}
		For each $a, b \in\mathbb{N}$, the arithmetic function $F$ defined by $F(n)=n^aJ_{2}(n)^b$ for all $n\in\mathbb{N}$ is injective.
	\end{theorem}
	\begin{proof}
		Since the proof of this theorem follows the same argument as in Theorem \ref{thmPsi}, we skip some details. Let $m,n\in\mathbb{N}$ and $F(m)=F(n)$. It is easy to see that $m=1$ if and only if $n=1$. So we assume that $m,n\geq2$. Let
		\begin{center}
			$m=p_1^{m_1}p_2^{m_2}\cdots p_k^{m_k}$ and $n=q_1^{n_1}q_2^{n_2}\cdots q_\ell^{n_{\ell}}$,
		\end{center}
		where $p_1&lt;p_2&lt;\cdots&lt;p_k$, $q_1&lt;q_2&lt;\cdots&lt;q_{\ell}$ are primes and $m_i, n_j$ are positive integers for all $i,j$. Then
		\begin{equation}\label{Jeq1}
			\prod_{i=1}^{k}p_i^{am_i+2b(m_i-1)} \prod_{i=1}^{k}(p_i^2-1)^b = \prod_{i=1}^{\ell}q_i^{an_i+2b(n_i-1)}\prod_{i=1}^{\ell}(q_i^2-1)^b. 
		\end{equation}
		For convenience, we write LHS and RHS to denote the left-hand side and the right-hand side of \eqref{Jeq1}, respectively. If $p_{k}\leq 3$, then Lemma \ref{lemofJ2} implies that $m=n$. So assume that $p_{k}&gt;3$. Suppose that $p_k\geq q_{\ell}$. Since the exponent of $p_k$ in LHS is at least $3m_i-2\geq 1$, we see that $p_k$ divides LHS. So $p_k$ divides RHS too. Since $p_k$ does not divide $q_i-1$, $q_i+1$, and $q_j$ for any $i=1,2,\ldots,\ell$ and $j=1,2,\ldots,\ell-1$, we see that $p_k$ divides $q_\ell$. So $p_k=q_{\ell}$. Similarly, if $p_k\leq q_{\ell}$, then this leads to $p_k=q_{\ell}$ and  $m_k=n_{\ell}$. Thus \eqref{Jeq1} reduces to 
		\begin{equation}\label{Jeq2}
			\prod_{i=1}^{k-1}p_i^{am_i+2b(m_i-1)} \prod_{i=1}^{k-1}(p_i^2-1)^b = \prod_{i=1}^{\ell-1}q_i^{an_i+2b(n_i-1)}\prod_{i=1}^{\ell-1}(q_i^2-1)^b. 
		\end{equation}
		
		We observe that \eqref{Jeq2} is obtained from \eqref{Jeq1} by the change of $k$ to $k-1$ and $\ell$ to $\ell-1$. So we can repeat this process like the proof of Theorem \ref{thmPsi} to obtain  $m=n$, as required.
	\end{proof}
	
	When $s\geq 3$, it seems that the function $n\mapsto n^aJ_s(n)^b$ is injective for any $a,b\in\mathbb{N}$, but we do not have a proof. In the following theorem, we need to restrict ourselves to the case $a\geq sb$, but we hope to solve the case $a&lt;sb$ in the future. Please see also our comments and the list of other problems in Section \ref{SecQ}.
	
	\begin{theorem}
		For each $a,b,s\in\mathbb{N}$, if $a\geq sb$, then the arithmetic function $F$ defined by $F(n)=n^aJ_s(n)^b$ for all $n\in\mathbb{N}$ is injective.
	\end{theorem}
	\begin{proof}
		For each $n\in\mathbb{N}$, let $P_n$ be the set of all prime factors of $n$. Let $a,b,s\in\mathbb{N}$ and $a\geq sb$. We remark that we do not need to use the inequality $a\geq sb$ until the calculation in \eqref{eq1000}. To show that $F$ is injective, let $m,n\in\mathbb{N}$ and $F(m)=F(n)$. It is easy to see that $m=1$ if and only if $n=1$. So we assume that $m,n\geq2$. We will first show that $P_m=P_n$. So suppose, by way of contradiction, that $P_m\neq P_n$.

\bigskip
\noindent		
\textbf{Case 1:} $P_m\subseteq P_n$ and  $P_n\nsubseteq P_m$. Let
		\begin{center}
			$m=p_1^{m_1}p_2^{m_2}\cdots p_k^{m_k}$ and $n=p_1^{n_1}p_2^{n_2}\cdots p_k^{n_k}q_1^{n_{k+1}}q_2^{n_{k+2}}\cdots q_\ell^{n_{k+\ell}}$,
		\end{center}
		where $p_1&lt;p_2&lt;\cdots&lt;p_k$ and $q_1&lt;q_2&lt;\cdots&lt;q_{\ell}$ are primes, $k,\ell\geq 1$, $p_i\neq q_j$, and $m_i, n_j$ are positive integers for all $i,j$. After dividing both sides of the equation $F(m)=F(n)$ by $\prod\limits_{i=1}^{k}(p_i^s-1)^b$, we obtain
		\begin{equation}\label{Geq12}
			\prod_{i=1}^{k}p_i^{am_i+sb(m_i-1)} =\prod_{i=1}^{k}p_i^{an_i+sb(n_i-1)}\prod_{i=1}^{\ell}q_i^{an_{k+i}+sb(n_{k+i}-1)}\prod_{i=1}^{\ell}(q_i^s-1)^b. 
		\end{equation}
		Then $q_1$ divides the right-hand side of \eqref{Geq12} but does not divide the left-hand side. So this case leads to a contradiction.


\bigskip
\noindent \textbf{Case 2:} $P_n\subseteq P_m$ and  $P_m\nsubseteq P_n$. Similar to Case 1, this leads to a contradiction. 

\bigskip\noindent
\textbf{Case 3:} $P_m\nsubseteq P_n$ and $P_n\nsubseteq P_m$. Let
		\begin{center}
			$m=p_1^{m_1}p_2^{m_2}\cdots p_k^{m_k}\left(\prod\limits_{i=1}^{t}w_i^{u_i}\right)$ and $n=q_1^{n_1}q_2^{n_2}\cdots q_\ell^{n_{\ell}}\left(\prod\limits_{i=1}^{t}w_i^{v_i}\right)$,
		\end{center}
		where $p_1&lt;p_2&lt;\cdots&lt;p_k$, $q_1&lt;q_2&lt;\cdots&lt;q_{\ell}$, and $w_1&lt;w_2&lt;\cdots&lt;w_t$ are primes, $p_i, q_j, w_x$ are distinct for all $i,j,x$, and $m_i, n_j,u_x,v_y$ are positive integers for all $i,j,x,y$. In addition, if $P_m\cap P_n=\emptyset$, then we take $t=0$ and define the empty product to be $1$ as usual; if $P_m\cap P_n\neq\emptyset$, then $t\geq 1$. By the fact that $F(m)=F(n)$, we obtain
		\begin{equation}\label{Geq1}
			\prod_{i=1}^{t}w_i^{au_i+sb(u_i-1)}\prod_{i=1}^{k}p_i^{am_i+sb(m_i-1)} \prod_{i=1}^{k}(p_i^s-1)^b =\prod_{i=1}^{t}w_i^{av_i+sb(v_i-1)} \prod_{i=1}^{\ell}q_i^{an_i+sb(n_i-1)}\prod_{i=1}^{\ell}(q_i^s-1)^b. 
		\end{equation}
		
	Let\ \  $L_1=\prod_{i=1}^{k}p_i^{am_i+sb(m_i-1)}$,\ \ \  $L_2=\prod_{i=1}^{k}(p_i^s-1)^b$,\ \ \  $R_1=\prod_{i=1}^{\ell}q_i^{an_i+sb(n_i-1)}$,\ \ \ and $R_2=\prod_{i=1}^{\ell}(q_i^s-1)^b$.
	 Since $p_i$, $q_j$, and $w_x$ are all distinct, \eqref{Geq1} implies that $L_1\vert R_2$ and $R_1\vert L_2$. Recall that $a\geq sb$. Then
	 \begin{equation}\label{eq1000}
	 	L_1\leq R_2&lt;\prod_{i=1}^{\ell}q_i^{sb}\leq R_1\leq L_2&lt;\prod_{i=1}^{k}p_i^{sb}\leq L_1,
	 \end{equation}
	which is a contradiction.
		
		Therefore we can conclude that $P_m=P_n$. Let 
		\begin{center}
			$m=p_1^{m_1}p_2^{m_2}\cdots p_k^{m_k}$ and $n=p_1^{n_1}p_2^{n_2}\cdots p_k^{n_{k}}$,
		\end{center}
		where $p_1&lt;p_2&lt;\cdots&lt;p_k$ are primes and $m_i, n_j$ are positive integers for all $i,j$. By the fact that $F(m)=F(n)$, we obtain
		\[
		\prod_{i=1}^{k}p_i^{am_i+sb(m_i-1)} =\prod_{i=1}^{k}p_i^{an_i+sb(n_i-1)}. 
		\]
		By the unique factorization, we obtain $m_i=n_i$ for all $i$, and so $m=n$, as required.
	\end{proof}
	
	\section{Noninjectivity and a connection with other concepts}
	\label{sec3}
	
	Not every arithmetic function has the property like $\varphi$, $\psi$, and $J_s$. In this section, we give various examples to show that if we replace $\varphi$ by other arithmetic functions $f$, the function $n\mapsto nf(n)$ may not be injective. We also give some connections  to other problems such as the existence or nonexistence of infinitely many Mersenne primes, amicable pairs, and primitive solutions to certain equations.
	
	\begin{example}\label{ExA}
		Let $A(n)=n\sigma(n)$ for all $n\in\mathbb{N}$. A straightforward calculation shows that $A(12)=A(14)=2^4\cdot 3\cdot 7$, and so $A$ is not injective. In fact, we can generate infinitely many $x,y\in\mathbb{N}$ such that $A(x)=A(y)$ using the equality $A(12)=A(14)$. Let $x=12d$ and $y=14d$ where $(d,42)=1$. It is easy to see that $A$ is multiplicative and  $(d,14)=(d,12)=1$, and so $A(x)=A(12d)=A(12)A(d)=A(14)A(d)=A(14d)=A(y)$. Since there are infinitely many $d\in\mathbb{N}$ with $(d,42)=1$, we obtain infinitely many $x,y\in\mathbb{N}$ such that $A(x)=A(y)$ too. From this, it is easy to see that if we can find another pair of integers $x_0,y_0$ such that $A(x_0)=A(y_0)$, then we can generate infinitely many such pairs by following the above method.
		
