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\begin{center}
\vskip 1cm{\LARGE\bf The Lie Bracket and the Arithmetic\\
\vskip0.3cm Derivative}
\vskip 1cm
\large
Jun Fan and Sergey Utev \\
Department of Mathematics \\
University of Leicester \\ 
Leicester LE1 7RH \\
United Kingdom \\
\href{jf249@leicester.ac.uk}{\tt jf249@leicester.ac.uk} \\
\href{su35@leicester.ac.uk}{\tt su35@leicester.ac.uk} \\
\ 
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\vskip .2 in
\begin{abstract}
We apply the Lie bracket approach to characterize the semi-derivations on the positive integers. The approach is motivated by the Stroock-Lie bracket identity commonly used in Malliavin calculus.
\end{abstract}
\section{Introduction}

The Malliavin derivative $D$, the Skorohod integral $\delta$, and the associated Malliavin calculus are powerful tools for the analysis of stochastic processes. The Malliavin calculus, named after P. Malliavin, is also called the 
{\it stochastic calculus of variations} \cite[p.\ VII, p.\ 1]{MT}. The definition of the Malliavin derivative and the Skorohod integral can be found, for example, in \cite[p.\ 25]{DRB}, \cite[pp.\ 20, 27]{GBF}, and \cite[pp.\ 25, 40]{DN}, respectively. The Stroock-Lie bracket type identity $[D,\delta]=D\delta-\delta D=I$ is a common notion in the Malliavin calculus \cite[p.\ 355]{YH}, which is also referred to as the fundamental theorem of calculus \cite[Thm.\ 3.18, p.\ 37]{GBF}. The Malliavin derivative and the Skorohod integral (the adjoint operator) act in the space of random processes, which are treated as functions of a Gaussian process. 

On the one hand, many random structures satisfy the functional Gaussian approximation \cite{MPU}. 
On the other hand, the Lie bracket is a powerful tool in the study of differential equations, in particular, quantum stochastic calculus \cite{P}. 
This motivated us to apply the Lie bracket in a totally different environment. More exactly, we treat $D$ and $\delta$ as operators acting on integer sequences or dynamical systems on the natural numbers $\mathbb{N}$. Following Barbeau \cite{EJB}, Ufnarovski and \AA hlander \cite{UB}, and Kovi\v c \cite{JK}, we study a partial number derivative operator $D_A$. The operator is introduced and characterized  as a solution to the modified Stroock-Lie bracket type identity. The Barbeau  arithmetic derivative $D$ is then characterized as the dynamics on the positive integers, which satisfies the Stroock-Lie bracket identity $[D,\ell]=I$, and  which holds for all linear functions $\ell=\ell_p=pn$, where $p$ is a prime number. Moreover, the Stroock-Lie bracket characterization is 
illustrated by examples on several commutative rings without zero divisors.  Finally, arithmetic type differential equations driven by $D_A$ are 
briefly analyzed.

