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\begin{center}
\vskip 1cm{\LARGE\bf Can the Arithmetic Derivative be Defined on \\
\vskip .1in
a Non-Unique Factorization Domain?}
\vskip 1cm
\large
Pentti Haukkanen, Mika Mattila, and Jorma K. Merikoski \\
School of Information Sciences \\
FI-33014 University of Tampere\\ 
Finland\\
\href{mailto:pentti.haukkanen@uta.fi}{\tt pentti.haukkanen@uta.fi} \\
\href{mailto:mika.mattila@uta.fi}{\tt mika.mattila@uta.fi} \\
\href{mailto:jorma.merikoski@uta.fi}{\tt jorma.merikoski@uta.fi} \\
\ \\
Timo Tossavainen\\
School of Applied Educational Science and Teacher Education\\ 
University of Eastern Finland\\
P.O. Box 86 \\
FI-57101 Savonlinna \\
Finland \\
\href{mailto:timo.tossavainen@uef.fi}{\tt timo.tossavainen@uef.fi}
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\begin{abstract}
Given $n\in\mathbb{Z}$, its arithmetic derivative $n'$ is defined as follows: (i)~$0'=1'=(-1)'=0$. (ii)~If $n=up_1\cdots p_k$, where $u=\pm 1$ and $p_1,\dots,p_k$ are primes (some of them possibly equal), then
$$
n'=n\sum_{j=1}^k\frac{1}{p_j}=u\sum_{j=1}^kp_1\cdots p_{j-1}p_{j+1}\cdots p_k.
$$
An analogous definition can be given in any unique factorization domain. What about the converse? Can the arithmetic derivative be (well-)defined on a non-unique factorization domain? In the general case, this remains to be seen, but we answer the question negatively for the integers of certain quadratic fields. We also give a sufficient condition under which the answer is negative.
\end{abstract}

\section{The arithmetic derivative}

Let $n\in\mathbb{Z}$. Its arithmetic derivative $n'$ 
(\seqnum{A003415} in \cite{Sl}) 
is defined~\cite{Ba,UA} as follows: 

\medskip
\noindent
(i)~$0'=1'=(-1)'=0$. 

\medskip
\noindent
(ii)~If $n=up_1\cdots p_k$, where $u=\pm 1$ and $p_1,\dots,p_k\in\mathbb{P}$, the set of primes, (some of them possibly equal), 
then
\begin{eqnarray}
\label{zder}
n'=n\sum_{j=1}^k\frac{1}{p_j}=u\sum_{j=1}^kp_1\cdots p_{j-1}p_{j+1}\cdots p_k.
\end{eqnarray}
If $k=1$,  we set $p_1\cdots p_{k-1}p_{k+1}\cdots p_k=1$ in the last expression. 

\medskip
A few basic properties of $n'$ follow: 
\begin{eqnarray*}
\forall p\in\mathbb{P}: p'=1,\qquad\qquad
\\
\forall n\in\mathbb{Z}: (-n)'=-n',\qquad
\\
\forall m,n\in\mathbb{Z}: (mn)'=m'n+mn'.
\end{eqnarray*}
The third equality is called the Leibniz rule.
Moreover, $f(n)=n'$ is the only mapping $\mathbb{Z}\to\mathbb{Z}$ having these properties. For details, see~\cite[Theorems~1 and~13]{UA}.

Kovi\v c~\cite[Proposition~1]{Ko} studied how to extend $f$ to $\mathbb{Q}(i)=\{a+bi\,|\,a,b\in\mathbb{Q}\}$.
Ufnarovski and \AA hlander~\cite[Section~10]{UA} outlined how to define the arithmetic derivative on a unique factorization domain (UFD from now on). We begin by performing this task in detail. To that end, we follow the terminology of~\cite[Section~6.5]{Co}. 

Let $D$ be a UFD.  First, we must decide what atoms (irreducible elements) are \lq\lq positive\rq\rq. Write $\cal P$ for a set of atoms of~$D$ such that every atom of~$D$ is associated with one and only one element of~$\cal P$. Call $\cal P$ the set of positive atoms. Further, denote by~$\cal U$ the set of units of~$D$. 

