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\begin{center}
\vskip 1cm{\LARGE\bf On Arithmetic Partial Differential Equations}
\vskip 1cm
\large
Pentti Haukkanen 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: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}
\end{center}

\vskip .2 in

\begin{abstract}
Kovi\v c, and implicitly Ufnarovski and \AA hlander, defined a notion of
arithmetic partial derivative.
We generalize the definition for rational numbers and study several
arithmetic partial differential equations of the first and second order. For some equations, 
we give a complete solution, and for others, we extend previously known results. 
For example, we determine under which conditions two consecutive partial derivations are commutative.  
\end{abstract}

\section{Introduction}

Let the symbols $\mathbb{Z}$, $\mathbb{Q}$, and~$\mathbb{R}$ have their
ordinary meaning. We also
write $\mathbb{N}=\{0,1,2,\dots\}$, $\mathbb{Z}_+=\{1,2,3,\dots\}$, and
$\mathbb{P}$ for the set of primes.

Let $a\in\mathbb{Q}\setminus\{0\}$. There
exists a unique sequence of integers (with only finitely many nonzero terms)
$$
(\nu_p(a))_{p\in\mathbb{P}}
$$
such that
\begin{eqnarray}
\label{a}
a=(\mathrm{sgn}\,a)\prod_{p\in\mathbb{P}}p^{\nu_p(a)},
\end{eqnarray}
where sgn denotes the sign function. If $\nu_p(a)\ne 0$, we say that $a$
\emph{is divisible by}~$p$. We use the notation $p\mid a$ (respectively, $p\nmid a$)
when $a$ is divisible by~$p$ (respectively, $a$ is not divisible by~$p$).
The formula (\ref{a}) is also valid for~$a=0$ as we define $\nu_p(0)=0$ for all $p\in\mathbb{P}$.

We define the {\em arithmetic derivative} of~$a\in\mathbb{Q}\setminus\{0\}$ by
$$
a'=
a\sum_{p\in\mathbb{P}}\frac{\nu_p(a)}{p}=\sum_{p\in\mathbb{P}}a'_p,
$$
where
$$
a'_p=\frac{\nu_p(a)}{p}a
$$
is the {\em arithmetic partial derivative} of~$a$ with respect to~$p$.
In particular, $(\pm 1)'=(\pm 1)'_p=0$ for all~$p\in\mathbb{P}$. We also set
$0'=0$ and $0'_p=0$ for all~$p\in\mathbb{P}$. Further, we define
the {\em second arithmetic partial derivative} of $a\in\mathbb{Q}$ with respect
to $p\in\mathbb{P}$ and $q\in\mathbb{P}$ to be
$$
a''_{pq}=(a'_p)'_q.
$$

If $a\in\mathbb{Q}\setminus\{0\}$, there are unique
$\alpha\in\mathbb{Z}$ and $\tilde{a}\in\mathbb{Q}\setminus\{0\}$ such that
\begin{eqnarray}
\label{acp}
a=\tilde{a}p^\alpha
\end{eqnarray}
and $p\nmid\tilde{a}$; in fact, $\alpha=\nu_p(a)$. Then
$$
a'_p=\tilde{a}\alpha p^{\alpha-1}.
$$
For the second partial derivative,
$$
a''_{pp}=\tilde{a}\alpha(\alpha-1)p^{\alpha-2}
$$
if and only if $p\nmid\alpha$.

The starting point for the present study was set by Barbeau~\cite{Ba}, who defined the arithmetic derivative for $a\in\mathbb{N}$.
Ufnarovski and \AA hlander~\cite{UA} generalized this concept for
$a\in\mathbb{Q}$. Among other things, they solved certain arithmetic
differential equations. The present authors and Mattila~\cite{HMMT} studied
whether the arithmetic derivative can be defined on a non-unique factorization domain. 
Mistri and Pandey~\cite{MP} defined the derivative of an ideal (``ideal derivative'')
on a number ring and studied its connections with the arithmetic derivative.
Among other things, they solved certain ideal differential equations.

According to our knowledge, the idea of the arithmetic partial derivative is due to
Kovi\v c~\cite{Ko} for $a\in\mathbb{Z}_+$; yet it was already implicitly used
by Ufnarovski and \AA hlander. Our definition
extends this notion to the set of rational numbers.

The question when
\begin{eqnarray}
\label{chngord}
a''_{pq}=a''_{qp}
\end{eqnarray}
arises. Kovi\v c gave a sufficient condition for $a\in\mathbb{Z}_+$
to satisfy this. In Section~\ref{secndorder}, we will solve the arithmetic partial differential
equation $x''_{pq}=x''_{qp}$, obtaining a necessary and sufficient condition
for $a\in\mathbb{Q}$ to satisfy~(\ref{chngord}).
We will also solve several other partial differential equations:  $x'_p=ax^n$ ($n\in\mathbb{Z}$)
in Sections~\ref{equation} and~\ref{othereq}, and,
extending Kovi\v c's results, the equations
$x'_p=x'_q$, $x''_{pq}=x''_{qp}$,
$x''_{pp}=x$, and $x''_{pp}+x''_{qq}=x$
in Section~\ref{secndorder}. 
In Section~\ref{equationex}, we will discuss some
examples. We will complete
our paper by investigating the cardinality and density of solution sets in
Section~\ref{density} and by making a few concluding remarks in
Section~\ref{remarks}. 