\section{Lie bracket analysis of linear functions of positive integers}
\begin{definition}
For two functions $F,U: \mathbb{Z}_+ \to \mathbb{Z}_+$, we define the 
{\it Lie bracket\/} or the {\it commutator\/} $[F,U]$ by
\[ [F,U] = F \circ U - U \circ F, \text{ where } F \circ U(n) = F(U(n)). \]
Hence $[F,U](n) = F(U(n)) - U(F(n))$.
\end{definition}
Let ${\cal L} = \{\ell_x: \ell_x(n) = nx,\; n \in \mathbb{Z}_+\}$ denote
the set of linear functions on $\mathbb{Z}_+$. Notice that ${\cal L}$
is a commutative semiring with unity $\ell_1=I$ and zero $\ell_0=0$
with respect to the multiplication and addition operations.
Then $\ell_x \circ \ell_y = \ell_{xy}$,
$\ell_x + \ell_y = \ell_{x+y}$,
$\ell_x \circ \ell_1 =\ell_1 \circ \ell_x = \ell_x$, and
$\ell_x + \ell_0 = \ell_0 + \ell_x = \ell_x$. By construction, the semiring ${\cal L}$ is isomorphic to the semiring $(\mathbb{Z}_+, +, \cdot)$ of non-negative integers. 
\begin{definition}
Consider $D:\mathbb{Z}_+ \to \mathbb{Z}_+$ and define the 
{\it Lie bracket linearity set\/} of $D$ by
\begin{equation}\label{linset} W_D = \{x: \text{ there exists $ y = y_x $ such that } [D,\ell_x] = \ell_y \}. \end{equation}
\end{definition}
\begin{lemma} \label{semigroup} 
$W_D$ is a multiplicative semigroup in $(\mathbb{Z}_+, \cdot)$ that includes $1$.
\end{lemma}
\begin{proof} 
Let $x,z \in W_D$. By the Lie bracket definition and algebraic manipulations,
\begin{displaymath}
[D, \ell_{xz}] (n) = D \circ \ell_{xz}(n) - \ell_{xz} \circ D(n)
= [D, \ell_x] (zn) + \ell_x \circ [D, \ell_z](n). 
\end{displaymath}
Now we apply to the last line, first, the definition of $W_D$ and, then, the semiring properties of linear functions. This leads to $[D, \ell_{xz}] (n) = \ell_{y_x} \circ \ell_z (n) + \ell_x \circ \ell_{y_z} (n) = \ell_{y_x z + xy_z} (n)$.
\end{proof}
\begin{lemma} \label{Leibnitz}
The following statements are equivalent: 
\begin{itemize}
\item[(i)] $D$ satisfies the Leibnitz rule $D(mn) = mD(n) + nD(m)$;\\
\item[(ii)] $D(1) = 0$ and $W_D = \mathbb{Z}_+$.
\end{itemize}
Moreover, $[D, \ell_m] = \ell_{D(m)}$.
\end{lemma}
\begin{proof}
(i) $\Longrightarrow$ (ii) follows by direct calculations. 

\bigskip

\noindent (ii) $\Longrightarrow$ (i): Fix $m$. Then, for any $n\in \mathbb{Z}_+$, we have
\[ y_m n = \ell_{y_m}(n) = [D,\ell_m](n) = D \circ \ell_m(n) - \ell_m \circ D (n) = D(mn) - m D(n). \]
Finally, take $n=1$. Then $y_m = D(m) - mD(1) = D(m)$. Therefore, $[D,\ell_m] = \ell_{D(m)}$ and $D(mn) = mD(n) + nD(m)$, proving the lemma.
\end{proof} 
\begin{remark}
Let $x$  be a linear function $l_x$, and let the function composition $\circ $ be replaced by the usual multiplication. 
Then the main Lie bracket characteristic becomes the Pincherle derivative  $f^\prime=f\cdot x - x\cdot f = [f,l_x]$, as introduced in \cite{SP}. Tempesta \cite{PT} applied the Pincherle derivative and the associated Lie bracket approach  in quantum calculus.
\end{remark}
We now apply Lemmas  \ref{semigroup} and 
\ref{Leibnitz} to characterize the arithmetic type derivative $D_A$ as the dynamics on the positive integers, which satisfies the Stroock-Lie bracket type identity $[D_A,\ell_x]=y_xI$.

Let ${\cal P}$ denote the set of all primes. We say that sets $A, B$ are orthogonal, if for any $x \in A$ and $y \in B$, we have $\gcd (x,y) = 1$,  i.e., the sets $A$ and $B$ do not have any common divisors. Notice that disjoint subsets in ${\cal P}$ are orthogonal.
\begin{lemma} \label{charD}
Consider the nonempty subset of primes $A \subset {\cal P}$. Let $D_A : \mathbb{Z}_+ \to \mathbb{Z}_+$ be such that $D_A(1) = 0$. The following properties are equivalent:
\begin{itemize}
\item[(i)] $[D_A, \ell_p] = I$, for $p\in A$, and $[D_A, \ell_p] = 0$, for $p\in\bar{A} = {\cal P}-A$;