Given $a\in D$, we define its arithmetic derivative $a'$ as follows: If $a=0$ or $a\in\cal U$, then $a'=0$. Otherwise, there are unique (up to the ordering) $p_1,\dots,p_k\in\cal P$ (some of them possibly equal) and $u\in\cal U$ such that
$$
a=up_1\cdots p_k.
$$
Then
\begin{eqnarray}
\label{ufdder}
a'=u\sum_{j=1}^kp_1\cdots p_{j-1}p_{j+1}\cdots p_k.
\end{eqnarray}

To be precise, we should actually write $a'_{\cal P}$ (or something like that) for the
derivative of $a$, since $a'$ depends on $\cal P$. However, for the simplicity of notation, we will  omit this practice if there is no need to emphasize~$\cal P$. 

The given definition implies the analogous equalities as above:
\begin{eqnarray*}
\forall p\in{\cal P}: p'=1,\qquad
\\
\forall v\in{\cal U},a\in D: (va)'=va',
\\
\forall a,b\in D: (ab)'=a'b+ab'.
\end{eqnarray*}
Again, $f(x)=x'$ is the only mapping $D\to D$ with these properties.

\begin{example}
Let~$D=\mathbb{Z}$. If ${\cal P_\text{1}}=\mathbb{P}$, we obtain the ordinary arithmetic derivative defined above. For example, because $30=2\cdot 3\cdot 5$, we have $30'_{\cal P_\text{1}}=30'=3\cdot 5+2\cdot 5+2\cdot 3=31$. 
Obviously, another selection of positive atoms results in a different derivative function.
For instance, if ${\cal P_\text{2}}=\{2,-3,5,-7,11,\dots\}$, then $30=(-1)\cdot 2\cdot(-3)\cdot 5$ and $30'_{\cal P_\text{2}}=(-1)\cdot[(-3)\cdot 5+2\cdot 5+2\cdot(-3)]=11$.
\end{example}

\begin{example}
Let $D$ be an arbitrary field~$F$. Since all nonzero elements of~$F$ are units, then ${\cal P}=\emptyset$ and, hence, $a'=0$ for all $a\in F$.
\end{example}

\begin{example}
To give an example of a nontrivial derivative on the field~$\mathbb{Q}$, we define \cite[Theorem~14]{UA} 
\begin{eqnarray}
\label{ratioder}
\Big(\frac{m}{n}\Big)'=\frac{m'n-mn'}{n^2}.
\end{eqnarray}
Here $m,n\in\mathbb{Z}$, $n\ne 0$, and $m'$ and $n'$ are ordinary arithmetic derivatives on~$\mathbb{Z}$. An analogous definition can be given in the division field of any UFD.
\end{example}

Let us summarize the above discussion.

\begin{proposition} ${ }$

\medskip
\noindent
$(\mathrm{i})$ Let $D$ be a {\rm UFD}. The mapping $f(a)=a'_{\cal P}$ defined on~$D$ by~{\rm(\ref{ufdder})} is an arithmetic derivative. It depends on the chosen set~$\cal P$ of positive atoms.

\medskip
\noindent
$(\mathrm{ii})$ The mapping $g(a)=a'$ defined on~$\mathbb{Q}$ by~{\rm(\ref{ratioder})} is an extension of the mapping $f(a)=a'$ defined on~$\mathbb{Z}$ by~{\rm(\ref{zder})}. Similarly, {\rm(\ref{ufdder})} can be extended to the division field of~$D$.
\end{proposition}


\section{ A problem and its partial answers}

If a factorization domain (FD in the sequel) is not a UFD, we call it a non-unique factorization domain (NUFD in the sequel). We saw above that the arithmetic derivative can be defined on any UFD. What about the converse?

\begin{problem}
\label{probl}
Is it possible to define an arithmetic derivative on some {\rm NUFD}?
\end{problem}
\noindent
The next theorem gives a partial answer which is negative. In the following,
we mostly apply the same terminology and notation as in \cite[Chapter~4]{ST}.

\begin{theorem}
\label{thm}
Let $D_m$ be the integral domain of integers of~$\mathbb{Q}(\sqrt{m})$, where $m\in\mathbb{Z}\backslash\{1\}$ is squarefree
{\rm(\seqnum{A005117}}  in \rm{\cite{Sl})}.
{\it If $m$ satisfies
\begin{eqnarray}
\label{m}
m\not\equiv 1\,(\mathrm{mod}\,4)\,\mathrm{and}\,m<-2,
\end{eqnarray}
then either $1-m$ or $4-m$  does not have a well-defined derivative as an element of~$D_m$}.
\end{theorem}

\begin{proof}
Clearly, $D_m$ is an FD, yet it is not a UFD, see \cite[p.~93]{ST} or \cite[Theorem (actually, in Finnish: Lause)~4.23]{Va}. We modify and enhance the argument used in the latter reference.