		Moser \cite[p.~102]{Guy} asked whether there is an infinite number of primitive solutions to the equation $A(x)=A(y)$, that is, the integers $x,y\geq 1$ such that $A(x)=A(y)$ and $A(x/d)\neq A(y/d)$  for any common divisor $d&gt;1$ of $x$ and $y$. A conditional answer is known: if $2^p-1$ and $2^q-1$ are distinct Mersenne primes, then $x=2^{p-1}(2^q-1)$ and $y=2^{q-1}(2^p-1)$ is a primitive solution to $A(x)=A(y)$. To see this, recall that if $2^n-1$ is a prime, then $n$ is a prime too. Therefore we have $p,q,2^p-1,2^q-1$ are primes. Without loss of generality, assume that $p&gt;q$. Then
		\begin{align*}
			A(x)&amp;=2^{p-1}(2^q-1)\sigma(2^{p-1})\sigma(2^q-1)\\
			&amp;=2^{p-1}(2^q-1)(2^p-1)(2^q)\\
			&amp;=2^{p+q-1}(2^p-1)(2^q-1).
		\end{align*}
		By a similar calculation, we see that $A(y)=A(x)$. Since $2^p-1$ and $2^q-1$ are distinct odd primes, the greatest common divisor of $x$ and $y$ is $2^{q-1}$. So if $d&gt;1$, $d\mid x$, and $d\mid y$, then $d=2^{\ell}$ for some $\ell=1,2,\ldots,q-1$. By a similar calculation, we obtain
		\begin{align*}
			A\left(\frac{x}{d}\right)=2^{p+q-\ell-1}(2^{p-\ell}-1)(2^q-1)\ \  \text{and \ }
			A\left(\frac{y}{d}\right)=2^{p+q-\ell-1}(2^{q-\ell}-1)(2^p-1).
		\end{align*}
		From this, we see that $2^p-1$ divides $A(\frac{y}{d})$ but does not divide $A(\frac{x}{d})$. So $A(\frac{x}{d})\neq A(\frac{y}{d})$.  This shows that $x,y$ is indeed a primitive solution to the equation $A(x)=A(y)$. Nevertheless, since we do not know whether or not there are infinitely many Mersenne primes, this is only a conditional solution to Moser's problem. Without restricting to primitive solutions, Erd\H{o}s \cite{Erd} showed that the number of $m,n\in\mathbb{N}$ satisfying $m&lt;n&lt;x$ and $m\sigma(m)=n\sigma(n)$ is asymptotic to $cx$ as $x\rightarrow\infty$, where $c$ is a positive constant.
		For more information on the equation $A(x)=A(y)$, we refer the reader to Guy's book \cite[Section B11]{Guy}. The sequence $(A(n))_{n\geq 1}$ is registered in the OEIS as the sequence \seqnum{A064987}. Moreover, the sequence of $n\in\mathbb{N}$ such that $A(x)=n$ has more than one solution is \seqnum{A337873} in the OEIS. Some such integers $n$ and distinct $x_1, x_2$ such that $A(x_1)=A(x_2)=n$ are shown in Table \ref{TableA}. We remark that $n$ and $x_2$ in our table are not listed in an increasing order, but the integer $x_1$ is listed in an increasing order. The reader can also find more related information in the sequence  \seqnum{A212490} and our comments in Section \ref{SecQ}.
	\end{example}

	\begin{example}\label{ExB}
		Let $s(n)=\sigma(n)-n$ be the sum of proper positive divisors of $n$ and let $B(n)=ns(n)$ for all $n\in\mathbb{N}$. It is not difficult to check that $s(6)=6$ and $s(9)=4$, and so $B(6)=B(9)$. So $B$ is not injective. In general, if $x,y\in\mathbb{N}$, $s(x)=y$, and $s(y)=x$, then we have $B(x)=B(y)$. For example, since $s(220)=284$ and $s(284)=220$, we have that $B(220)=220\cdot 284=B(284)$. A pair $(x,y)$ satisfying $s(x)=y$ and $s(y)=x$ is called an amicable pair, and mathematicians have found millions such pairs. It is not known whether there are infinitely many amicable pairs, but it is believed that there are. If this is true, then $B(x)=B(y)$ for an infinite number of $x,y$. For more information on amicable pairs and related concept, we refer the reader to Guy's book \cite[Sections B4--B8]{Guy} and the sequences \seqnum{A063990}, \seqnum{A002025}, and \seqnum{A002046} in the OEIS \cite{Slo}. We remark that \seqnum{A063990} gives the list of amicable numbers in an increasing order, but the adjacent numbers are not necessarily the amicable pairs $(x,y)$. The sequences \seqnum{A002025} and \seqnum{A002046} give the list of $x$ and $y$ in the amicable pairs, respectively. The sequence that gives amicable pairs in an increasing order is \seqnum{A259180} in the OEIS. Moreover, the sequence of $n\in\mathbb{N}$ such that $B(x)=n$ has more than one solution is also registered in the OEIS as the sequence \seqnum{A212327}. Some values of such $n$ and distinct $x_1,x_2\in\mathbb{N}$ such that $B(x_1)=B(x_2)=n$ are shown in Table \ref{TableB}.
	\end{example}

\begin{example}\label{ExD}
	Let $D(n)=nd(n)$ for all $n\in\mathbb{N}$, where $d(n)$ is the number of positive divisors of $n$. Let $x=18a$ and $y=27a$ where $a\in\mathbb{N}$ and $(a,6)=1$. Since $D(18)=108=D(27)$ and $D$ is multiplicative, we obtain $D(x)=D(18)D(a)=D(27)D(a)=D(y)$. So it may be more interesting to consider only the primitive solutions to the equation $D(x)=D(y)$ where the primitive solutions are defined in a similar way as in Example \ref{ExA}. We leave this problem to the interested reader. The sequence $(D(n))_{n\geq 1}$ is \seqnum{A038040} in the OEIS. The sequence of $n\in\mathbb{N}$ such that $D(x)=n$ has more than one solution is the sequence \seqnum{A338382} in the OEIS. Some values of such integers $n$, and $x_1,x_2$ such that $D(x_1)=D(x_2)=n$ are shown in Table \ref{TableD}. 
\end{example}

\begin{example}\label{ExW}
	Let $W_1(n)=n\omega(n)$ and $W_2(n)=n\Omega(n)$ for all $n\in\mathbb{N}$. Let $x=30\cdot 5^k$ and $y=45\cdot5^k$ where $k\in\mathbb{N}$. Then $W_1(x)=30\cdot5^k\cdot3=45\cdot5^k\cdot2=W_1(y)$. So $W_1$ is not injective. In general, if $W_1(m)=W_1(n)$ and $(m,n)&gt;1$, then we can generate infinitely many $x,y\in\mathbb{N}$ such that $W_1(x)=W_1(y)$, namely, $x=m\cdot p^k$ and $y=n\cdot p^k$ where $k$ is any positive integer and $p$ is any prime divisor of $(m,n)$. For $W_2$, we observe that if $p$ is any odd prime, then $W_2(16p)=(16p)(5)=(20p)(4)=W_2(20p)$. So $W_2$ is not injective and there are infinitely many $m,n\in\mathbb{N}$ such that $W_2(m)=W_2(n)$. Some values of $n\in\mathbb{N}$ such that $W_1(x)=n$ or $W_2(y)=n$ have more than one solution are shown in Table \ref{TableW1} and Table \ref{TableW2}, respectively.
\end{example}

\begin{example}\label{ExH}
	For each positive integer $b\geq 2$, let $S_{b}(n)$ be the sum of digits of $n$ when $n$ is written in base $b$, and let $H_b(n)=nS_{b}(n)$ for all $n\in\mathbb{N}$. If $b=2$, then it is easy to see that $H_2(22)=66=H_2(33)$. For $b&gt;2$, we have
	\[H_b(b^3+1)=2(b^3+1)=(b^2+(b-2)b+2)(b+1)=H_b(b^2+(b-2)b+2)=H_b(2(b^2-b+1)),\]
	and $b^3+1\neq 2(b^2-b+1)$. So $H_b$ is not injective for any $b\geq2$. In general, if $H_b(m)=H_b(n)$, then there are infinitely many $x,y\in\mathbb{N}$ satisfying the equation $H_b(x)=H_b(y)$, namely, $x=b^tm$ and $y=b^tn$ where $t$ is an arbitrary positive integer. Some values of $n\in\mathbb{N}$ such that $H_{10}(x)=n$ has more than one solution are shown in Table \ref{TableH}.
\end{example}

	\section{Restricted injectivity}
	\label{sec4}

	Since the functions defined in Example \ref{ExA} to Example \ref{ExH} are not injective on $\mathbb{N}$, it is natural to consider the injectivity of these functions on other infinite proper subsets of $\mathbb{N}$. In 1959, Erd\H{o}s \cite{Erd} observed that although the function $n\mapsto n\sigma(n)$ is not injective on $\mathbb{N}$, it is injective on the set of squarefree integers. In fact, Erd\H{o}s' observation is a special case of the next theorem.
	
	\begin{theorem}\label{thmAsqf}
		For each $a,b\in\mathbb{N}$, let $F$ be defined by $F(n)=n^a\sigma(n)^b$ for all $n\in\mathbb{N}$. Then $F$ is injective on squarefree integers. That is, if $m,n\in\mathbb{N}$ are squarefree and $m^a\sigma(m)^b=n^a\sigma(n)^b$, then $m=n$. In particular, the function $A$ in Example \ref{ExA} is injective on squarefree integers.
	\end{theorem}
	\begin{proof}
		If $m$ and $n$ are squarefree, then $m^a\sigma(m)^b=m^a\psi(m)^b$ and $n^a\sigma(n)^b=n^a\psi(n)^b$. So the assumption that $m^a\sigma(m)^b=n^a\sigma(n)^b$ implies $m^a\psi(m)^b=n^a\psi(n)^b$, and so $m=n$ by Theorem \ref{thmPsi}.
	\end{proof}
	
	We remark that the integer $a$ in Theorem \ref{thmAsqf} cannot be zero since $\sigma$ is not injective on squarefree integers. For instance, we have $\sigma(6)=\sigma(11)$ and 6, 11 are squarefree. However, if we replace $\sigma$ by $d$ in Theorem \ref{thmAsqf}, the resulting function is also injective on squarefree integers.
	
	\begin{theorem}\label{thmDsqf}
		For each $a,b\in\mathbb{N}$, let $F$ be defined by $F(n)=n^ad(n)^b$ for all $n\in\mathbb{N}$. Then $F$ is injective on squarefree integers. In particular, the function $D$, defined in Example \ref{ExD}, is injective on the set of squarefree integers. 
	\end{theorem}
	\begin{proof}
		Let $m,n\in\mathbb{N}$ be squarefree and $F(m)=F(n)$. It is easy to see that $m=1$ if and only if $n=1$, so we assume that $m,n&gt;1$. Let $m=\prod\limits_{i=1}^{k}p_i$ and $n=\prod\limits_{i=1}^{\ell}q_i$, where $p_1,p_2,\ldots,p_k$ and $q_1,q_2,\ldots,q_\ell$ are distinct primes. Then we have 
		\begin{equation}\label{Eq9}
			2^{kb}\prod\limits_{i=1}^{k}p_i^a=2^{\ell b}\prod\limits_{i=1}^{\ell}q_i^a.
		\end{equation}
		We denote the left-hand side and the right-hand side of (\ref{Eq9}) by LHS and RHS, respectively. If $k\geq \ell+1$, then after dividing both sides of (\ref{Eq9}) by $2^{\ell b}$, LHS has at least $k$ distinct prime factors while RHS has $\ell\leq k-1$ distinct prime factors, a contradiction. Similarly, the inequality $\ell&gt; k$  leads to a contradiction. So $k=\ell$, and (\ref{Eq9}) reduces to $m=n$, as required.
	\end{proof}
	
	Not every arithmetic function has the property like $\sigma$ and $d$ in Theorems \ref{thmAsqf} and \ref{thmDsqf}. This is shown in the following examples.
	
	\begin{example}\label{Ex14}
		Let $B$, $W_1$, $W_2$, and $H_b$ be the functions defined in Example \ref{ExB}, \ref{ExW}, and \ref{ExH}. We show that these functions are not injective on the set of squarefree integers. For the function $B$, we have $1955$ and $2093$ are squarefree, but
		\begin{center}
			$B(1955)=1955s(1955)=1955\cdot 637=2093\cdot 595=2093s(2093)=B(2093)$.
		\end{center}  
		
		For $W_1$ and $W_2$, let $p_1,p_2,\ldots,p_9$ be distinct primes and $(2,p_i)=(5,p_i)=(11,p_i)=1$ for all $1\leq i\leq 9$. Let 
		\begin{center}
			$a=11\prod\limits_{i=1}^{9}p_i$ \ \  and \ \  $b=10\prod\limits_{i=1}^{9}p_i$.
		\end{center}
		Then $W_1(a)=W_1(b)=W_2(a)=W_2(b)$. Therefore  $W_1$ and $W_2$ are not injective on the set of squarefree integers.
		