\item[(ii)] $D_A = \sum_{p \in A} D_p$, where $D_p(n) = jp^{j-1}m$, for $n = p^jm$ with $m\perp p$. Moreover, $D_A$ satisfies the Leibnitz rule and has a representation
\begin{equation} \label{rep} 
D_A (n) = n \sum_{p\in A} \dfrac{n_p}{p}, \text{ where } n = \prod_{p\in {\cal P}} p^{n_p}.\end{equation}
\end{itemize}
\end{lemma}
\begin{proof} 
(ii) $\Longrightarrow$ (i) follows by direct calculations.

\bigskip

\noindent (i) $\Longrightarrow$ (ii): Consider the linearity set $W_{D_A}$. 
Notice that $\{ 0, 1 \} \subseteq W_{D_A}$ and
the smallest multiplicative semigroup, containing all primes,
is $\mathbb{Z}_+ - \{0,1\}$. By Lemma \ref{semigroup}, it then follows 
that $W_{D_A} = \mathbb{Z}_+$. Thus, by Lemma \ref{Leibnitz}, the function $D$ satisfies the Leibnitz rule. Moreover, $D_A(p) = 1$, for $p \in A$, and $D_A(p) = 0$, for $p \not\in A$. Now we apply the  argument from Ufnarovski and \AA hlander \cite{UB}. Consider the log transform of $D_A$ defined by $L_A(n) = D_A/n$. By the Leibnitz rule, we see that $L_A$ is a homomorphism of the multiplicative semigroup to the additive semigroup on $\mathbb{Z}_+$, which shows that
\[ L_A (n) = \sum_{p\in {\cal P} } \dfrac{n_p}{p}  D_A(p) =  \sum_{p\in A} \dfrac{n_p}{p}, \text{ where } n = \prod_{p\in {\cal P}} p^{n_p}. \]
This implies representation \eqref{rep}. Take $A = \{p\}$. We then derive $D_p(p^{j} m) = jp^{j-1}m$, for $m \perp p$. This proves the representation $D_A = \sum_{p\in A} D_p$. \end{proof} 
\begin{corollary} \label{Chararithm}
Let $D : \mathbb{N} \rightarrow \mathbb{N}$ such that $D (1) = 0$. Assume that for each linear function $\ell = \ell_p = p n$, where $p$ is a fixed prime, the following Stroock-Lie bracket identity
\begin{equation} \label{Liebracket}
[D,\ell] = I, \text{ i.e., } D \ell (n) = n + \ell D (n), \end{equation}
holds for all $n \in \mathbb{N}$. Then $D$ is an arithmetic derivative, i.e.,
\[ D \left( n \right) = n ^ \prime = n \sum\limits_{i = 1}^k \dfrac{n_i}{p_i},
\text{ where } n = \prod_{i = 1}^k p_i^{n_i}. \]
Moreover, $[D, \ell_m ] = D(m) I$. In particular,
we have the following characterization of the Stroock-Lie bracket identity
\begin{equation}\label{lmchar}
[D, \ell_m ] = I \text{ if and only if $m$ is a prime}.
\end{equation}
\end{corollary}
\begin{proof} 
Let us prove the last statement.
The Stroock-Lie bracket equation states that 
$$D \ell_m (n) = n + \ell_m D (n) = n + m D(n) = n + mn'.$$ 
The left-hand side of the former equation is computed by 
$$D \ell_m (n) = (mn)' = D (mn) = m^\prime n + m n^\prime.$$ 
Hence, by equating it to the right-hand side of the same equation, we derive $m^\prime n + m n^\prime = n + mn'$. Clearly the last equation holds if and only if $m^\prime = 1$.  Therefore, $m$ is a prime, as proved in \cite{UB}.
\end{proof}