\medskip
\noindent
{\it Case~1}. $m\equiv 3\,(\mathrm{mod}\,4)$. Then $m$ is odd and $m\le -5$. Since $1-m$ is even and greater than five, it is a composite number (when considered as a positive integer) and expressible as
\begin{eqnarray}
\label{factor11}
1-m=p_1\cdots p_k,
\end{eqnarray}
where $k\ge 2$ and $p_1,\dots,p_k\in\mathbb{P}$ with $2=p_1\le\dots\le p_k$. On the other hand,
\begin{eqnarray}
\label{factor12}
1-m=(1-\sqrt{m})(1+\sqrt{m}).
\end{eqnarray}
We will see later that the factors of the right-hand sides of both~(\ref{factor11}) and~(\ref{factor12}) are atoms in~$D_m$. 

Next, if some of the $p_j$'s in~(\ref{factor11}) are equal, we omit their repetition; let $\{p_{j_1},\dots,p_{j_h}\}$ be the set obtained so. Since the only units of~$D_m$ are $\pm 1$, see \cite[Proposition~4.2]{ST} or \cite[Lause~4.8]{Va}, the atoms $p_{j_1},\dots,p_{j_h}$, $1-\sqrt{m}$, $1+\sqrt{m}$ are pairwise non-associated. 

Assume first that $\cal P$ is such that
\begin{eqnarray}
\label{p}
p_{j_1},\dots,p_{j_h},1-\sqrt{m},1+\sqrt{m}\in\cal P.
\end{eqnarray}
 If $(1-m)'$ exists, then, by~(\ref{factor12}),
$$
(1-m)'=1+\sqrt{m}+1-\sqrt{m}=2.
$$
On the other hand, (\ref{factor11}) implies that
$$
(1-m)'=p_2\cdots p_k+\dots+p_1\cdots p_{k-1}\ge p_2+p_1\ge 2+2=4
$$
which contradicts the previous conclusion. So, $(1-m)'$ is not well-defined under~(\ref{p}).

Second, if (\ref{p}) does not hold, we anyway have 
$$
\pm p_{j_1},\dots,\pm p_{j_h},\pm(1-\sqrt{m}),\pm(1+\sqrt{m})\in\cal P
$$
with an appropriate selection of signs. Hence a simple modification of the above argument is
sufficient to show that the derivative of $1-m$ is not well-definable.

\medskip
\noindent
{\it Case~2}. $m\equiv 2\,(\mathrm{mod}\,4)$. Now $m$ is even and $m\le -6$. Thus, $4-m$ is also an even composite number such that $4-m\ge 10$. So, again
\begin{eqnarray}
\label{factor21}
4-m=p_1\cdots p_k,
\end{eqnarray}
where $k$ and $p_1,\dots,p_k$ are as described above. On the other hand,
\begin{eqnarray}
\label{factor22}
4-m=(2-\sqrt{m})(2+\sqrt{m}).
\end{eqnarray}
The factors of the right-hand sides of~(\ref{factor21}) and~(\ref{factor22}) are atoms in~$D_m$, see below. 

We continue similarly as in Case~1 only replacing $1\pm\sqrt{m}$ with $2\pm\sqrt{m}$. 
So, assume first that 
\begin{eqnarray}
\label{p_2}
p_{j_1},\dots,p_{j_h},2-\sqrt{m},2+\sqrt{m}\in\cal P.
\end{eqnarray}
If $(4-m)'$ exists, then
$
(4-m)'=2+\sqrt{m}+2-\sqrt{m}=4
$
by~(\ref{factor22}).
However, we have $k>2$ or $p_k>2$, since otherwise $4-m=2\cdot 2=4$ contradicting the assumption $m\le -6$. By~(\ref{factor21}), we encounter a dilemma in both cases; if $k>2$, then
$$
(4-m)'\ge p_2p_3+p_1p_3+p_1p_2\ge 4+4+4=12,
$$ 
and
$$
(4-m)'\ge
p_{k-1}+p_k\ge 2+3=5
$$
if $p_k>2$. Consequently, $(1-m)'$ is not well-defined under~(\ref{p_2}). If $\cal P$ does not satisfy this condition, an analogous argument as at the end of Case~1 applies again.