		It is shown in Example \ref{ExH} that $H_2$ is not injective on the set of squarefree integers. Moreover, we have
		\begin{center}
		$H_{3}(51)=255=H_{3}(85)$,\quad $H_{4}(26)=130=H_{4}(65)$,\quad $H_{5}(21)=105=H_{5}(35)$,
		
		$H_{6}(26)=156=H_{6}(39)$,\quad
		$H_{7}(55)=385=H_{7}(77)$,\quad $H_{8}(26)=130=H_{8}(65)$, 
		
		$H_{9}(15)=105=H_{9}(21)$,\quad and \quad $H_{10}(15)=90=H_{10}(30)$.
	\end{center}
	
		
		\noindent So $H_b$ is not injective on squarefree integers for  $2\leq b\leq 10$. One can use a computer to verify that $H_b$ is not injective for other values of $b$ too. We believe that $H_b$ is not injective for any $b\geq 11$ but we do not have a proof.
	\end{example}
	
	
	Some products of a polynomial and two arithmetic functions are also injective on the set of squarefree integers. This can be proved by applying our theorems as follows.
	
	\begin{corollary}
		For each $a,b\in\mathbb{N}$, the functions $F_1$ and $F_2$ defined by 
		\begin{center}
			$F_1(n)=n^a\sigma(n)^b\varphi(n)^b$ and $F_2(n)=n^a\sigma(n)^b\psi(n)^b$ for all $n\in\mathbb{N}$
		\end{center}
		 are injective on the set of squarefree integers.
	\end{corollary}
	\begin{proof}
		Let $n\in\mathbb{N}$ be squarefree. We observe that $\sigma(n)=\psi(n)$ and it is easy to see that $\psi(n)\varphi(n)=J_2(n)$. So $F_1(n)=n^aJ_2(n)^b$ and $F_2(n)=n^a\psi(n)^{2b}$. Therefore $F_1$ and $F_2$ are injective on the set of squarefree integers by Theorems \ref{thmJ2} and \ref{thmPsi}, respectively.
	\end{proof}
	
	We remark that $F_1$ and $F_2$ are not injective on $\mathbb{N}$. For example, when $a=b=1$, we have $F_1(56)=F_1(60)$ and $F_2(12)=F_2(14)$.
	
	\section{Other results on  noninjectivity}\label{Sect5}
	\label{sec5}
	In this section, we study the product of a different simple polynomial and $\varphi$. So for each nonnegative integer $c$, let $f_c$ be the arithmetic function given by
	\begin{center}
		$f_c(n)=(n+c)\varphi(n)$ for all $n\in\mathbb{N}$.
	\end{center}
	Although $f_0$ is injective, we believe that $f_c$ is not injective for any $c\geq1$, and we will provide some supporting evidence. If $c$ is fixed and is given explicitly, we can always use a computer to search for distinct positive integers $a,b$ such that $f_c(a)=f_c(b)$. For each $c=1,2,3$, Tables \ref{Tablef1}, \ref{Tablef2} and \ref{Tablef3} show distinct positive integers $x_1,x_2\leq2000$ such that $f_c(x_1)=f_c(x_2)$. Therefore we immediately see that $f_1,f_2,f_3$ are not injective. However, this only gives us a small number of $c$ for which $f_c$ is not injective. 
	
	In what follows, we will develop a tool and use it together with Table \ref{Tablef1} to generate an infinite number of $c$ such that $f_c$ is not injective, and then use the results that we obtain to show that $f_c$ is not injective for any positive integer $c\leq1000$, and that there are at least 98 percent of $c\leq N$ such that $f_c$ is not injective when $N$ is any large positive integer. To do so, we define for each $c,n\in\mathbb{N}$, the product
	$$\alpha(c,n)=\prod_{p\mid c \text{ and }p\nmid n} \left(1-\frac{1}{p}\right)$$ 
	where the product is taken over all primes $p$ that are a factor of $c$ and do not divide $n$. As usual, the empty product is defined to be $1$. So if $c=1$ or every prime divisor of $c$ is a divisor of $n$, then $\alpha(c,n)=1$.
	
	The following lemmas are simple but they are the key to the construction of an infinite number of $c\in\mathbb{N}$ such that $f_c$ is not injective. 
	
	\begin{lemma} \label{lemmaA1}
		Let $c$ and $n$ be positive integers. Then the following statements hold. 
		\begin{itemize}
			\item [\rm{(i)}]  $\varphi(cn)=c\varphi(n)\alpha(c,n)$.
			\item[\rm{(ii)}] $\varphi(cn)=c\varphi(n)$ if and only if $c=1$ or every prime divisor of $c$ is a divisor of $n$.
		\end{itemize}
	\end{lemma}
	\begin{proof}
		For (i), we apply the well known formula to obtain
		$$\frac{\varphi(cn)}{\varphi(n)}=\frac{cn\prod_{p\mid cn}(1-\frac{1}{p})}{n\prod_{p\mid n}(1-\frac{1}{p})}=c\alpha(c,n),$$
		which implies (i). Then (ii) follows immediately from (i).
	\end{proof}

	\begin{lemma} \label{lemA2}
		The value of the finite product of the form $\prod\limits_{p} (1-\frac{1}{p})^{a_p}$ uniquely determines the set of primes in the product. More precisely, if $p_1&lt;p_2&lt;\cdots&lt;p_m$ and $q_1&lt;q_2&lt;\cdots&lt;q_k$ are primes, $a_1,a_2,\ldots,a_m,b_1,b_2,\ldots,b_k$ are positive integers, and
		\begin{equation}\label{lemA2eq}
			\prod_{i=1}^{m}\left(1-\frac{1}{p_i}\right)^{a_i}=\prod_{i=1}^k \left(1-\frac{1}{q_i}\right)^{b_i},
		\end{equation}
		then $m=k$, $p_i=q_i$, and $a_i=b_i$ for every $i=1,2,\ldots,m$.
	\end{lemma}
	\begin{proof}
		The idea of proof is the same as that of Theorem \ref{thm1}. The equality \eqref{lemA2eq} leads to
		\begin{equation}\label{lem_1}
			\prod_{i=1}^{k}q_i^{b_i} \prod_{i=1}^{m}(p_i-1)^{a_i}=\prod_{i=1}^{m}p_i^{a_i} \prod_{i=1}^{k} (q_i-1)^{b_i}.
		\end{equation}
		Let LHS and RHS denote the left-hand side and the right-hand side of \eqref{lem_1}. If $q_k\geq p_m$, then we start with LHS, which is divisible by $q_k$, and so $q_k\mid \text{RHS}$, which implies $q_k=p_m$. Similarly, if $p_m\geq q_k$, then we start with $p_m\mid \text{RHS}$, which eventually leads to $p_m=q_k$. By the unique factorization, the exponents of $p_m$ and $q_k$ are also equal, that is, $a_m=b_k$.
		Therefore \eqref{lem_1} reduces to an equation that is similar to \eqref{lem_1} but $k$ becomes $k-1$ and $m$ becomes $m-1$. So we can repeat this process like the proof of Theorem \ref{thm1} to obtain the desired result.
	\end{proof}

	\begin{lemma}
		Let $a,b,c$ be positive integers and $\varphi(a)=\varphi(b)$. Then the following statements are equivalent.
		\begin{itemize}
			\item [\rm{(i)}] $\varphi (ca)=\varphi(cb)$.
			\item [\rm{(ii)}] $\alpha(c,a)=\alpha(c,b)$.
			\item [\rm{(iii)}] $\{p\in\mathbb{N} : p$ is prime, $p\mid c$, and $p\nmid a\}$=$\{p\in\mathbb{N} : p$ is prime, $p\mid c$, and $p\nmid b\}$.
			\item[\rm{(iv)}] $\{p\in\mathbb{N} : p$ is prime, $p\mid c$, and $p\mid a\}$=$\{p\in\mathbb{N} : p$ is prime, $p\mid c$, and $p\mid b\}$.
		\end{itemize}
	\end{lemma}
	\begin{proof}
		By Lemma \ref{lemmaA1}, we see that (i) and (ii) are equivalent. Lemma \ref{lemA2} implies that (ii) and (iii) are equivalent. Clearly, the sets in (iv) are the complement of the corresponding sets in (iii) with respect to the set of prime divisors of $c$. So (iii) and (iv) are equivalent. This completes the proof.
	\end{proof}




	\begin{lemma}\label{lemmaA4}
		Let $a,b,c,d$ be positive integers. Then the following statements hold.
		\begin{itemize}
			\item [\rm{(i)}] If $d\mid c$ and $f_\frac{c}{d}(a)=f_\frac{c}{d}(b)$, then $f_c(da)=f_c(db)$ if and only if $\alpha(d,a)=\alpha(d,b)$.
			\item [\rm{(ii)}] If $f_1(a)=f_1(b)$, then $f_c(ca)=f_c(cb)$ if and only if $\alpha(c,a)=\alpha(c,b)$.
			\item [\rm{(iii)}] If $f_1(a)=f_1(b)$ and $(c,ab)=1$, then $f_c(ca)=f_c(cb)$.
		\end{itemize}
	\end{lemma}
	\begin{proof}
		For (i), suppose that $d\mid c$ and $f_\frac{c}{d}(a)=f_\frac{c}{d}(b)$.
		By Lemma \ref{lemmaA1}, we obtain $$f_c(da)=(da+c)\varphi(da)=d^2\left(a+\frac{c}{d}\right)\varphi(a)\alpha(d,a)=d^2f_{\frac{c}{d}}(a)\alpha(d,a).$$ Similarly, we have $f_c(db)=d^2f_{\frac{c}{d}}(b)\alpha(d,b)$. From this, we immediately obtain (i). Then (ii) follows from (i) by the substitution $d=c$. In addition, if $(c,ab)=1$, then $\alpha(c,a)=\alpha(c,b)$, and so (iii) follows from (ii). This completes the proof.
	\end{proof}

	
	We are now ready to show that there are infinitely many $c\in\mathbb{N}$ such that $f_c$ is not injective.
	
	\begin{theorem} \label{theorem3}
		Let $c$ be a positive integer. Then the following statements hold.
		\begin{itemize}
			\item[\rm{(i)}] If $(c,130)=1$, then $f_c$ is not injective.
			\item[\rm{(ii)}] If the set of prime divisors of $c$ is a subset of $\{2,5,13\}$, then $f_c$ is not injective.
			\item[\rm{(iii)}] If $c=p^k$ where $p$ is a prime and $k$ is a positive integer, then $f_c$ is not injective.
		\end{itemize}
	\end{theorem}
	\begin{proof}
		For (i), let $(c,130)=1$. Let $a=13$ and $b=20$. From Table 1, we know that $f_1(a)=f_1(b)=168$. Since $(c,ab)=1$, we obtain by Lemma \ref{lemmaA4} that $f_c(ca)=f_c(cb)$. So $f_c$ is not injective, and so (i) is proved.
		
		For (ii), let $c=2^{c_1}5^{c_2}13^{c_3}$ where $c_1,c_2,c_3$ are nonnegative integers. Let $a=649$ and $b=753$. We know from Table 1 that $f_1(a)=f_1(b)=377000$ and $(c,ab)=1$. By Lemma \ref{lemmaA4}, we obtain $f_c (ca)=f_c (cb)$, and so $f_c$ is not injective.
		