\section{The equation $x'_p=a$}
\label{equation}

Given $p\in\mathbb{P}$ and $a\in\mathbb{Q}$, we consider the equation
\begin{eqnarray}
\label{pdiffeq}
x'_p=a,
\end{eqnarray}
where $x\in\mathbb{Q}$ is unknown. For $a=0$, this holds if and only if
$\nu_p(x)=0$, i.e., $p\nmid x$.

Assuming that $a\ne 0$, we factorize $a=\tilde{a}p^\alpha$
as in~(\ref{acp}). Because $\nu_p(\tilde{a})=0$, we have
$$
(\tilde{a}x)'_p=\tilde{a}x'_p.
$$
So, the equation
\begin{eqnarray}
\label{pdifeq}
x'_p=p^\alpha
\end{eqnarray}
is equivalent to
$$
(\tilde{a}x)'_p=a.
$$
We can therefore solve~(\ref{pdiffeq}) by solving~(\ref{pdifeq}) and multiplying
the solution by~$\tilde{a}$.

Let
\begin{eqnarray}
\label{xx}
x=\tilde{x}p^\xi,
\end{eqnarray}
where
$$
\xi\in\mathbb{Z},\quad p\nmid\tilde{x}\in\mathbb{Q}\setminus\{0\}.
$$
Then
(\ref{pdifeq}) is equivalent to
$$
\tilde{x}\xi p^{\xi-1}=p^\alpha.
$$
Here $\xi\ne 0$, since otherwise $0=p^\alpha$. Therefore,
\begin{eqnarray}
\label{y}
\tilde{x}=\frac{p^{\alpha-\xi+1}}{\xi}
\end{eqnarray}
under the condition
\begin{eqnarray}
\label{ycond}
p\nmid\tilde{x}.
\end{eqnarray}
This condition is satisfied if and only if
\begin{eqnarray}
\label{betacond}
\xi=\beta p^{\alpha-\xi+1},
\end{eqnarray}
where
\begin{eqnarray}
\label{pnmid}
p\nmid\beta\in\mathbb{Z}\setminus\{0\}.
\end{eqnarray}
We determine the applicable values of~$\xi$. Since $\xi\in\mathbb{Z}$,
(\ref{pnmid}) implies that $\alpha-\xi+1\ge 0$, i.e.,
$$
\xi\le\alpha+1.
$$

\medskip

\emph{Case~1.} $\alpha\ge 0$. We study~(\ref{y}) under~(\ref{ycond}).

\emph{Subcase~1a.} $1\le\xi\le\alpha$. 
We go through all possible values of~$\xi$ by
writing
$$
\xi=\alpha-i,\quad i=0,1,\dots,\alpha-1;
$$
then (\ref{betacond}) reads
\begin{eqnarray}
\label{icond}
\alpha-i=\beta p^{i+1}.
\end{eqnarray}
(Actually
$\beta\in\mathbb{Z}_+$ there, but we also want to cover the case
$\alpha&lt;0$.)
We therefore obtain
\begin{eqnarray}
\label{x}
x=\tilde{x}p^\xi=\frac{p^{\alpha-\xi+1}}{\xi}p^\xi=
\frac{p^{\alpha-(\alpha-i)+1}}{\alpha-i}p^{\alpha-i}=
\frac{p^{\alpha+1}}{\alpha-i}.
\end{eqnarray}
Given $i\in\{0,1,\dots,\alpha-1\}$, this is a solution of~(\ref{pdifeq}) if
and only if $i$ satisfies~(\ref{icond}).
Denoting by $d\parallel m$ that $d\mid m$ and $\gcd{(m,m/d)}=1$,
we rewrite~(\ref{icond}) as
$$
p^{i+1}\parallel (\alpha-i).
$$

\emph{Subcase~1b.} $\xi=\alpha+1$. Then $\tilde{x}=1/\xi$
satisfies~(\ref{ycond}) if and only if $p\nmid\xi$. Thus,
$$
x=\tilde{x}p^{\alpha+1}=\frac{p^{\alpha+1}}{\alpha+1}
$$
is a solution of~(\ref{pdifeq}) if and only if $p\nmid(\alpha+1)$.