\medskip

To complete the proof, we still have to verify that the factors of the right-hand sides of  (\ref{factor11}), (\ref{factor12}), (\ref{factor21}) and (\ref{factor22}) are atoms. We do so by using the norm function.
We begin by noticing that
$D_m=\mathbb{Z}(\sqrt{m})=\{x+y\sqrt{m}\,|\,x,y\in\mathbb{Z}\}$, see \cite[Theorem~3.2]{ST} or \cite[Theorem~4.2]{Va}. An element $a=x+y\sqrt{m}\in D_m$ is rational
if $y=0$ and irrational if $y\ne 0$. If $a$ is irrational, then, recalling that $m$ is negative, we have
$$
N(a)=x^2-my^2=x^2+|m|y^2\ge |m|.
$$
If also $b\in D_m$ is irrational, then
\begin{eqnarray}
\label{ab}
N(ab)=N(a)N(b)\ge m^2.
\end{eqnarray}

Let $c\in D_m$ so that $c\ne 0,\pm 1$. If $c=ab$ where $a$ and $b$ are irrational, then $N(c)\ge m^2$ by~(\ref{ab}). Therefore, $c$ is an atom if the following two conditions are satisfied: (i)~$c$ has no rational atom divisor (except possibly $\pm c$) and (ii)~$N(c)<m^2$.

Now choose any $p_j$ from~(\ref{factor11}) or~(\ref{factor21}). It clearly satisfies the condition~(i). 
Since $|m|\ge 5$, we have
$$
p_j\le p_k=\frac{1}{2}\cdot 2p_k\le\frac{1}{2}\,p_1\cdots p_k
\le \frac{1}{2}(4-m)<\frac{1}{2}(5+|m|)\le\frac{1}{2}\cdot 2|m|=|m|.
$$
Hence $N(p_j)=p_j^2<m^2$ and, consequently, also (ii) is satisfied. In other words, $p_j$ is an atom in~$D_m$.

Next, consider the factors of the right-hand sides of (\ref{factor12}) and~(\ref{factor22}), i.e., $q=t\pm\sqrt{m}$, where $t\in\{1, 2\}$.
If $r,x,y\in\mathbb{Z}$ so that $t\pm\sqrt{m}=r(x+y\sqrt{m})$, then $ry=\pm 1$ implying also that $r=\pm 1$. Consequently, $q$ satisfies~(i).
Also the condition~(ii) is now satisfied because
$$
N(1\pm\sqrt{m})<N(2\pm\sqrt{m})=4-m<5+|m|\le 2|m|<m^2.
$$
So, these numbers are atoms also.
\end{proof}

We conclude this section by stating a sufficient condition under which the answer to Problem~\ref{probl} is negative.

\begin{theorem}
\label{twofact}
Let $D$ be an {\rm NUFD}. Assume that $a\in D$ can be factorized so that
$$
a=p_1p_2=q_1q_2,
$$
where $p_1,p_2,q_1,q_2$ are atoms with $\{p_1,p_2\}\ne\{q_1,q_2\}$. Then $a$ does not have a well-defined derivative.
\end{theorem}

\begin{proof}
If $a'$ exists, then, by the Leibniz rule,
$$
a'=p_1+p_2=q_1+q_2.
$$
But two elements are uniquely defined by their sum and product. To show this, simply note that
\begin{eqnarray*}
\{p_1,p_2\}=\{q_1, q_2\}\iff
\forall x\in D:(x-p_1)(x-p_2)=(x-q_1)(x-q_2)
\\
\iff
\forall x\in D:x^2-(p_1+p_2)x+p_1p_2=x^2-(q_1+q_2)x+q_1q_2\qquad
\\
\iff
p_1+p_2=q_1+q_2\land
p_1p_2=q_1q_2.\qquad\qquad
\end{eqnarray*}
(To show the forward implication in the last equivalence, substitute $x=0$ and $x=1$.) Therefore $p_1+p_2\ne q_1+q_2$ and our claim follows.
\end{proof}

Now we encounter another problem arising from Theorem~{\rm\ref{twofact}}.
\begin{problem}
\label{probltwofact}
Does every {\rm NUFD} contain an element~$a$ satisfying the assumption of Theorem~{\rm\ref{twofact}}?
\end{problem}
\noindent
If the answer to this question is positive, then the answer to Problem~\ref{probl} is negative.