		For (iii), let $c=p^k$ where $p$ is a prime and $k$ is a positive integer. If $p\notin\{2,5,13\}$, then the result follows from (i). If $p\in\{2,5,13\}$, then the result can be obtained from (ii). So the proof is complete.
	\end{proof}
	
	By a similar method, we can generate more $c\in\mathbb{N}$ such that $f_c$ is not injective. We give one more similar theorem and then use it to show that $f_c$ is not injective for any positive integers $c\leq1000$. We remark that the integers $2, 157, 443, 17, 47, \ldots, 331$ appearing in the statement of the next theorem are prime numbers.
	
	\begin{theorem}\label{theorem4}
		Let $c$ be a positive integer. Then the following statements hold.
		\begin{itemize}
			\item [\rm{(i)}] If $(c,2)=(c,157)=(c,443)=1$, then $f_c$ is not injective.
			\item [\rm{(ii)}] If $(c,2)=(c,17)=(c,47)=1$, then $f_c$ is not injective.
			\item [\rm{(iii)}] If $(c,13)=(c,71)=(c,881)=1$, then $f_c$ is not injective.
			\item [\rm{(iv)}] If $(c,2)=(c,23)=1$, then $f_c$ is not injective.
			\item [\rm{(v)}] If $(c,5)=(c,61)=(c,271)=1$, then $f_c$ is not injective.
			\item [\rm{(vi)}] If $(c,3)=(c,13)=(c,31)=1$, then $f_c$ is not injective.
			\item [\rm{(vii)}] If $(c,5)=(c,37)=(c,41)=(c,331)=1$, then $f_c$ is not injective.
		\end{itemize}
	\end{theorem}
	\begin{proof}
		The proof of this theorem is similar to that of Theorem \ref{theorem3}. We only need to choose an appropriate choice of $a,b\in\mathbb{N}$. For (i), we choose $a=443$ and $b=628$ to obtain from Table \ref{Tablef1} and Lemma \ref{lemmaA4} that $f_1(a)=f_1(b)=196248$, $(c,ab)=1$, and $f_c(ca)=f_c(cb)$. In the same way, for (ii), (iii), (iv), (v), (vi), and (vii), we choose $(a,b)=(47,68),(881,923),(23,32),(271,305),(31,39)$, and $(1517,1655)$, respectively, to \ \  obtain that $f_c(ca)=f_c(cb)$. This completes the proof.
	\end{proof}
	\begin{corollary}
		For each positive integer $c\leq1000$, the function $f_c$ is not injective.
	\end{corollary}
	\begin{proof}
		Let $1\leq c\leq 1000$ be a positive integer. If $c$ is odd and is divisible by neither $157$ nor $443$, then the result follows from Theorem \ref{theorem4}(i). Suppose $c$ is odd and is divisible by $157$ or $443$. Since $c\leq1000$, the possible values of $c$ are $c=157,157\cdot3,157\cdot 5,443$. If $c\neq 157\cdot 5$, then $(c,130)=1$ and the result follows from Theorem \ref{theorem3}(i). If $c=157\cdot 5$, we apply Theorem \ref{theorem4}(ii) to obtain the desired result. 

	Therefore it remains to consider the case that $c$ is even. Let $c=2^kd$ where $k\geq1$ and $d$ is odd. If $(d,13)=(d,71)=d(881)=1$, then the result follows from Theorem \ref{theorem4}(iii). So we only need to consider the case that $13\mid d$, $71\mid d$, or $881\mid d$. Since $c\leq1000$, we see that $d\leq500$. So $881\mid d $ is not possible.

\bigskip\noindent
	\textbf{Case 1:} $71\mid d$. Then $d=71, 71\cdot 3, 71\cdot 5, 71\cdot 7$. If $d\neq71\cdot 5$, then the result can be obtained from Theorem \ref{theorem4}(v). If $d=71\cdot 5$, then we apply Theorem \ref{theorem4}(vi) to obtain the desired result.

\bigskip\noindent
	\textbf{Case 2:} $13\mid d$. Then $d=13, 13\cdot 3, 13\cdot 5, \ldots, 13\cdot 37$. If $d\neq13\cdot 5, 13\cdot 15, 13\cdot 25, 13\cdot 35$, then we use Theorem \ref{theorem4}(v); if $d=13\cdot 5, 13\cdot 25$, then we apply Theorem \ref{theorem3}(ii) to obtain the desired result. So it remains to consider the case $d=13\cdot 15$ or $d=13\cdot 35$. 
	
	Suppose $d=13\cdot 35$. Then we use Table \ref{Tablef2} to solve it as follows. Let $a=173$ and $b=213$. Then $f_2(a)=f_2(b)$. Since $d=13\cdot 35$, we have $c=2^k\cdot  5\cdot  7\cdot  13$. Since $c\leq 1000$, we have $k=1$ and $c=2\cdot  5\cdot  7\cdot  13$. Then $\left(\dfrac{c}{2},a\right)=\left(\dfrac{c}{2},b\right)=1$ and \[f_c\left(\dfrac{c}{2}a\right)=\left(\dfrac{c}{2}a+c\right)\varphi\left(\dfrac{c}{2}a\right)=\dfrac{c}{2}(a+2)\varphi\left(\dfrac{c}{2}\right)\varphi(a)=\dfrac{c}{2}\varphi\left(\dfrac{c}{2}\right)f_2(a).\]
	Similarly, we have $f_c\left(\dfrac{c}{2}b\right)=\dfrac{c}{2}\varphi\left(\dfrac{c}{2}\right)f_2(b)$, and so $f_c\left(\dfrac{c}{2}a\right)=f_c\left(\dfrac{c}{2}b\right)$. This shows that $f_c$ is not injective.
	
	Finally, let $d=13\cdot 15$. Then $c=2^k\cdot  3\cdot  5\cdot  13$ where $k=1,2$. We first consider the case $k=1$. Let $a=391=17\cdot  23$ and $b=526=2\cdot  263$. From Table \ref{Tablef2}, we know that $f_2(a)=f_2(b)$. In addition, we have  $\left(\dfrac{c}{2},a\right)=\left(\dfrac{c}{2},b\right)=1$, and by the same calculation as above, we obtain $f_c\left(\dfrac{c}{2}a\right)=f_c\left(\dfrac{c}{2}b\right)$, as desired. Next, let $k=2$. Let $a=301=7\cdot  43$ and $b=339=3\cdot  113$. By Table \ref{Tablef3}, we have $f_3(a)=f_3(b)$. In addition, the equality $\left(\dfrac{c}{3},a\right)=\left(\dfrac{c}{3},b\right)=1$ also holds. Therefore
	\[f_c\left(\dfrac{c}{3}a\right)=\dfrac{c}{3}\varphi\left(\dfrac{c}{3}\right)f_3(a)=\dfrac{c}{3}\varphi\left(\dfrac{c}{3}\right)f_3(b)=f_c\left(\dfrac{c}{3}b\right).\]
	Hence $f_c$ is not injective, as required. This completes the proof.
	\end{proof}
	
	Finally, we show that there are at least 98 percent of positive integers $c\leq N$ such that $f_c$ is not injective for every large positive integer $N$.
	
	\begin{theorem}
		For each $N\in\mathbb{N}$, let $A=A(N)$ be the set of all positive integers $c\leq N$ such that $f_c$ is not injective. Then
		\[\vert A(N)\vert\geq (0.981728)N+O(1).\]
	\end{theorem}
	\begin{proof}
		For each $d\in\mathbb{N}$, let $A_d=A_d(N)$ be the set of all positive integers $c\leq N$ such that $(c,d)=1$. Let
		\begin{center}
			$B_1=A_5\cap A_{61}\cap A_{271}$,\  $B_2=A_{13}\cap A_{71}\cap A_{881}$,\  $B_3=A_5 \cap A_{37}\cap A_{41}\cap A_{331}$.
		\end{center}
	Therefore $B_1, B_2$, and $B_3$ are the set of positive integers $c\leq N$ satisfying conditions (v), (iii), and (vii) in Theorem \ref{theorem4}, respectively. Therefore $B_1\cup B_2\cup B_3\subseteq A$. By the inclusion-exclusion principle, we obtain
	\begin{equation}\label{981}
		\vert A\vert \geq \vert B_1\vert+\vert B_2\vert+\vert B_3\vert-\vert B_1\cap B_2\vert-\vert B_1\cap B_3\vert-\vert B_2\cap B_3\vert+\vert B_1\cap B_2\cap B_3\vert.
	\end{equation}
	Let $N$ be a large positive integer. Each block of integers 
	\[[1,d],\left(d,2d\right],\left(2d,3d\right],\ldots,\left((k-1)d,kd\right],\] 
	where $k=\left\lfloor N/d\right\rfloor$, contains exactly $\varphi(d)$ integers $c$ such that $(c,d)=1$. Therefore
	\[A_d=\sum\limits_{\substack{c\leq N\\ (c,d)=1}}1=\varphi(d)\left\lfloor\dfrac{N}{d}\right\rfloor+r_d,\]
	where $0\leq r_d&lt;d$. Thus
	\begin{equation}\label{982}
		A_d\geq \dfrac{\varphi(d)}{d}N+O_d(1),
	\end{equation}
	where the implied constant depends at most on $d$ but not on $N$.
	
	We also observe that $A_{d_1}\cap A_{d_2}=A_{d_1d_2}$. Therefore $B_1=A_x$, $B_2=A_y$, $B_3=A_z$, and $B_1\cap B_2\cap B_3=A_{xyz}$, where $x=5\cdot  61\cdot  271$, $y=13\cdot  71\cdot  881$, and $z=5\cdot  37\cdot  41\cdot  331$. By \eqref{981} and \eqref{982}, we obtain
	\begin{equation}\label{983}
		\vert A\vert\geq \left(\dfrac{\varphi(x)}{x}+\dfrac{\varphi(y)}{y}+\dfrac{\varphi(z)}{z}-\dfrac{\varphi(xy)}{xy}-\dfrac{\varphi(xz)}{xz}-\dfrac{\varphi(yz)}{yz}+\dfrac{\varphi(xyz)}{xyz}\right)N+O(1),
	\end{equation}
	where the implied constant depends at most on $x,y,z$ but not on $N$. We have
	
	\[\dfrac{\varphi(x)}{x}=\prod\limits_{p\vert x}\left(1-\dfrac{1}{p}\right)=\left(1-\dfrac{1}{5}\right)\left(1-\dfrac{1}{61}\right)\left(1-\dfrac{1}{271}\right)\geq 0.783981,\]
	
	\[\dfrac{\varphi(y)}{y}=\prod\limits_{p\vert y}\left(1-\dfrac{1}{p}\right)=\left(1-\dfrac{1}{13}\right)\left(1-\dfrac{1}{71}\right)\left(1-\dfrac{1}{881}\right)\geq 0.909042.\]
	
	\noindent Similarly, we also have
	\begin{center}
		$\dfrac{\varphi(z)}{z}\geq 0.757099$,\quad $\dfrac{\varphi(xy)}{xy}\leq 0.712673$,\quad
		$\dfrac{\varphi(xz)}{xz}\leq 0.741940$, \quad
		
		$\dfrac{\varphi(yz)}{yz}\leq 0.688236$,\quad and \quad
		$\dfrac{\varphi(xyz)}{xyz}\geq 0.674455$.
	\end{center}
	 
	\noindent Applying these estimates in \eqref{983}, we obtain the desired result. This completes the proof.
	\end{proof}
	
	\section{Notes on some related results}
	\label{sec6}
	We also obtain a result that looks interesting and seem to be related to the Euler function. We record it here for a possibility of future reference. We observe that
	
	\begin{center}
		$\dfrac{1}{2}=1-\dfrac{1}{2}$, \hspace{0.25cm} $\dfrac{1}{3}=\dfrac{1}{2}\left(1-\dfrac{1}{3}\right)=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)$,
		