\emph{Subcase~1c.} $\xi\le -1$. Then $\gamma=-\xi\ge 1$, and
$$
\tilde{x}=\frac{p^{\alpha-\xi+1}}{\xi}=-\frac{p^{\alpha+\gamma+1}}{\gamma}.
$$
Clearly, $p^{\alpha+\gamma+1}&gt;\gamma$, and thus
$p^{\alpha+\gamma+1}\nmid\gamma$. Therefore, $p\mid\tilde{x}$, and
no new solution is obtained. 

\medskip

\emph{Case~2.} $\alpha&lt;0$. For $\xi\le\alpha$, we proceed as in
Subcase~1a. Again, (\ref{ycond}) holds if and only if
(\ref{betacond}) is satisfied. In order to go through all possible values of~$\xi$, we write
$$
\xi=\alpha-i,\quad i=0,1,2,\dots;
$$
then (\ref{icond}) is equivalent to~(\ref{betacond}). Given
$i\in\mathbb{N}$, (\ref{x}) is a solution
if and only if $i$ satisfies~(\ref{icond}). There are only finitely
many such $i$'s, because $p^{i+1}\nparallel (\alpha-i)$ when $i$ is large.
For $\xi=\alpha+1$, proceed as in Subcase~1b.

\medskip

We summarize the above reasoning in the following theorem.

\begin{theorem}
\label{athm}
Let $p\in\mathbb{P}$ and $\alpha\in\mathbb{Z}$. 
We write $I_0=\{0,1,\dots,\alpha-1\}$ for $\alpha&gt;0$, $I_0=\emptyset$
for $\alpha=0$, and $I_0=\mathbb{N}$
for $\alpha&lt;0$. Let also
$$
I=\big\{i\in I_0\,:\,p^{i+1}\parallel (\alpha-i)\big\}.
$$
Then
\begin{eqnarray}
\label{c1a}
x=\frac{p^{\alpha+1}}{\alpha-i}
\end{eqnarray}
is a solution of
$$
x'_p=p^\alpha
$$
for each $i\in I$.
If $p\nmid(\alpha+1)$, then also
\begin{eqnarray}
\label{c1b}
x=\frac{p^{\alpha+1}}{\alpha+1}
\end{eqnarray}
is a solution. All solutions are obtained in this way. In particular {\rm (}$\alpha=0${\rm )}, the only solution of
$x'_p=1$ is $x=p$. The equation
$$
x'_p=0
$$
holds if and only if $p\nmid x$.
\end{theorem}

Next, let us take a closer look at~$I$. In the following, we assume that $\alpha&gt;0$. A simple modification applies to $\alpha&lt;0$.
The case $\alpha=0$ does not require any further work. 

Let
\begin{eqnarray}
\label{alpha}
\alpha=np+r,\quad 0\le r&lt;p.
\end{eqnarray}
\medskip
Then the solution candidates~(\ref{c1a}) are
$$
x=\frac{p^{np+r+1}}{np+r-i},\quad i=0,1,\dots,np+r-1. 
$$ 
To study the validity of~(\ref{icond}), we write
$$
i=mp+k,
$$
where $m=0,1,\dots,n$. If $m&lt;n$, then $k=0,1,\dots,p-1$, and if $m=n$,
then $k=0,1,\dots,r-1$. (So, here $m$ and~$k$ are variables while $p$, $n$,
and~$r$ are constants.) Now
$$
\alpha-i=np+r-(mp+k)=(n-m)p+r-k.
$$
By~(\ref{icond}),
$p$ must divide $(n-m)p+r-k$. Hence, $r=k$, and 
$$
\alpha-i=(n-m)p.
$$

It is convenient to replace $\beta$ with~$h$ in~(\ref{icond}); then
\begin{eqnarray}
\label{mccond}
(n-m)p=hp^{mp+r+1}.
\end{eqnarray}
Since $h\in\mathbb{Z}_+$, $m=n$ cannot hold; Thus, $m&lt;n$, and
the candidates are of the form
$$
x=\frac{p^{np+r+1}}{(n-m)p}=\frac{p^{np+r+1}}{hp^{mp+r+1}}=
\frac{p^{(n-m)p}}{h}.
$$
They are solutions if and only if $m$ and $h$ satisfy~(\ref{mccond}), i.e.,
\begin{eqnarray}
\label{mccnd}
n-m=hp^{mp+r}.
\end{eqnarray}

It turns out that the cases $m=0$ and $m=n-1$ are easy to solve,
but the case $1\le m\le n-2$ is remarkably more difficult. Namely, if $m=0$, then (\ref{mccnd}) implies that $p^r\mid n$ and $p^{r+1}\nmid n$.
Then $h=n/p^r$, and
$$
x=\frac{1}{h}\,p^{np}=\frac{p^r}{n}p^{np}=\frac{p^\alpha}{n}
$$
is a solution. In the case of $r=0$, we can eliminate~$n$ by writing
$$
x=\frac{p^{np}}{n}=\frac{p^{np+1}}{np}=\frac{p^{\alpha+1}}{\alpha}.
$$
For $m=n-1$, (\ref{mccnd}) implies that $hp^{(n-1)p+r}=1$, which is
impossible unless $n=1$, $r=0$, and $h=1$. Then
$\alpha=p$, see Example~\ref{pp}.