\section{Lie bracket properties}
\subsection{Lie bracket properties for the arithmetic derivative}
From Corollary \ref{Chararithm}, we derive
\begin{corollary} \label{mprime}
Let $m_1 + \cdots + m_k = m$. Then 
\ $[D, \ell_{m_1} + \cdots + \ell_{m_k}] = I\;$ 
holds if and only if $m$ is a prime. 
\end{corollary} 
\begin{remark} 
\begin{itemize}
\item[(i)] In particular,
$[D, \ell_{p}+\ell_{2}] = I$ if and only if $p$ and $p+2$ are twin primes.
\item[(ii)] According to the Goldbach weak conjecture, every prime number greater than $5$ can be expressed as the sum of three primes. Then, for each such triple of primes $p_1, p_2, p_3$ with $p_1 + p_2 + p_3$ being a prime, we have
$[D, \ell_{p_1} + \ell_{p_2} + \ell_{p_3}] = I$.
\end{itemize}
\end{remark}
Barbeau \cite{EJB} proved that if the natural number $n$ is not a prime or unity, then $n' \geqslant 2 \sqrt{n}$. The equality holds if and only if $n = p^2$, where $p$ is a prime. In particular, the equation $m'=2$ does not have solutions in positive integers. Therefore, we obtain the following lemmas. 
\begin{lemma} 
For any primes $p_1$ and $p_2$,
\[ \left[ D, \ell_{p_1} + \ell_{p_2} \right] \neq \left[ D, \ell_{p_1} \right] + \left[ D, \ell_{p_2} \right]. \]
\end{lemma}
\begin{proof} 
By definition, the right-hand side of the former equation equals to $2n$, $n \in \mathbb{N}$. The left-hand side of the equation is $\left[ D, \ell_{p_1} + \ell_{p_2} \right] \left(n \right) = D \left( \left( p_1 + p_2 \right) n \right) - \left( p_1 + p_2 \right) D (n) = \left( p_1 + p_2 \right)^\prime n$. It remains to notice that the equation $\left( p_1 + p_2 \right)^\prime = 2$ does not have solutions. 
\end{proof}
\begin{lemma} 
For any primes $p_1$ and $p_2$, 
\[ \left[ D + \ell_{p_1}, \ell_{p_2} \right] = I. \]
\end{lemma}
\begin{proof} 
The left-hand side of the former equation is $\left[ D + \ell_{p_1}, \ell_{p_2} \right] = \left[ D, \ell_{p_2} \right] + \left[ \ell_{p_1}, \ell_{p_2} \right] = I + \left[ \ell_{p_1}, \ell_{p_2} \right]$. Clearly $\left[ \ell_{p_1}, \ell_{p_2} \right]= 0$, which completes the proof. 
\end{proof}
Next lemma shows that there are infinitely many non-linear dynamics or integer sequences satisfying the Stroock-Lie bracket lemma for the arithmetic derivative.
\begin{lemma} \label{nonlinear} 
Let $p_1$, $p_2$ and $q$ be three different prime numbers. And let $\sigma(q^q) = p_1 q^q$, $\sigma(n) = p_2 n$, for $n \neq q^q$. Then \ $[D, \sigma ] = I$ and $\sigma$ is not linear.
\end{lemma}
\begin{proof} 
For $n \neq q^q$, 
$$D \sigma(n) = (p_2 n)' = n + p_2 (n') = n + \sigma(n).$$ 
On the other hand, by definition we have $D (q^q) = q^q$ if $q$ is a prime. Hence, for $n = q^q$, $D \sigma(q^q) = (p_1 q^q)' = q^q + p_1 (q^q)' = q^q + \sigma(q^q)$,  proving that $[D, \sigma]=I$. Clearly $\sigma$ is not linear, since $\sigma ( 2q^q ) =p_2(2q^q)\neq 2q^q=2\sigma (q^q)$.
\end{proof}