		$\dfrac{1}{4}=\left(\dfrac{1}{2}\right)^2=\left(1-\dfrac{1}{2}\right)^2$, \hspace{0.25cm} 
		$\dfrac{1}{5}=\dfrac{1}{4}\left(1-\dfrac{1}{5}\right)=\left(1-\dfrac{1}{2}\right)^2\left(1-\dfrac{1}{5}\right)$, and so on.
	\end{center}
	 In general, we have the following result.
	 
	 \begin{theorem} \label{1/n}
	 	For each integer $n\geq 2$, there exists a unique set of primes $p_1&gt;p_2&gt;\cdots&gt;p_k$ and positive integers $a_1,a_2,\ldots,a_k$ such that
	 	\begin{equation} \label{Eq1/n}
	 		\dfrac{1}{n}=\prod\limits_{i=1}^{k}\left(1-\dfrac{1}{p_i}\right)^{a_i}.
	 	\end{equation}
	 \end{theorem}
 \begin{proof}
 	We call the product in the form of the right-hand side of (\ref{Eq1/n}) a good form, and we observe that if $1/a$ and $1/b$ are written in a good form, then $1/ab=(1/a)(1/b)$ can also be written in a good form. We use this observation and a strong induction on $n$ to prove this theorem. It is easy to check that the result holds when $n=2$. So assume that $n\geq 3$ and the result holds for $2,3,\ldots,n-1$. If $n$ is a prime, then we write
 	\[\dfrac{1}{n}=\left(1-\dfrac{1}{n}\right)\left(\dfrac{1}{n-1}\right),\] 
 	and then write $1/(n-1)$, by the induction hypothesis, in a good form to obtain a good form for $1/n$. So assume that $n$ is a composite. Then we write $n=q_1^{n_1}q_2^{n_2}\cdots q_k^{n_k}$ where $q_1,q_2,\ldots q_k$ are distinct primes and $n_1,n_2,\ldots,n_k$ are positive integers. Then $2\leq q_i\leq n-1$ for every $1\leq i\leq k$. By the induction hypothesis, the number $1/q_i$ can be written in a good form, and so $1/q_i^{n_i}=(1/q_i)^{n_i}$ can also be written in a good form for every $i$. This gives a good form for $1/n$, as required.
 \end{proof}

	\begin{remark}\label{rem}
	Let $a(1)=0$, and for $n\geq 2$, let $a(n)$ be the number of factors counted with multiplicity in writing $1/n$ in the form of \eqref{Eq1/n}. Then it is easy to see that $a(p)=1+a(p-1)$ for every prime $p$ and $a(mn)=a(m)+a(n)$ for every $m,n\in\mathbb{N}$. In fact, the sequence $(a(n))_{n\geq 1}$ is the same as \seqnum{A064097} in the OEIS. So, perhaps, Theorem \ref{1/n} is known, but as far as we are aware, it is not widely known. We did not know about this before  Ruankong sent us the statement of Theorem \ref{1/n} sometime ago. He did not give a proof and did not publish the result either, but he allowed us to include it in this paper. So our idea and proof may be different from what he had in mind.
	\end{remark}

	\section{Open questions}\label{SecQ}
	\label{sec7}
	
	In this section, we propose some problems related to our results. We do not claim that these problems are difficult or interesting. They are not important and may even be trivial. However, we would merely like to record them for ourselves and to share them among interested readers. We do not plan to solve them soon and we do not mind if the readers solve them.

\begin{question}
	We show that the function $n\mapsto(n+c)\varphi(n)$ is not injective for positive integers $c\leq1000$, and also for more than 98 percent of positive integers $c\leq N$ when $N$ is large. Can one show that the function is not injective for any $c\in\mathbb{N}$?
\end{question}

\begin{question}
	By Ford's result \cite{For1}, we know that if $k\geq2$ is a fixed positive integer, there exists $m\in\mathbb{N}$ such that the equation $\varphi(x)=m$ has exactly $k$ solutions. Since $n\mapsto n\varphi(n)$ is injective, the equation $x\varphi(x)=m$ has at most one solution. What are the answers if we replace $\varphi(n)$ by $\sigma(n)$ or other arithmetic functions. For example, if $m\in\mathbb{N}$ is given, how many solutions in $x\in\mathbb{N}$ to the equations $x\sigma(x)=m$,  $x\psi(x)=m$, and $xd(x)=m$? Makowski \cite[p.~102]{Guy} observed that if $M_1=2^{p_1}-1, M_2=2^{p_2}-1, \ldots, M_k=2^{p_k}-1$ are distinct Mersenne primes, 
	\begin{center}
		$M=\prod\limits_{i=1}^{k}M_i$, and $n_i=\dfrac{M}{M_i}$ for each $i=1,2,\ldots,k$,
	\end{center}
	 then $n_i\sigma(n_i)$ is a constant. This shows that if $k$ is less than or equal to the number of Mersenne primes, then there exists $m\in\mathbb{N}$ such that the equation $x\sigma(x)=m$ has at least $k$ solutions in $x\in\mathbb{N}$. What are the answers if we replace $\sigma(x)$ by other arithmetic functions?
\end{question}
	 
\begin{question}
	 For each $k\geq2$, does there exist $m\in\mathbb{N}$ for which the equation $x\sigma(x)=m$ has exactly $k$ solutions? For example, $x\sigma(x)=6$, $x\sigma(x)=336$, and $x\sigma(x)=333312$ have exactly one, two, and three solutions, respectively, namely, $x=2$ for the first equation, $x=12, 14$ for the second equation, and $x=336, 372, 434$ for the third equation, respectively. The smallest $m$ such that $x\sigma(x)=m$ has exactly $n$ solutions is the sequence \seqnum{A212490} in the OEIS. It should be observed that $6\mid 336$ and $336\mid 333312$. If $a_n$ is the $n$th term of the sequence \seqnum{A212490}, is it true that $a_{n+1}$ is always divisible by $a_{n}$?
\end{question}

\begin{question}
	 Let $B$ be the function defined in Example \ref{ExB} by $B(x)=xs(x)$. Recall that if $(x,y)$ is an amicable pair, then $B(x)=B(y)$. Nevertheless, if $B(x)=B(y)$, then $(x,y)$ may or may not be an amicable pair. For instance, we know from Table \ref{TableB} that $B(6)=B(9)=36$, $B(320)=B(340)=141440$, and $B(1280)=B(1504)=2286080$, but they are not amicable pairs. Are there infinitely many $x,y\in\mathbb{N}$ such that $B(x)=B(y)$? Are there infinitely many such $x,y\in\mathbb{N}$ that are not an amicable pair?
\end{question}

\begin{question}
	 We show in Examples \ref{ExH} and \ref{Ex14}  that the function $H_b$, which is defined by $H_b(n)=nS_{b}(n)$, is not injective on $\mathbb{N}$ for any $b\geq 2$ and is not injective on squarefree integers for $2\leq b\leq 10$. Can one show that $H_b$ is not injective on squarefree integers for any $b\geq 11$?
\end{question}

\begin{question}
	We show that the function $n\mapsto n^a J_s(n)^b$ is injective if $s=2$ or  $a\geq sb$. Is the function $n\mapsto n^a J_s(n)^b$ injective if $s\geq 3$ and $a&lt;sb$?
\end{question}

\begin{question}
	It is not difficult to show that an analogue of Lemma \ref{lemA2} where $1-\frac{1}{p}$ is replaced by $1+\frac{1}{p}$ also holds. That is, the product $\prod\limits_{p}\left(1+\frac{1}{p}\right)^{a_p}$ uniquely determines the primes $p$ and the exponents $a_p$ in the product. However, it is not clear how an analogue of Theorem \ref{1/n} should look like. Can one determine the set of all rational numbers $q$ that can be written as
	\[q=\prod\limits_{p}\left(1+\dfrac{1}{p}\right)^{a_p}\] 
	for some primes $p$ and positive integers $a_p$?
\end{question}


\begin{question}
	Other questions stated in Guy's book \cite{Guy} are the following:

\begin{itemize}
\item[(i)] Among all $m,n\in\mathbb{N}$ such that $m\sigma(m)=n\sigma(n)$, is $m/n$ bounded?
	
\item[(ii)] Are there relatively prime positive integers $m$ and $n$ satisfying $m\sigma(n)=n\sigma(m)$?
\end{itemize}
\end{question}

\begin{question}
	Let $(a(n))_{n\geq1}$ be the sequence \seqnum{A064097} in the OEIS. This sequence is related to Theorem \ref{1/n} and is also mentioned in Remark \ref{rem}. Some conjectures regarding $a(n)$ stated in the OEIS are as follows.

\begin{itemize}
\item[(i)] (Cloitre) $\log n&lt;a(n)&lt;(5/2)\log n$ for $n\geq 2$, and there exists a positive constant $c$ such that
	\[\sum\limits_{1\leq k\leq n}a(k)\sim cn\log n.\] 
	
\item[(ii)] (Wilson) $\lfloor \log 2n\rfloor&lt;a(n)&lt;(5/2)\log n$ for $n\geq 2$.
\end{itemize}

\end{question}
	
\section{Acknowledgment}
	
	\indent\indent Prapanpong Pongsriiam's research project is funded jointly by the Faculty of Science Silpakorn University and the National Research Council of Thailand (NRCT), grant number NRCT5-RSA63021-02. He is also supported by the Tosio Kato Fellowship given by the Mathematical Society of Japan during his visit at Nagoya University in July 2022 to July 2023. We thank Pongpol Ruankong for sending us the statement of Theorem \ref{1/n} and making us aware of the sequence \seqnum{A064097} in the OEIS. We are also grateful to the anonymous referee for reading our paper carefully and giving us suggestions which improve the quality of this article.


\section{Tables}
	\begin{table}[H]
	\begin{center}
	\medskip
		\begin{tabular}{|c|r|r|}\hline
			\textbf{$B(x_1)=B(x_2)$}  &amp; \multicolumn{1}{|c|}{\textbf{$x_1$}} &amp; \multicolumn{1}{|c|}{\textbf{$x_2$}}  \\
			\hline
			36 &amp;	6&amp;	9\\
			62480&amp;	220&amp;	284\\
			141440&amp;	320&amp;	340\\
			1432640&amp;	1184&amp;	1210\\
			2286080&amp;	1280&amp;	1504\\
			1245335&amp;	1955&amp;	2093\\
			6680960&amp;	2080&amp;	2288\\
			7660880&amp;	2620&amp;	2924\\
			27931280&amp;	5020&amp;	5564\\
			39685376&amp;	6232&amp;	6368\\
			\hline
		\end{tabular}
	\end{center}
	\caption{Distinct integers $x_1,x_2\in(1,10000]$ such that $B(x_1)=B(x_2)$ where $B$ is defined   in Example \ref{ExB} by $B(x)=xs(x)$ for all $x\in\mathbb{N}$. The values of $x_1$ are increasing, but $x_2$ and $B(x_1)=B(x_2)$ are not listed in an increasing order.} 
	\label{TableB}
\end{table}