If $1\le m\le n-2$, it seems
difficult to find explicit solutions in general. In this paper, we only note that,
if $n&lt;p^p$ and $m&gt;0$, then
$$
hp^{mp}=h(p^p)^m\ge p^p&gt;n.
$$
Hence (\ref{mccnd}) does not hold, and no solution is obtained.

We also record that the candidate~(\ref{c1b}),
$$
x=\frac{p^{np+r+1}}{np+r+1},
$$
is a solution if and only if $p\nmid (np+r+1)$, i.e., $r\ne p-1$.

Including the case $\alpha&lt;0$, we now obtain the following summary.

\begin{theorem}
Let $p\in\mathbb{P}$ and $\alpha\in\mathbb{Z}\setminus\{0\}$.
Express $\alpha$ as in~{\rm (\ref{alpha})}.
If $h\in\mathbb{Z}\setminus\{0\}$ and $m\in\mathbb{Z}$ satisfy~{\rm (\ref{mccnd})} and
\begin{eqnarray}
\label{m1}
h&gt;0\,\,and\,\,0\le m\le n-1\,\,for\,\,\alpha&gt;0,
\\
\nonumber
h&lt;0\,\,and\,\,m\ge 0\,\,for\,\,\alpha&lt;0,\qquad
\end{eqnarray}
then
\begin{eqnarray}
\label{x1}
x=\frac{p^{(n-m)p}}{h}
\end{eqnarray}
is a solution of
$$
x'_p=p^\alpha.
$$
If $r\ne p-1$,
then also
\begin{eqnarray}
\label{x2}
x=\frac{p^{\alpha+1}}{\alpha+1}
\end{eqnarray}
is a solution. All solutions are obtained in this way.
\end{theorem}

\section{The equation $x'_p=ax^n$}
\label{othereq}

Given $p\in\mathbb{P}$, $a\in\mathbb{Q}$, and $n\in\mathbb{Z}$, we now study the
equation
$$
x'_p=ax^n.
$$ 
If $n\ge 1$, then it has
a trivial solution~$x=0$. We have already settled the cases $a=0$ and $n=0$ in
Theorem~\ref{athm}. Next, we consider the case $n=1$.

\begin{theorem}
\label{axthm}
Let $p\in\mathbb{P}$ and $a\in\mathbb{Q}$. The equation
$$
x_p'=ax
$$
has a nontrivial solution if and only if $ap\in\mathbb{Z}$. Then all nontrivial solutions
are of the form
$$
x=cp^{ap},
$$
where $p\nmid c\in\mathbb{Q}\setminus\{0\}$. Conversely, all numbers of this
form are nontrivial solutions.
\end{theorem}
\begin{proof}
Let $x=\tilde{x}p^\xi$ be as in~(\ref{xx}).
Then
$$
x_p'=ax\iff\tilde{x}\xi p^{\xi-1}=a\tilde{x}p^\xi\iff\xi=ap,
$$
and the theorem follows (as we write $c=\tilde{x}$).
\end{proof}

\begin{corollary}
Let $p_1,p_2,\ldots,p_m\in\mathbb{P}$ be different, and let
$a_1,a_2,\ldots,a_m\in\mathbb{Q}$. 
The system of equations 
$$
x_{p_1}'=a_1x,\quad x_{p_2}'=a_2x,\quad\ldots\,,\quad x_{p_m}'=a_mx
$$
has a nontrivial solution if and only if $a_1p_1,a_2p_2,\ldots,a_mp_m\in\mathbb{Z}$.
All nontrivial solutions are then of the form
$$
x=cp_1^{a_1p_1}p_2^{a_2p_2}\cdots p_m^{a_mp_m},
$$
where $p_1,p_2,\ldots, p_m\nmid c\in\mathbb{Q}\setminus\{0\}$.
Conversely, all numbers of this form are nontrivial solutions.
\end{corollary}
\begin{proof}
The claim follows by induction. 
\end{proof}

We tackle the remaining values of~$n$ in the following theorem. Therein
``nontrivial'' is relevant only for $n\ge 2$ because $x=0$
is not a solution whenever $n\le -1$.

\begin{theorem}
\label{axnthm}
Let $p\in\mathbb{P}$, $a\in\mathbb{Q}\setminus\{0\}$, and
$n\in\mathbb{Z}\setminus\{0,1\}$. 
The equation
\begin{eqnarray}
\label{axn}
x_p'=ax^n
\end{eqnarray}
has a nontrivial solution if and only if the following conditions are satisfied:

\smallskip
{\rm (i)}~There is $\xi\in\mathbb{Z}\setminus\{0\}$ such that
\begin{eqnarray}
\label{t5.4a}
\nu_p(\xi)=\nu_p(a)+(n-1)\xi+1 
\end{eqnarray}
and
\begin{eqnarray}
\label{t5.5}
\frac{\nu_q(\xi)-\nu_q(a)}{n-1}\in\mathbb{Z}
\end{eqnarray}
for all $q\in\mathbb{P}\setminus\{p\}$.