\subsection{Lie bracket properties for other derivatives}
Following Ufnarovski and \AA hlander \cite{UB}, we define 
the generalized arithmetic derivative by
\[ D \left( x \right) = x \sum\limits_{i = 1}^k \dfrac{x_i D \left( p_i \right)}{p_i}\;, \text{ where } 
x = \prod_{i = 1}^k p_i^{x_i}. \]
Note that from this definition $D \left( p \right) = 1$ is no longer true for prime $p$,  in general. Then the restriction on the linear function $\ell_m$ in  \eqref{lmchar} can be weakened to $m$ such that $D(m) = 1$. 
\subsubsection{Partial arithmetic derivative $D_p$}
The function $D_p$, defined in Lemma \ref{charD},  is originated in Kovi\v c \cite{JK} and referred to as the partial arithmetic derivative. Notice that $D_p=D_{\{p\}}=D$ is a generalized arithmetic derivative defined by $D(p) = 1$ and $D(q) = 0$ for all other primes $q$. 
\begin{lemma} \label{curioslie}
Let $p$, $q$ be two different primes. Then
\begin{itemize}
\item[(i)] $[ D_p, \ell_p ] = I$;
\item[(ii)] $[ D_p, \ell_q ] = 0$;
\item[(iii)] $[ D_p, \ell_p + \ell_q ] = 0$;
\item[(iv)] $[ D_p+D_q, \ell_p + \ell_q ] = 0$. 
\end{itemize}
\end{lemma}
\begin{proof} 
The first two properties follow from Lemma \ref{charD} (i). By the Leibnitz rule for $D_p$ and since $p \nmid p+q$, we have $[ D_p, \ell_p + \ell_q ] = 0$. This shows that $D_p$ is not linear and proves (iii). Finally, the last property (iv) follows by the linearity of the Lie bracket. 
\end{proof}
\subsubsection{General arithmetic derivative $D_A$}
Now consider the general arithmetic derivative $D_A$ defined in Lemma \ref{charD}. Notice that for $A\neq \emptyset$, the derivative
$D_A$ is not linear. Motivated by Haukkanen et al.~\cite{HJ}, we derive the following properties on $D_A$.
\begin{lemma} \label{simple} 
Let $A, A_i$, $i = 1, 2, \ldots$ be nonempty subsets of primes. Then 
\begin{itemize}
\item[(i)] $[D_{A_1}, D_{A_2}] = 0$ if and only if $A_1 = A_2$;

\item[(ii)] Let $n = \prod_{p\in \cap A_i} p^{n_p}$. Then $D_{A_i} (n) = D_{A_j} (n)$, for all $i,j$.

\item[(iii)] Let $n = p^j$ where $p$ is a prime. Assume that all prime divisors  of $2, 3, \ldots, j$ are not in $\bigcup_i A_i$. Then, for any positive integers $k$ and $i_1,\ldots, i_k$, we have
\[ D_{A_{i_1}}  \cdots D_{A_{i_k}} (n) =D_p^k (n). \]
\end{itemize}
\end{lemma} 
\begin{proof} 
It follows by direct calculations.
\end{proof}
By inspecting the proof of Lemma \ref{curioslie}, we extend  it to the general arithmetic derivatives.
\begin{lemma} 
Let $p \in A$, $q \in B$ and $A \perp B$. Then
\begin{itemize}
\item[(i)] $[ D_A, \ell_p ] = I$;
\item[(ii)]  $[ D_A, \ell_q ] = 0$;
\item[(iii)] $[ D_A, \ell_p + \ell_q ] = 0$;
\item[(iv)] $[ D_A + D_B, \ell_p + \ell_q ] = 0$.
\end{itemize}
\end{lemma}
\section{Extensions of the Stroock-Lie bracket lemma to other rings}
Motivated by Ufnarovski and \AA hlander \cite{UB} and Haukkanen et al.~\cite{H}, we discuss the extensions of Lemma \ref{charD} to commutative rings with the unique factorization property.