		\begin{table}[H]
		\begin{center}
		\medskip
		\medskip
		\begin{tabular}{|c|r|r|}\hline
			\textbf{$A(x_1)=A(x_2)$}  &amp; \multicolumn{1}{|c|}{\textbf{$x_1$}} &amp; \multicolumn{1}{|c|}{\textbf{$x_2$}}  \\
			\hline
			336 &amp; 12 &amp; 14 \\
			5952 &amp; 48 &amp; 62 \\
			10080 &amp; 60 &amp; 70 \\
			27776 &amp; 112 &amp; 124 \\
			44352 &amp; 132 &amp; 154  \\
			61152 &amp; 156 &amp; 182 \\
			60480 &amp; 160 &amp; 189 \\
			97536 &amp; 192 &amp; 254 \\
			102816 &amp; 204 &amp; 238 \\
			127680 &amp; 228 &amp; 266 \\
			178560 &amp; 240 &amp; 310 \\
			185472 &amp; 276 &amp; 322 \\
			260400 &amp; 300 &amp; 350 \\
			196560 &amp; 315 &amp; 351 \\
			333312 &amp; 336 &amp; 372 \\
			333312 &amp; 336 &amp; 434 \\
			292320 &amp; 348 &amp; 406 \\
			333312 &amp; 372 &amp; 434 \\
			472416 &amp; 444 &amp; 518 \\
			455168 &amp; 448 &amp; 508 \\
			578592 &amp; 492 &amp; 574 \\
			635712 &amp; 516 &amp; 602 \\
			785664 &amp; 528 &amp; 682 \\
			833280 &amp; 560 &amp; 620 \\
			758016 &amp; 564 &amp; 658 \\
			1083264 &amp; 624 &amp; 806 \\
			1179360 &amp; 630 &amp; 702 \\
			961632 &amp; 636 &amp; 742 \\
			1330560 &amp; 660 &amp; 770 \\
			1189440 &amp; 708 &amp; 826 \\
			1270752 &amp; 732 &amp; 854 \\
			1530816 &amp; 804 &amp; 938 \\
			1717632 &amp; 852 &amp; 994 \\
			\hline
		\end{tabular}
		\quad\quad\quad
		\begin{tabular}{|c|r|r|}\hline
			\textbf{$A(x_1)=A(x_2)$}  &amp; \multicolumn{1}{|c|}{\textbf{$x_1$}} &amp; \multicolumn{1}{|c|}{\textbf{$x_2$}}  \\
			\hline
			1834560 &amp; 780 &amp; 910 \\
			1821312 &amp; 816 &amp; 1054 \\
			1815072 &amp; 876 &amp; 1022 \\
			2261760 &amp; 912 &amp; 1178 \\
			2123520 &amp; 948 &amp; 1106 \\
			2926080 &amp; 960	&amp; 1270\\
			2342592 &amp; 996 &amp; 1162 \\
			3249792&amp;	1008&amp;	1116\\
			3084480 &amp; 1020 &amp;	1190\\
			2691360 &amp;1068 &amp;1246\\
			3285504&amp;	1104&amp;	1426\\
			3830400&amp;	1140&amp;	1330\\
			3194016 &amp;	1164&amp;	1358\\
			4612800&amp;	1200&amp;	1550\\
			3461472&amp;	1212&amp;	1414\\
			3666432&amp;	1232&amp;	1364\\
			3599232&amp;	1236&amp;	1442\\
			5503680&amp;	1260&amp;	1404\\
			3882816&amp;	1284&amp;	1498\\
			4028640&amp;	1308&amp;	1526\\
			5462016&amp;	1344&amp;	1524\\
			5462016&amp;	1344&amp;	1778\\
			4328352&amp;	1356&amp;	1582\\
			5564160&amp;	1380&amp;	1610\\
			5178240&amp;	1392&amp;	1798\\
			5407248&amp;	1452&amp;	1694\\
			5055232&amp;	1456&amp;	1612\\
			6552000&amp;	1500&amp;	1750\\
			5462016&amp;	1524&amp;	1778\\
			5810112&amp;	1572&amp;	1834\\
			6352416&amp;	1644&amp;	1918\\
			6538560&amp;	1668&amp;	1946\\
			9999360&amp;	1680&amp;	1860\\
			\hline
		\end{tabular}
	\end{center}
		\caption{Distinct integers $x_1,x_2\in(1,2000]$ such that $A(x_1)=A(x_2)$ where $A$ is defined   in Example \ref{ExA} by $A(x)=x\sigma(x)$ for all $x\in\mathbb{N}$. The values of $x_1$ are increasing, but $x_2$ and $A(x_1)=A(x_2)$ are not listed in an increasing order.}
		\label{TableA}
\end{table}
	

	\begin{table}[H]
	\begin{center}
	\medskip
	\medskip
		\begin{tabular}{|c|r|r|}\hline
			\textbf{$D(x_1)=D(x_2)$}  &amp; \multicolumn{1}{|c|}{\textbf{$x_1$}} &amp; \multicolumn{1}{|c|}{\textbf{$x_2$}}  \\
			\hline
			108&amp;	18&amp;	27\\	
			192&amp;	24&amp;	32\\	
			448&amp;	56&amp;	64\\	
			1080&amp;	90&amp;	135\\	
			1920&amp;	120&amp;	160\\	
			1512&amp;	126&amp;	189\\	
			2688&amp;	168&amp;	192\\
			2688&amp;	168&amp;	224\\
			2688&amp;	192&amp;	224\\
			2376&amp;	198&amp;	297\\	
			2808&amp;	234&amp;	351\\	
			4224&amp;	264&amp;	352\\	
			4480&amp;	280&amp;	320\\	
			3672&amp;	306&amp;	459\\	
			4992&amp;	312&amp;	416\\	
			4104&amp;	342&amp;	513\\	
			8640&amp;	360&amp;	432\\	
			6000&amp;	400&amp;	500\\	
			6528&amp;	408&amp;	544\\	
			4968&amp;	414&amp;	621\\
			8100&amp;	450&amp;	675\\	
			7296&amp;	456&amp;	608\\
			12096&amp;	504&amp;	576\\
			6264&amp;	522&amp;	783\\
			12960&amp;	540&amp;	648\\			
			8832&amp;	552&amp;	736\\	
			6696&amp;	558&amp;	837\\
			14400&amp;	600&amp;	800\\			
			9856&amp;	616&amp;	704\\	
			15120&amp;	630&amp;	945\\	
			7992&amp;	666&amp;	999\\	
			11136&amp;	696&amp;	928\\	
			11648&amp;	728&amp;	832\\	
			8856&amp;	738&amp;	1107\\	
			\hline
		\end{tabular}
		\quad\quad\quad
		\begin{tabular}{|c|r|r|}\hline
			\textbf{$D(x_1)=D(x_2)$}  &amp; \multicolumn{1}{|c|}{\textbf{$x_1$}} &amp; \multicolumn{1}{|c|}{\textbf{$x_2$}}  \\
			\hline
			11904&amp;	744&amp;	992\\	
			9288&amp;	774&amp;	1161\\	
			26880&amp;	840&amp;	960\\
			26880&amp;	840&amp;	1120\\
			10152&amp;	846&amp;	1269\\	
			15876&amp;	882&amp;	1323\\	
			14208&amp;	888&amp;	1184\\	
			15232&amp;	952&amp;	1088\\	
			11448&amp;	954&amp;	1431\\	
			26880&amp;	960&amp;	1120\\
			15744&amp;	984&amp;	1312\\	
			23760&amp;	990&amp;	1485\\	
			16512&amp;	1032&amp;	1376\\	
			12744&amp;	1062&amp;	1593\\	
			17024&amp;	1064&amp;	1216\\	
			13176&amp;	1098&amp;	1647\\	
			18048&amp;	1128&amp;	1504\\
			28080&amp;	1170&amp;	1755\\	
			28224&amp;	1176&amp;	1568\\	
			36000&amp;	1200&amp;	1500\\	
			14472&amp;	1206&amp;	1809\\	
			20352&amp;	1272&amp;	1696\\	
			15336&amp;	1278&amp;	1917\\	
			20608&amp;	1288&amp;	1472\\	
			32400&amp;	1296&amp;	1350\\	
			15768&amp;	1314&amp;	1971\\	
			42240&amp;	1320&amp;	1760\\	
			33600&amp;	1400&amp;	1600\\	
			22656&amp;	1416&amp;	1888\\	
			20412&amp;	1458&amp;	1701\\	
			23424&amp;	1464&amp;	1952\\	
			48384&amp;	1512&amp;	1728\\	
			25984&amp;	1624&amp;	1856\\	
			27776&amp;	1736&amp;	1984\\	
			\hline
		\end{tabular}
	\end{center}
	\caption{Distinct integers $x_1,x_2\in(1,2000]$ such that $D(x_1)=D(x_2)$ where $D$ is defined   in Example \ref{ExD} by $D(x)=xd(x)$ for all $x\in\mathbb{N}$. The values of $x_1$ are increasing, but $x_2$ and $D(x_1)=D(x_2)$ are not listed in an increasing order.}
	\label{TableD}
	\end{table}
	
	\begin{table}[H]
	\begin{center}
	\medskip
	\begin{tabular}{|c|r|r|}\hline
		\textbf{$W_1(x_1)=W_1(x_2)$}  &amp; \multicolumn{1}{|c|}{\textbf{$x_1$}} &amp; \multicolumn{1}{|c|}{\textbf{$x_2$}}  \\
		\hline
		90&amp;	30&amp;	45\\
		126&amp;	42&amp;	63\\
		198&amp;	66&amp;	99\\
		234&amp;	78&amp;	117\\
		270&amp;	90&amp;	135\\
		306&amp;	102&amp;	153\\
		342&amp;	114&amp;	171\\
		378&amp;	126&amp;	189\\
		414&amp;	138&amp;	207\\
		450&amp;	150&amp;	225\\
		522&amp;	174&amp;	261\\
		558&amp;	186&amp;	279\\
		594&amp;	198&amp;	297\\
		840&amp;	210&amp;	280\\
		666&amp;	222&amp;	333\\
		702&amp;	234&amp;	351\\
		738&amp;	246&amp;	369\\
		774&amp;	258&amp;	387\\
		810&amp;	270&amp;	405\\
		846&amp;	282&amp;	423\\
		882&amp;	294&amp;	441\\
		918&amp;	306&amp;	459\\
		954&amp;	318&amp;	477\\
		1320&amp;	330&amp;	440\\
		1026&amp;	342&amp;	513\\
		1062&amp;	354&amp;	531\\
		1098&amp;	366&amp;	549\\
		\hline
	\end{tabular}
	\quad\quad\quad
	\begin{tabular}{|c|r|r|}\hline
		\textbf{$W_1(x_1)=W_1(x_2)$}  &amp; \multicolumn{1}{|c|}{\textbf{$x_1$}} &amp; \multicolumn{1}{|c|}{\textbf{$x_2$}}  \\
		\hline
		1134&amp;	378&amp;	567\\
		1560&amp;	390&amp;	520\\
		1206&amp;	402&amp;	603\\
		1242&amp;	414&amp;	621\\
		1680&amp;	420&amp;	560\\
		1278&amp;	426&amp;	639\\
		1314&amp;	438&amp;	657\\
		1350&amp;	450&amp;	675\\
		1848&amp;	462&amp;	616\\
		1422&amp;	474&amp;	711\\
		1494&amp;	498&amp;	747\\
		2040&amp;	510&amp;	680\\
		1566&amp;	522&amp;	783\\
		1602&amp;	534&amp;	801\\
		2184&amp;	546&amp;	728\\
		1674&amp;	558&amp;	837\\
		2280&amp;	570&amp;	760\\
		1746&amp;	582&amp;	873\\
		1782&amp;	594&amp;	891\\
		1818&amp;	606&amp;	909\\
		1854&amp;	618&amp;	927\\
		1926&amp;	642&amp;	963\\
		1962&amp;	654&amp;	981\\
		2640&amp;	660&amp;	880\\
		1998&amp;	666&amp;	999\\
		2760&amp;	690&amp;	920\\
		2856&amp;	714&amp;	952\\
		\hline
	\end{tabular}
	\end{center}
	\caption{Distinct integers $x_1,x_2\in(1,1000]$ such that $W_1(x_1)=W_1(x_2)$ where $W_1$ is defined   in Example \ref{ExW} by $W_1(x)=x\omega(x)$ for all $x\in\mathbb{N}$. The values of $x_1$ are increasing, but $x_2$ and $W_1(x_1)=W_1(x_2)$ are not listed in an increasing order.}
	\label{TableW1}
	\end{table}