\smallskip
{\rm (ii)}~For odd~$n$, $\mathrm{sgn}\,\xi=\mathrm{sgn}\,a$.

\smallskip
Then all nontrivial solutions are of the form
\begin{eqnarray}
\label{axnsol}
x=\Big(\frac{\xi}{ap}\Big)^\frac{1}{n-1},
\end{eqnarray}
where $\xi$ satisfies {\rm (i)} and~{\rm (ii)}.
Conversely, all numbers of this form are nontrivial solutions.
\end{theorem}
\begin{proof}
To solve~(\ref{axn}), we write again $x=\tilde{x}p^\xi$
as in~(\ref{xx}).
Then $\xi\ne 0$ because $x=\tilde{x}$ is not a solution. 
Since
$$
x_p'=ax^n\iff\tilde{x}\xi p^{\xi-1}=a\tilde{x}^np^{n\xi}
\iff\xi=a\tilde{x}^{n-1}p^{(n-1)\xi+1},
$$
the equation
\begin{eqnarray}
\label{betaeq}
\xi=a\tilde{x}^{n-1}p^{(n-1)\xi+1}
\end{eqnarray}
is equivalent to~(\ref{axn}).

If $\xi$ satisfies (i) and~(ii), our task is to find $\tilde{x}$ satisfying~(\ref{betaeq}),
i.e.,
$$
\tilde{x}^{n-1}=\frac{\xi}{ap^{(n-1)\xi+1}}=:y.
$$
By~(i), $(n-1)\,|\,\nu_q(y)$ for all $q\in\mathbb{P}$.
By~(ii), $y&gt;0$ if $n$ is odd. Therefore,
$$
\tilde{x}=y^\frac{1}{n-1}\in\mathbb{Q}\setminus\{0\},
$$
and
$$
x=\tilde{x}p^\xi=\Big(\frac{\xi}{ap^{(n-1)\xi+1}}\Big)^\frac{1}{n-1}p^\xi=
\Big(\frac{\xi}{ap}\Big)^\frac{1}{n-1}
$$
is a nontrivial solution of~(\ref{axn}).

Conversely, if $x$ satisfies~(\ref{axn}), then (\ref{betaeq}) holds.
Hence,
$$
\nu_p(\xi)=\nu_p(a)+(n-1)\xi+1,
$$
i.e.,~(\ref{t5.4a}) follows. For $q\in\mathbb{P}\setminus\{p\}$,
$$
\nu_q(\xi)=\nu_q(a)+(n-1)\nu_q(\tilde{x}),
$$
and
$$
\nu_q(\tilde{x})=\frac{\nu_q(\xi)-\nu_q(a)}{n-1}.
$$
Since $\nu_q(\tilde{x})\in\mathbb{Z}$, (\ref{t5.5}) follows. If $n$ is odd, then
$\tilde{x}^{n-1}&gt;0$. Hence, by~(\ref{betaeq}), $\xi$ and~$a$ have the same
signs, which verifies~(ii).
\end{proof}

\begin{corollary}
\label{axncor1}
Let $p\in\mathbb{P}$, $p\nmid a\in\mathbb{Q}\setminus\{0\}$, and
$n\in\mathbb{Z}\setminus\{0,1\}$. The equation
$$
x_p'=ax^n
$$
has a nontrivial solution if and only if $n=2$. The solution is
\begin{eqnarray}
\label{xxxxx}
x=-\frac{1}{ap}.
\end{eqnarray}
\end{corollary}
\begin{proof}
By Theorem~\ref{axnthm},   
\begin{eqnarray}
\label{xxxx}
x=\Big(\frac{\xi}{ap}\Big)^\frac{1}{n-1},
\end{eqnarray}
where $\xi\in\mathbb{Z}\setminus\{0\}$ satisfies (i) and (ii).
Let
\begin{eqnarray}
\label{bgamma}
\xi=\tilde{\xi}p^\gamma,
\end{eqnarray}
where
\begin{eqnarray}
\label{gammapxi}
\gamma\in\mathbb{N},\quad
p\nmid\tilde{\xi}\in\mathbb{Z}\setminus\{0\}.
\end{eqnarray}
Then (\ref{t5.4a}) reduces to
$$
\gamma=(n-1)\tilde{\xi}p^\gamma+1 
$$
or, equivalently,
$$
\tilde{\xi}=\frac{\gamma-1}{(n-1)p^\gamma}.  
$$
To complete the proof, we study which values of~$\gamma$ apply.