\subsection{Extensions to polynomial rings}

Consider a polynomial ring $K [\mathbb{C}]$, which is a unique factorization domain. By $F_i$ denote monic irreducible polynomials, i.e.,
single-variable polynomials with leading coefficients $1$. Any  $F \in K [\mathbb{C}]$ admits  the  unique factorization $F = z \prod_{i = 1}^{k} F_i^{n_i}$, where $n_i \in \mathbb{N}$ and $z \in \mathbb{C}$. Following Ufnarovski and \AA hlander \cite{UB}, define the derivative of polynomials as
\[ D F = F \sum\limits_{i = 1}^k \dfrac{n_i}{F_i} \]
and $D \left( z \right) = 0$, for $z \in \mathbb{C}$. Since $D \left( z F \right) = z D \left( F \right) = z F \sum_{i = 1}^k \left( n_i / F_i \right)$, it 
follows that $D$ satisfies the Leibnitz rule. Notice that $D \left( G \right) = 1$ if and only if $G$ is a monic irreducible polynomial. Consider a linear functional $\ell = \ell_G (H) = GH$, where $G,H \in K [\mathbb{C}]$. Then, by direct calculations similar to (\ref{lmchar}), the Stroock-Lie bracket identity holds $[ D, \ell_G ] = I$ if and only if $D (G) = 1$, i.e., $G$ is the monic irreducible polynomial. Moreover, the sum of $k$ monic irreducible polynomials is an irreducible polynomial with the leading coefficient  $k$.  Therefore, similar to Corollary \ref{mprime}, we then derive the following result. 
\begin{corollary}
Let $F_i,i=1,\ldots, k$, be the monic irreducible polynomials in the polynomial ring $K [\mathbb{C}]$. 
Then $ [D, \ell_{F_1} + \cdots + \ell_{F_k} ] = kI$.
\end{corollary}

\subsection{Extensions to integers and rational numbers}
Following Ufnarovski and \AA hlander \cite{UB}, we define 
$$D \left( x \right) = x \sum_{i = 1}^k \frac{x_i}{p_i},$$ 
where $0<x = \prod_{i = 1}^k p_i^{x_i} \in \mathbb{Q}$,\  $p_i$ are different primes, and $x_i$ are integers.  Then, for $0>x\in\mathbb{Q}$, we take $D(x)=-D(x)$.

The function $D$ is a map from $\mathbb{Q}$ to $\mathbb{Q}$. As proved in  Ufnarovski and \AA hlander \cite{UB}, the map $D$ satisfies the
Leibnitz rule. In particular, the Stroock-Lie bracket  identity $[D,\ell_x]=I$ holds if and only if $D(x)=1$. In this case, $x$ does not need to be a prime, for example one can take $x=-5/4$ (see \cite{UB}).



\section{Arithmetic type differential equations}
\subsection{First order linear arithmetic type differential equations}
We begin by considering several cases of the arithmetic type differential equations.
\begin{lemma} \label{extra1} 
Let $A$ be a nonempty set of primes and $x$ be a positive integer. Then 
\begin{itemize}
\item[(i)] $D_A(x)=0$ if and only if $x \perp A$;

\item[(ii)] $D_A(x)=1$ if and only if $x$ is a prime in $A$;

\item[(iii)] $D_A(x)=x$ if and only if $x=p^pk$ for 
$p\in A$ and a positive integer $k \perp A$;

\item[(iv)] $pD_A(x)=x$,  where $p\in A$, if and only if $x=pk$ for a positive integer $k\perp A$.
\end{itemize}
\end{lemma}
\begin{proof} 
\begin{itemize}
\item[(i)] Notice that $D_A=\sum_{p\in A}D_p$, $D_p(x)\geq 0$. Haukkanen et 
al.~\cite[Theorem 1]{HJ} proved that $D_p(x)=0$ if and only if $p \nmid x$. Thus, $D_A=0$ if and only if $x\perp A$.

\item[(ii)] For the above representation, $D_A(x)=1$ if and only if there exists $p\in A$ such that $D_p(x)=1$, which is equivalent to $x=p$.