	\begin{table}[H]
	\begin{center}
	\medskip
	\begin{tabular}{|c|r|r|}\hline
		\textbf{$W_2(x_1)=W_2(x_2)$}  &amp; \multicolumn{1}{|c|}{\textbf{$x_1$}} &amp; \multicolumn{1}{|c|}{\textbf{$x_2$}}  \\
		\hline
		160&amp;	32&amp;	40\\
		240&amp;	48&amp;	60\\
		360&amp;	72&amp;	90\\
		400&amp;	80&amp;	100\\
		540&amp;	108&amp;	135\\
		560&amp;	112&amp;	140\\
		600&amp;	120&amp;	150\\
		840&amp;	168&amp;	210\\
		880&amp;	176&amp;	220\\
		900&amp;	180&amp;	225\\
		1344&amp;	192&amp;	224\\
		1000&amp;	200&amp;	250\\
		1040&amp;	208&amp;	260\\
		1260&amp;	252&amp;	315\\
		1320&amp;	264&amp;	330\\
		1360&amp;	272&amp;	340\\
		1400&amp;	280&amp;	350\\
		2016&amp;	288&amp;	336\\
		1500&amp;	300&amp;	375\\
		1520&amp;	304&amp;	380\\
		1560&amp;	312&amp;	390\\
		1840&amp;	368&amp;	460\\
		1960&amp;	392&amp;	490\\
		1980&amp;	396&amp;	495\\
		2040&amp;	408&amp;	510\\
		2100&amp;	420&amp;	525\\
		3024&amp;	432&amp;	504\\
		2200&amp;	440&amp;	550\\
		2280&amp;	456&amp;	570\\
		\hline
	\end{tabular}
	\quad\quad\quad
	\begin{tabular}{|c|r|r|}\hline
		\textbf{$W_2(x_1)=W_2(x_2)$}  &amp; \multicolumn{1}{|c|}{\textbf{$x_1$}} &amp; \multicolumn{1}{|c|}{\textbf{$x_2$}}  \\
		\hline
		2320&amp;	464&amp;	580\\
		2340&amp;	468&amp;	585\\
		3360&amp;	480&amp;	560\\
		2480&amp;	496&amp;	620\\
		2500&amp;	500&amp;	625\\
		4608&amp;	512&amp;	576\\
		2600&amp;	520&amp;	650\\
		2760&amp;	552&amp;	690\\
		2940&amp;	588&amp;	735\\
		2960&amp;	592&amp;	740\\
		3060&amp;	612&amp;	765\\
		3080&amp;	616&amp;	770\\
		4536&amp;	648&amp;	756\\
		3280&amp;	656&amp;	820\\
		3300&amp;	660&amp;	825\\
		4704&amp;	672&amp;	784\\
		3400&amp;	680&amp;	850\\
		3420&amp;	684&amp;	855\\
		3440&amp;	688&amp;	860\\
		3480&amp;	696&amp;	870\\
		3500&amp;	700&amp;	875\\
		5040&amp;	720&amp;	840\\
		3640&amp;	728&amp;	910\\
		3720&amp;	744&amp;	930\\
		3760&amp;	752&amp;	940\\
		3800&amp;	760&amp;	950\\
		6912&amp;	768&amp;	864\\
		3900&amp;	780&amp;	975\\
		&amp; &amp; \\
		\hline
	\end{tabular}
	\end{center}
	\caption{Distinct integers $x_1,x_2\in(1,1000]$ such that $W_2(x_1)=W_2(x_2)$ where $W_2$ is defined   in Example \ref{ExW} by $W_2(x)=x\Omega(x)$ for all $x\in\mathbb{N}$. The values of $x_1$ are increasing, but $x_2$ and $W_2(x_1)=W_2(x_2)$ are not listed in an increasing order.}
	\label{TableW2}
	\end{table}

	\begin{table}[H]
		\begin{center}
			\medskip
			\begin{tabular}{|c|r|r|}\hline
				\textbf{$H_{10}(x_1)=H_{10}(x_2)$}  &amp; \multicolumn{1}{|c|}{\textbf{$x_1$}} &amp; \multicolumn{1}{|c|}{\textbf{$x_2$}}  \\
				\hline
				36&amp;	6&amp;	12\\
				90&amp;	15&amp;	30\\
				160&amp;	32&amp;	40\\
				280&amp;	28&amp;	35\\
				306&amp;	51&amp;	102\\
				684&amp;	57&amp;	114\\
				360&amp;	60&amp;	120\\
				640&amp;	64&amp;	80\\
				792&amp;	66&amp;	132\\
				900&amp;	75&amp;	150\\	
				900&amp;	75&amp;	300\\
				1105&amp;	85&amp;	221\\
				1204&amp;	86&amp;	301\\
				1408&amp;	88&amp;	128\\
				1440&amp;	96&amp;	240\\
				520&amp;	104&amp;	130\\
				630&amp;	105&amp;	210\\
				1360&amp;	136&amp;	170\\
				900&amp;	150&amp;	300\\
				1872&amp;	156&amp;	312\\
				1980&amp;	165&amp;	330\\
				2520&amp;	168&amp;	420\\
				1720&amp;	172&amp;	215\\
				2992&amp;	187&amp;	272\\
				2080&amp;	208&amp;	260\\
				2440&amp;	244&amp;	305\\
				\hline
			\end{tabular}
			\quad\quad\quad
			\begin{tabular}{|c|r|r|}\hline
				\textbf{$H_{10}(x_1)=H_{10}(x_2)$}  &amp; \multicolumn{1}{|c|}{\textbf{$x_1$}} &amp; \multicolumn{1}{|c|}{\textbf{$x_2$}}  \\
				\hline
				3060&amp;	255&amp;	510\\
				4140&amp;	276&amp;	345\\
				2800&amp;	280&amp;	350\\
				4576&amp;	286&amp;	416\\
				1600&amp;	320&amp;	400\\
				4732&amp;	338&amp;	364\\
				5220&amp;	348&amp;	435\\
				3520&amp;	352&amp;	440\\
				5760&amp;	384&amp;	480\\
				6160&amp;	385&amp;	560\\
				4240&amp;	424&amp;	530\\
				6370&amp;	455&amp;	490\\
				6840&amp;	456&amp;	570\\
				7744&amp;	484&amp;	704\\
				7380&amp;	492&amp;	615\\
				7920&amp;	528&amp;	660\\
				8460&amp;	564&amp;	705\\
				8008&amp;	572&amp;	616\\
				11305&amp;	595&amp;	665\\
				11920&amp;	596&amp;	745\\
				6400&amp;	640&amp;	800\\
				13360&amp;	668&amp;	835\\
				10080&amp;	672&amp;	840\\
				11160&amp;	744&amp;	930\\
				14212&amp;	748&amp;	836\\
				15520&amp;	776&amp;	970\\
				\hline
			\end{tabular}
		\end{center}
			\caption{Distinct integers $x_1,x_2\in(1,1000]$ such that $H_{10}(x_1)=H_{10}(x_2)$ where $H_{10}$ is defined   in Example \ref{ExH} by $H_{10}(x)=xS_{10}(x)$ for all $x\in\mathbb{N}$. The values of $x_1$ are increasing, but $x_2$ and $H_{10}(x_1)=H_{10}(x_2)$ are not listed in an increasing order.}
			\label{TableH}
	\end{table}

	\begin{table}[H]
	\begin{center}
			\medskip
			\begin{tabular}{|c|r|r|r|r|}\hline
				\textbf{$f_1(x_1)=f_1(x_2)$}  &amp; \multicolumn{1}{|c|}{\textbf{$x_1$}} &amp; \multicolumn{1}{|c|}{\textbf{$x_2$}} &amp; 
				\multicolumn{1}{|c|}{Factorization of $x_1$} &amp;
				\multicolumn{1}{|c|}{Factorization of $x_2$} \\
				\hline
				168 &amp;	13&amp;	20 &amp; 13 &amp; $2^2\cdot 5$\\
				528&amp;	23&amp;	32&amp; 23 &amp; $2^5$\\
				960&amp;	31&amp;	39&amp; 31&amp; $3\cdot 13$\\
				1368&amp;	37&amp;	56&amp; 37&amp; $2^3\cdot 7$\\
				2208&amp;	47&amp;	68&amp; 47&amp; $2^2\cdot 17$\\
				5040&amp;	71&amp;	104&amp; 71&amp; $2^3\cdot 13$\\
				18720&amp;	155&amp;	194&amp; $5\cdot 31$&amp; $2\cdot 97$\\
				73440&amp;	271&amp;	305&amp; 271&amp; $5\cdot 61$\\
				78880&amp;	289&amp;	492&amp; $17^2$ &amp; $2^2\cdot 3\cdot 41$\\
				144072&amp;	413&amp;	666&amp; $7\cdot 59$ &amp; $2\cdot 3^2\cdot 37$\\
				196248&amp;	443&amp;	628&amp; 443 &amp; $2^2\cdot 157$\\
				131328&amp;	455&amp;	512&amp; $5\cdot 7 \cdot 13$ &amp; $2^9$\\
				212520&amp;	461&amp;	804&amp; 461&amp; $2^2 \cdot 3 \cdot 67$\\
				199080&amp;	473&amp;	710&amp; $11\cdot 43$&amp; $2\cdot 5 \cdot 71$\\
				210528&amp;	515&amp;	730&amp; $5 \cdot 103$ &amp; $2\cdot 5 \cdot 73$\\
				253440&amp;	527&amp;	575&amp; $17\cdot 31$&amp; $5^2 \cdot 23$\\
				256320&amp;	533&amp;	800&amp; $13\cdot 41$&amp; $2^5\cdot 5^2$\\
				226800&amp;	539&amp;	674&amp; $7^2 \cdot 11$&amp; $2\cdot 337$\\
				218120&amp;	573&amp;	664&amp; $3\cdot191$&amp; $2^3\cdot 83$\\
				361200&amp;	601&amp;	902&amp; 601 &amp; $2\cdot 11 \cdot 41$\\
				320544&amp;	635&amp;	741&amp; $5\cdot 127$&amp; $3\cdot 13 \cdot 19$\\
				377000&amp;	649&amp;	753&amp; $11 \cdot 59$&amp; $3\cdot 251$\\
				776160&amp;	881&amp;	923&amp; 881&amp; $13\cdot 71$\\
				863040&amp;	929&amp;	1239&amp; 929&amp; $3 \cdot 7 \cdot 59$\\
				820800&amp;	949&amp;	1025&amp; $13\cdot 73$&amp; $5^2\cdot 41$\\
				1585080 &amp;1259	&amp;	1784&amp; 1259 &amp; $2^3\cdot 223$\\
				708048&amp;	1340&amp;	1638&amp; $2^2\cdot 5 \cdot 67$&amp; $2\cdot 3^2 \cdot 7 \cdot 13$\\
				2185920 &amp;	1517&amp;	1655&amp; $37\cdot 41$&amp; $5\cdot 331$\\
				\hline
			\end{tabular}
		\end{center}
		\caption{Distinct integers $x_1,x_2\in(1,2000]$ such that $f_1(x_1)=f_1(x_2)$ where $f_1$ is the function defined in Section \ref{Sect5} by $f_1(x)=(x+1)\varphi(x)$ for all $x\in\mathbb{N}$. The values of $x_1$ are increasing, but $x_2$ and $f_1(x_1)=f_1(x_2)$ are not listed in an increasing order.}
			\label{Tablef1}
	\end{table}
	