\medskip

{\em Case~1.} $\gamma=0$. Then $\tilde{\xi}=-(n-1)^{-1}\in\mathbb{Z}$ if and
only if $n-1=\pm 1$. Since $n\ne 0,1$, necessarily $n=2$. Hence,
$\tilde{\xi}=-1$ implying also $\xi=-1$. To check~(\ref{t5.5}), now
$$
\frac{\nu_q(\xi)-\nu_q(a)}{n-1}=\frac{0-\nu_q(a)}{2-1}=-\nu_q(a)
\in\mathbb{Z}
$$
for all $q\in\mathbb{P}$.

\medskip

{\em Case~2.} $\gamma=1$. Then $\tilde{\xi}=0$,
contradicting~(\ref{gammapxi}).

\medskip

{\em Case~3.} $\gamma\ge 2$. Then
$$
|n-1|p^\gamma\ge p^\gamma&gt;\gamma-1&gt;0,
$$
so $0&lt;|\tilde{\xi}|&lt;1$, again contradicting~(\ref{gammapxi}).

\medskip

Substituting $n=2$ and $\xi=-1$ in~(\ref{xxxx}) yields~(\ref{xxxxx}).
\end{proof}

\begin{corollary} 
\label{axncor2}
Let $p\in\mathbb{P}\setminus\{2\}$ and
$n\in\mathbb{Z}\setminus\{0,1\}$. The equation
$$
x_p'=px^n
$$
has a nontrivial solution if and only if $n=2$ or $n=-1$. The solution is
$$
x=-\frac{2}{p^2}
$$
in the first-mentioned case and
\begin{eqnarray}
\label{xp}
x=p
\end{eqnarray}
in the second one. For $p=2$, only {\rm (\ref{xp})} holds.
\end{corollary}
\begin{proof}
With the above notation, (\ref{t5.4a}) reads $\gamma=1+(n-1)\tilde{\xi}p^\gamma+1$,
i.e.,
$$
\tilde{\xi}=\frac{\gamma-2}{(n-1)p^\gamma}.
$$

\medskip

{\em Case~1.} $\gamma=0$. Then $\tilde{\xi}=-2(n-1)^{-1}\in\mathbb{Z}$ if and
only if $n-1=\pm 1$ or $n-1=\pm 2$. Since $n\ne 0,1$, necessarily $n=2$, $n=3$, or $n=-1$.
We have $\tilde{\xi}=-2$ (for $p\ne 2$), $\tilde{\xi}=-1$, and $\tilde{\xi}=1$, respectively.
Further, $\xi=\tilde{\xi}$. The validity
of~(\ref{t5.5}) is easily verified. However, (ii) in
Theorem~\ref{axnthm} is violated for~$n=3$.

\medskip

{\em Case~2.} $\gamma=1$. Then
$$
\tilde{\xi}=-\frac{1}{(n-1)p}\notin\mathbb{Z}.
$$

\medskip

{\em Case~3.} $\gamma=2$. As Case~2 in the previous proof.

\medskip

{\em Case~4.} $\gamma\ge 3$. As Case~3 in the previous proof.

\medskip

Thus, $n=2$, $\xi=-2$ (for $p\ne 2$), and $n=-1$, $\xi=1$ remain. Substituting them
in~(\ref{axnsol}) yields the claims of the corollary.
\end{proof}

\section{Equations involving $x'_p$ and $x'_q$ or their derivatives}
\label{secndorder}

In this section, we extend Kovi\v c's results from $x\in\mathbb{Z}_+$
to $x\in\mathbb{Q}$ (and revise some of them). We begin by
\cite[Proposition~34]{Ko}.

\begin{theorem}
\label{xpxqthm}
Let $p,q\in\mathbb{P}$, $p\ne q$. All nontrivial solutions of
\begin{eqnarray}
\label{xpxqeq}
x'_p=x'_q
\end{eqnarray}
are of the form
$$
x=cp^{kp}q^{kq},
$$
where
$$
k\in\mathbb{Z},\quad p,q\nmid c\in\mathbb{Q}\setminus\{0\}.
$$
Conversely, all numbers of this form are nontrivial solutions.
\end{theorem}
\begin{proof}
Let
\begin{eqnarray}
\label{xxieta}
x=\tilde{x}p^\xi q^\eta,
\end{eqnarray}
where
$$
\xi,\eta\in\mathbb{Z},\quad
p,q\nmid\tilde{x}\in\mathbb{Q}\setminus\{0\}.
$$
Then (\ref{xpxqeq}) is equivalent to
$$
\tilde{x}\xi p^{\xi-1}q^{\eta}=\tilde{x}\eta p^{\xi}q^{\eta-1}, 
$$
i.e.,
\begin{eqnarray}
\label{xiq}
\xi q=\eta p.
\end{eqnarray}
Hence, $p\mid\xi$,
implying that $\xi=kp$ for some $k\in\mathbb{Z}\setminus\{0\}$. Similarly,
$\eta=hq$ for some $h\in\mathbb{Z}\setminus\{0\}$. Because
$kpq=hqp$ by~(\ref{xiq}), we have $h=k$, and
the theorem follows (as we write $c=\tilde{x}$).
\end{proof}