\section{Concluding remarks}

We defined an arithmetic derivative on a UFD and asked in Problem~\ref{probl} about the possibility to do so also in some NUFD.  If $m\not\equiv 1\,(\mathrm{mod}\,4)$ and $m<-2$, the FD of integers of~$\mathbb{Q}(\sqrt{m})$ is an NUFD. We proved in Theorem~\ref{thm} that the answer is negative in this case. 
If $m\equiv 1\,(\mathrm{mod}\,4)$ and $m<0$, then this FD is an NUFD if and only if 
$m\ne -3,-7,-11,-19,-43,-67,-163$, 
see \cite[p.~93]{ST}. 

Surveying these $m$'s might be the next  step. 
However, it is an essentially more laborious task. Namely, assuming 
$m\equiv 1\,(\mathrm{mod}\,4)$ 
implies (\cite[Theorem~3.2]{ST} or \cite[Theorem~4.2]{Va}) that $D_m=\{x+y(1+\sqrt{m})/2\,|\,x,y\in\mathbb{Z}\}$
which already 
complicates the situation compared to that in the proof of Theorem~\ref{thm}.
More difficulties arise as the simple condition $m<-2$ is replaced with
the condition excluding the mentioned values of $m$.
Studying the integers of $\mathbb{Q}(\sqrt{m})$ for $m>1$ may be even more difficult since, according to our knowledge, it is not completely understood which $m$'s yield a UFD and which do not.

Obviously, an alternative way to try to advance is to study NUFD's different from those described above.

\section{Acknowledgment}

We thank the referee for valuable remarks, in particular for those that led us to formulate
Theorem~\ref{twofact} and Problem~\ref{probltwofact}.

\begin{thebibliography}{9}

\bibitem{Ba} E.~J.~Barbeau, Remark on an arithmetic derivative, {\it Canad. Math. Bull.}~\textbf{4} (1961), 117--122.

\bibitem{Co} P.~M.~Cohn, {\it Algebra, Volume~1}, John Wiley, 1974.

\bibitem{Ko} J.~Kovi\v c, The arithmetic derivative and antiderivative, {\it J. Integer Seq.}~\textbf{15} (2012),
\href{https://cs.uwaterloo.ca/journals/JIS/VOL15/Kovic/kovic4.html}{Article~12.3.8}.

\bibitem{Sl} N. J. A. Sloane, {\it The On-Line Encyclopedia of Integer Sequences},
\href{http://oeis.org}{http://oeis.org}.

\bibitem{ST} I.~N.~Stewart and D.~O.~Tall, {\it Algebraic Number Theory}, Second Edition, Chapman and Hall, 1987. 

\bibitem{UA} V.~Ufnarovski and B.~\AA hlander, How to differentiate a number, {\it J. Integer Seq.}~\textbf{6} (2003), 
\href{http://www.cs.uwaterloo.ca/journals/JIS/VOL6/Ufnarovski/ufnarovski.html}{Article 03.3.4}.

\bibitem{Va} K.~V\"ais\"al\"a, {\it Lukuteorian ja korkeamman algebran alkeet} [in Finnish], Otava, 1950.

\end{thebibliography}



%******************
\bigskip
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\noindent 2010 {\it Mathematics Subject Classification:}
Primary 11A25; Secondary 11A51, 11R27.

\noindent {\it Keywords:} 
arithmetic derivative, unique factorization, non-unique factorization,
quadratic field.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences
\seqnum{A000040}, \seqnum{A003415} and \seqnum{A005117}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received  October 30 2012;
revised version received  January 1 2013.
Published in {\it Journal of Integer Sequences},  January 1 2013.

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\noindent
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\htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}.
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