	\begin{table}[H]
		\begin{center}
			\medskip
			\begin{tabular}{|c|r|r|r|r|}\hline
				\textbf{$f_2(x_1)=f_2(x_2)$}  &amp; \multicolumn{1}{|c|}{\textbf{$x_1$}} &amp; \multicolumn{1}{|c|}{\textbf{$x_2$}}  &amp;
				\multicolumn{1}{|c|}{Factorization of $x_1$} &amp;
				\multicolumn{1}{|c|}{Factorization of $x_2$}\\
				\hline
				1920&amp;	62&amp;	78 &amp; $2\cdot 31$ &amp; $2\cdot 3 \cdot 13$\\
				30100&amp;	173&amp;	213&amp;173&amp; $3\cdot 71$\\
				37440&amp;	193&amp;	310&amp; 193&amp; $2\cdot 5 \cdot 31$\\
				78480&amp;	325&amp;	434&amp; $5^2 \cdot 13$&amp; $2\cdot 7 \cdot 31$\\
				89040&amp;	369&amp;	422&amp; $3^2\cdot 41$&amp; $2\cdot 211$\\
				138336&amp;	391&amp;	526&amp; $17\cdot 23$ &amp; $2\cdot 263$\\
				126480&amp;	525&amp;	618&amp; $3\cdot 5^2 \cdot 7$&amp; $2\cdot 3\cdot 103$\\
				146880&amp;	542&amp;	610&amp; $2\cdot 271$&amp; $2\cdot 5\cdot 61$\\
				254016&amp;	565&amp;	754&amp; $5\cdot 113$&amp; $2\cdot 13 \cdot 29$\\
				363456&amp;	629&amp;	1260&amp; $17\cdot 37$&amp; $2^2\cdot 3^2 \cdot 5\cdot 7$\\
				219120&amp;	662&amp;	828&amp; $2\cdot 331$&amp; $2^2\cdot 3^2\cdot 23$\\
				288144&amp;	665&amp;	826&amp; $5\cdot 7 \cdot 19$&amp; $2\cdot 7 \cdot 59$\\
				453600&amp;	673&amp;	1078&amp; 673 &amp; $2\cdot 7^2\cdot 11$\\
				294528&amp;	765&amp;	942&amp; $3^2\cdot 5 \cdot 17$&amp; $2\cdot 3\cdot 157$\\
				290880&amp;	806&amp;	1008&amp; $2\cdot 13\cdot 31$&amp; $2^4\cdot 3^2 \cdot 7$\\
				320256&amp;	832&amp;	1110&amp; $2^6\cdot 13$&amp; $2\cdot 3\cdot 5\cdot 37$\\
				469440&amp;	976&amp;	1302&amp; $2^4\cdot 61$ &amp; $2\cdot 3\cdot 7\cdot 31$\\
				506880&amp;	1054&amp;	1150&amp; $2\cdot 17\cdot 31$&amp; $2\cdot 5^2\cdot 23$\\
				761376&amp;	1131&amp;	1234&amp; $3\cdot 13\cdot 29$ &amp; $2\cdot 617$ \\
				796320&amp;	1262&amp;	1420&amp; $2\cdot 631$ &amp; $2^2\cdot 5 \cdot 71$\\
				641088&amp;	1270&amp;	1482&amp; $2\cdot 5\cdot 127$&amp; $2\cdot 3\cdot 13\cdot 19$\\
				754000&amp;	1298&amp;	1506&amp; $2\cdot 11\cdot 59$&amp; $2\cdot 3\cdot 251$\\
				907200&amp;	1348&amp;	1510&amp; $2^2\cdot 337$&amp; $2\cdot 5\cdot 151$\\
				1552320&amp;	1762&amp;	1846&amp; $2\cdot 881$ &amp; $2\cdot 13 \cdot 71$\\
				\hline
			\end{tabular}
		\end{center}
			\caption{Distinct integers $x_1,x_2\in(1,2000]$ such that $f_2(x_1)=f_2(x_2)$ where $f_2$ is the function defined in Section \ref{Sect5} by $f_2(x)=(x+2)\varphi(x)$ for all $x\in\mathbb{N}$. The values of $x_1$ are increasing, but $x_2$ and $f_2(x_1)=f_2(x_2)$ are not listed in an increasing order.}
			\label{Tablef2}
	\end{table}

	\begin{table}[H]
		\begin{center}
			\medskip
			\begin{tabular}{|c|r|r|r|r|}\hline
				\textbf{$f_3(x_1)=f_3(x_2)$}  &amp; \multicolumn{1}{|c|}{\textbf{$x_1$}} &amp; \multicolumn{1}{|c|}{\textbf{$x_2$}} &amp;
				\multicolumn{1}{|c|}{Factorization of $x_1$} &amp;
				\multicolumn{1}{|c|}{Factorization of $x_2$} \\
				\hline
				60&amp;	7&amp;	12&amp; 7 &amp; $2^2\cdot 3$\\
				560&amp;	25&amp;	32&amp; $5^2$ &amp; $2^5$\\
				540&amp;	27&amp;	42&amp; $3^3$&amp; $2\cdot 3\cdot 7$\\
				1008&amp;	39&amp;	60&amp; $3\cdot 13$&amp; $2^2 \cdot 3\cdot 5$\\
				3168&amp;	69&amp;	96&amp; $3\cdot 23$&amp; $2^5 \cdot 3$\\
				7056&amp;	95&amp;	144&amp; $5\cdot 19$&amp; $2^4\cdot 3^2$\\
				6120&amp;	99&amp;	150&amp; $3^2\cdot 11$&amp; $2\cdot 3\cdot 5^2$\\
				8208&amp;	111&amp;	168&amp; $3\cdot 37$&amp; $2^3 \cdot 3\cdot 7$\\
				13248&amp;	141&amp;	204&amp; $3\cdot 47$&amp; $2^2\cdot 3\cdot 17$\\
				21360&amp;	175&amp;	264&amp; $5^2\cdot 7$&amp; $2^3\cdot 3 \cdot 11$\\
				30240&amp;	213&amp;	312&amp; $3\cdot 71$&amp; $2^3\cdot 3 \cdot 13$\\
				54000&amp;	247&amp;	297&amp; $13\cdot 19$&amp; $3^3 \cdot 11$\\
				76608&amp;	301&amp;	339&amp; $7\cdot 43$&amp; $3\cdot 113$\\
				112320&amp;	465&amp;	582&amp; $3\cdot 5\cdot 31$&amp; $2\cdot 3\cdot 97$\\
				186912&amp;	469&amp;	646&amp; $7\cdot 67$&amp; $2\cdot 17\cdot 19$\\
				324896&amp;	569&amp;	920&amp; 569&amp; $2^3\cdot 5 \cdot 23$\\
				376992&amp;	613&amp;	1068&amp; 613&amp; $2^2\cdot 3 \cdot 89$\\
				378000&amp;	753&amp;	872&amp; $3\cdot 251$&amp; $2^3 \cdot 109$\\
				396144&amp;	783&amp;	1176&amp; $3^3\cdot 29$&amp; $2^3 \cdot 3\cdot 7^2$\\
				580320&amp;	803&amp;	933&amp; $11\cdot 73$&amp; $3\cdot 311$\\
				440640&amp;	813&amp;	915&amp; $3\cdot 271$&amp; $3\cdot 5 \cdot 61$\\
				879840&amp;	937&amp;	1830&amp; 937&amp; $2\cdot 3\cdot 5\cdot 61$\\
				808128&amp;	973&amp;	1101&amp; $7\cdot 139$&amp; $3\cdot 367$\\
				822400&amp;	1025&amp;	1282&amp; $5^2\cdot 41$&amp; $2\cdot 641$\\
				1254960&amp;	1159&amp;	1491&amp; $19\cdot 61$&amp; $3\cdot 7 \cdot 71$\\
				1177488&amp;	1329&amp;	1884&amp; $3\cdot 443$&amp; $2^2\cdot 3\cdot 157$\\
				787968&amp;	1365&amp;	1536&amp; $3\cdot 5\cdot 7\cdot 13$ &amp; $2^9\cdot 3$\\
				1520640&amp;	1581&amp;	1725&amp; $3\cdot 17\cdot 31$&amp; $3\cdot 5^2\cdot 23$\\
				2537472&amp;	1649&amp;	1885&amp; $17\cdot 97$ &amp; $5\cdot 13 \cdot 29$\\
				\hline
			\end{tabular}
		\end{center}
			\caption{Distinct integers $x_1,x_2\in(1,2000]$ such that $f_3(x_1)=f_3(x_2)$  where $f_3$ is the function defined in Section \ref{Sect5} by $f_3(x)=(x+3)\varphi(x)$ for all $x\in\mathbb{N}$. The values of $x_1$ are increasing, but $x_2$ and $f_3(x_1)=f_3(x_2)$ are not listed in an increasing order.}
			\label{Tablef3}
	\end{table}


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		\bibitem{For1} K. Ford, The number of solutions of $\varphi(x)=m$, {\it Ann. Math.} {\bf 150} (1999), 1--29.
		
		\bibitem{FLP} K. Ford, F. Luca, and C. Pomerance, Common values of the arithmetic functions $\phi$ and $\sigma$, {\it Bull. Lond. Math. Soc.} {\bf 42} (2010), 478--488.
		
		\bibitem{FPo} K. Ford and P. Pollack, On common values of $\varphi(n)$ and $\sigma(n)$ II, {\it Algebra Number Theory} {\bf 6} (2012), 1669--1696.
		
		\bibitem{Guy} R. K. Guy,  {\it Unsolved Problems in Number Theory}, 3rd edition, Springer, 2004.
		
		\bibitem{Pom} C. Pomerance, Popular values of Euler's function, {\it Mathematika} {\bf 27} (1980), 84--89.
		
		\bibitem{Pong} P. Pongsriiam, Quasi-injectivity of some arithmetic functions, {\it J. Integer Sequences} {\bf 24} (2021), \href{https://cs.uwaterloo.ca/journals/JIS/VOL24/Pongsriiam/pong23.pdf}{Article 21.10.1}.
		
		\bibitem{Slo} N. J. A. Sloane et al., The On-line Encyclopedia of Integer Sequences, 2022. Available at \url{https://oeis.org/}.
	\end{thebibliography}
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\noindent 2020 {\it Mathematics Subject Classification}: Primary 11A25. Secondary 11B83, 11D45.

\noindent \emph{Keywords:\ } arithmetic function,\  Euler's totient function, the sum of divisors function, Mersenne prime, amicable pair, Dedekind's psi function, Jordan's totient function, prime omega function, sum-of-digits function.

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\noindent (Concerned with sequences
\seqnum{A000010}, \seqnum{A000396}, \seqnum{A000668}, \seqnum{A001065}, \seqnum{A002025}, \seqnum{A002046}, \seqnum{A007617}, \seqnum{A008888}, \seqnum{A038040}, \seqnum{A063900}, \seqnum{A063990}, \seqnum{A064097}, \seqnum{A064987}, \seqnum{A098007}, \seqnum{A212327}, \seqnum{A212490}, \seqnum{A259180}, \seqnum{A337873}, and  \seqnum{A338382}.)

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\vspace*{+.1in}
\noindent
Received December 13 2022; revised version received 
April 10 2023.
Published in {\it Journal of Integer Sequences},
November 2 2023.

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\noindent
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\htmladdnormallink{Journal of Integer Sequences home page}{https://cs.uwaterloo.ca/journals/JIS/}.
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\end{document}
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