\begin{corollary}
Let $p_1,p_2,\dots,p_m\in\mathbb{P}$ be different. Then all
nontrivial solutions of 
$$
x'_{p_1}=x'_{p_2}=\dots=x'_{p_m}
$$
are of the form
$$
x=cp_1^{kp_1}p_2^{kp_2}\cdots p_m^{kp_m},
$$
where $k\in\mathbb{Z}$ and
$p_1,p_2,\dots,p_m\nmid c\in\mathbb{Q}\setminus\{0\}$.
Conversely, all numbers of this form are nontrivial solutions.
\end{corollary}
\begin{proof}
The claim follows by induction.
\end{proof}

Let $p,q\in\mathbb{P}$. Kovi\v c~\cite[Proposition~33]{Ko} proved for
$a\in\mathbb{Z}_+$ that if
$\gcd{(\nu_p(a),q)}=\gcd{(\nu_q(a),p)}=1$, then
\begin{eqnarray}
\label{changord}
a''_{pq}=a''_{qp}.
\end{eqnarray}
We next show for $a\in\mathbb{Q}$ that (\ref{changord}) holds if and only if
$$
\nu_q(\nu_p(a))\nu_p(a)=\nu_p(\nu_q(a))\nu_q(a).
$$

\begin{theorem}
\label{xpqthm}
Let $p,q\in\mathbb{P}$, $p\ne q$. All nontrivial solutions of
\begin{eqnarray}
\label{xpqeq}
x''_{pq}=x''_{qp}
\end{eqnarray}
are of the form
$$
x=cp^hq^k,
$$
where
$$
h,k\in\mathbb{Z},\quad p,q\nmid c\in\mathbb{Q}\setminus\{0\}
$$
and
$$
h\nu_q(h)=k\nu_p(k).
$$
Conversely, all numbers of this form are nontrivial solutions.
\end{theorem}
\begin{proof}
Let $x$ be as in~(\ref{xxieta}) and
$$
\xi=\tilde{\xi}p^\alpha q^\beta,\quad
\eta=\tilde{\eta}p^\gamma q^\delta,
$$
where
$$
\alpha,\beta,\gamma,\delta\in\mathbb{N},\quad
p,q\nmid\tilde{\xi},\tilde{\eta}\in\mathbb{Z}\setminus\{0\}.
$$
Then
$$
x'_p=\tilde{x}\xi p^{\xi-1}q^\eta=\tilde{x}\tilde{\xi}p^\alpha q^\beta p^{\xi -1}q^\eta=
\tilde{x}\tilde{\xi}p^{\xi+\alpha-1}q^{\eta+\beta}
$$
and
\begin{eqnarray*}
x''_{pq}=\tilde{x}\tilde{\xi}(\eta+\beta)p^{\xi+\alpha-1}q^{\eta+\beta-1}=
\tilde{x}\tilde{\xi}(\tilde{\eta}p^\gamma q^\delta+\beta)p^{\xi+\alpha-1}q^{\eta+\beta-1}=
\\
\tilde{x}\tilde{\xi}\tilde{\eta}p^{\xi+\alpha+\gamma-1}q^{\eta+\beta+\delta-1}
+\tilde{x}\tilde{\xi}\beta p^{\xi+\alpha-1}q^{\eta+\beta-1},
\end{eqnarray*}
and similarly
$$
x''_{qp}=\tilde{x}\tilde{\xi}\tilde{\eta}p^{\xi+\alpha+\gamma-1}q^{\eta+\beta+\delta-1}
+\tilde{x}\tilde{\eta}\gamma p^{\xi+\gamma-1}q^{\eta+\delta-1}.
$$
Therefore, (\ref{xpqeq}) holds if and only if
$$
\tilde{\xi}\beta p^{\xi+\alpha-1}q^{\eta+\beta-1}=
\tilde{\eta}\gamma p^{\xi+\gamma-1}q^{\eta+\delta-1}.
$$
Substituting $\tilde{\xi}=\xi p^{-\alpha}q^{-\beta}$ and
$\tilde{\eta}=\eta p^{-\gamma}q^{-\delta}$, we obtain
$$
\xi\beta=\eta\gamma,
$$
and the theorem follows (as we write $c=\tilde{x}$, $h=\xi$, $k=\eta$,
$\nu_q(h)=\beta$, $\nu_p(k)=\gamma$).
\end{proof}

Next, we extend \cite[Proposition~35(i)]{Ko}. (In this proposition,
$n=p^{pk}c$ should actually read $n=p^pc$.)