\item[(iii)] We follow arguments in the proofs of \cite[Theorem 4 and 5]{UB}.  Assume that $x=p^jk$, where $p\nmid k$ and $p\in A$. Then $D_A(x)=p^{j-1}(jk+pD_A(k))$. And so, if  $0<j<p$, then $D_A(x)=p^{j-1}\tilde{k}$, where $p\nmid \tilde{k}$, implying that $D_A(x)\neq x$. On the other hand, assume that $x=p^p k$ with $k>1$. Then 
$$D_A(x)=p^{p}(k+D_A(k))=p^pk$$
if and only if $D_A(k)=0$, which holds by (i) if and only if $k\perp A$.

\item[(iv)] Clearly $p \vert x$. Let $x = np$, $n \in \mathbb{N}$. By assumption $p\in A$ and, hence, $D_A(p)=1$. By the Leibnitz rule, we obtain
\[ pD_A(x)=pD_A(np) = p(p D_A(n) + n) \geqslant pn = x, \]
where the equality holds if and only if $D_A(n) = 0$. By (i),  then $n\perp A$, proving (iv). 
\end{itemize}
\end{proof} 
\subsection{Lie bracket arithmetic type differential equations}
Fix a nonempty subset $A$ of primes and consider the following equation
\[ x=\left[ D_A, \ell_x \right] (a) = D_A(x) a, \]
where $a,x$ are positive integers. 

In the next lemma, we characterize those pairs $(a,x)$ that satisfy the equation.
\begin{lemma} \label{eqDA}
The arithmetic differential equation $a D_A(x) = x$ has a solution in natural numbers $a,x$ if and only if one of the following statements is satisfied:
\begin{itemize}
\item[(i)] $a = p$ and $x=kp$;

\item[(ii)] $a =1$ and $x=kp^p$,
\end{itemize}
where $p$ is a prime in $A$ and $k\perp A$.
\end{lemma}
\begin{proof} 
To identify the pair $(a,x)$, we first determine $a$, and then solve the equation $a D_A(x) = x$. First, assume $a=1$. The equation becomes $D_A(x) = x.$ According to property (iii) of Lemma \ref{extra1}, then $x=p^pk$, where $p$ is a prime in $A$ and $k\perp A$. 

Next, assume $a=p$ with $p\in A$. Then the equation becomes $pD_A(x) = x$. By property (iv) of Lemma \ref{extra1}, we then have $x=pk$,
where $p$ is a prime in $A$ and $k\perp A$.  

Now assume that $a \neq 1$. We show that  $a=p$, for some $p\in A$. Firstly, consider a subcase $a\perp A$. Then the equation becomes $aD_A(x)=x$. Hence, 
\[ x=a D_A(x)=aD_A[aD_A(x)]=a^2D_A^2(x)\;.\]
For any $j$, we then derive $x=a^jD_A^j (x)$, which is not possible. Hence, $a=pc$, for some $p\in A$ and $c\geq 1$. The equation becomes $pc D_A(x)=x$ and it remains to show that $c=1$. We proceed by absurd and assume that $c>1$. Let $D_A(x)=b$. Then
\[ pc D_A(x) = pcD_A(pcb)=pc(cb +p D_A(cb))>pcb=x, \]
because $c>1$ and $D_A(cb)\geq 0$.  Thus, the statement $x=pc D_A(x)$ is not possible. Hence, $c=1$ and the proof is complete. 
\end{proof}

\section{Acknowledgments}

We are grateful to the Editor-in-Chief and the referee for their valuable
comments and suggestions.

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\noindent
2000 Mathematics Subject Classification: Primary 11A25; Secondary
11A41, 11R27.

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\noindent {\it Keywords:} 
arithmetic derivative, unique factorization, prime number, Leibnitz rule, integer sequence.

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\vspace*{+.1in}
\noindent
Received July 26 2019;
revised version received February 19 2020.
Published in {\it Journal of Integer Sequences}, February 19 2020.

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