\begin{theorem}
\label{xppthm} 
Let $p\in\mathbb{P}$. All nontrivial solutions of
\begin{eqnarray}
\label{xppeq}
x_{pp}''=x
\end{eqnarray}
are of the form  
\begin{eqnarray}
\label{xppsol}
x=cp^{\pm p}, 
\end{eqnarray}
where $p\nmid c\in\mathbb{Q}\setminus\{0\}$. Conversely, all numbers of this
form are nontrivial solutions. 
\end{theorem}
\begin{proof}
Let $x=\tilde{x}p^\xi$ as in~(\ref{xx}).
Also, let $\xi=\tilde{\xi}p^\gamma$ as in~(\ref{bgamma}).
Then
$$
x_p'= \tilde{x}\xi p^{\xi-1}=\tilde{x}\tilde{\xi}p^{\xi+\gamma-1}
$$ 
and
\begin{eqnarray}
x_{pp}''
=\tilde{x}\tilde{\xi}(\xi+\gamma-1)p^{\xi+\gamma-2}=
xp^{-\xi}\tilde{\xi}(\tilde{\xi}p^\gamma+\gamma-1)p^{\xi+\gamma-2}=
x\tilde{\xi}p^{\gamma-2}(\tilde{\xi}p^\gamma+\gamma-1).
\label{xpp}
\end{eqnarray}
Therefore, (\ref{xppeq}) is equivalent to 
\begin{eqnarray}
\label{pgammab}
\tilde{\xi}p^\gamma(\tilde{\xi}p^\gamma+\gamma-1)=p^2
\end{eqnarray}
(or $x=0$). Since $\tilde{\xi}\mid p^2$ but $p\nmid\tilde{\xi}$, necessarily
$\tilde{\xi}=\pm 1$ and (\ref{pgammab}) reduces to
$$
p^\gamma[p^\gamma\pm(\gamma-1)]=p^2.
$$
If $\gamma=0$, then $1\mp 1=p^2$, which is impossible.
If $\gamma=2$, then $p=0$ or $p=\sqrt{2}$, which is again impossible.
If $\gamma\ge 3$, then $p^3\mid p^2$; this is also impossible.
But $\gamma=1$ works, giving us $\xi=\pm p$. 
Writing $c=\tilde{x}$, we obtain (\ref{xppsol}).
\end{proof}

Let $p,q\in\mathbb{P}$, $p\ne q$. According to Kovi\v c
\cite[Proposition~35(iii)]{Ko}, the only integer solutions of $x''_{pp}+x''_{qq}=x$
are $x=cp^p$ and $x=cq^q$, where $\gcd{(c,pq)}=1$. We extend
this theorem to~$\mathbb{Q}$, discovering that other integer solutions also exist.
The question about the complete solution, however, remains open.
 
\begin{theorem}
\label{xppxqqthm}
Let $p,q\in\mathbb{P}$, $p\ne q$. 
If 
$$
x\in\{cp^{\pm p}, cq^{\pm q},cp^{\pm p}q,cpq^{\pm q}\}, 
$$
where $p,q\nmid c\in\mathbb{Q}$, then
$$
x_{pp}''+x_{qq}''=x. 
$$
\end{theorem}
\begin{proof}
A simple calculation.
\end{proof}

Finally, we extend \cite[Proposition~35(ii)]{Ko}. (The proof of this
proposition is incorrect in \cite{Ko}. For example,
$(ep^{e-1}c)'_p=e(e-1)p^{e-2}c$ is not generally valid because $e$ may be divisible by~$p$.)

\begin{theorem}
Let $p\in\mathbb{P}$, $\alpha\in\mathbb{Z}_+$, and $c\in\mathbb{Q}$.
If $\alpha\ge p$ and $p\nmid c$ or $c=0$, then $x=cp^\alpha$
satisfies
$$
|x_{pp}''|\ge |x|. 
$$
\end{theorem}
\begin{proof}
Omitting the trivial case, we assume that $c\ne 0$. If $p\nmid\alpha$, then $\alpha\ge p$ implies $\alpha\ge p+1$,
and 
$$
|x_{pp}''|=|c|\alpha(\alpha-1)p^{\alpha-2}
\ge |c|(p+1)pp^{\alpha-2}&gt;|c|p^{\alpha}=|x|.
$$
If $p\mid\alpha$, we write $\alpha=\tilde{\alpha}p^\gamma$ where
$\gamma,\alpha\in\mathbb{Z}_+$
and $p\nmid\tilde{\alpha}$. Then 
$$
x_p'=c\alpha p^{\alpha-1}=c\tilde{\alpha}p^\gamma p^{\alpha-1}
=c\tilde{\alpha}p^{\gamma+\alpha-1} 
$$
and 
\begin{eqnarray*}
x_{pp}''=c\tilde{\alpha}(\gamma+\alpha-1)p^{\gamma+\alpha-2}=
c\alpha p^{-\gamma}(\gamma+\alpha-1)p^{\gamma+\alpha-2}
=
\\
c\alpha(\gamma+\alpha-1)p^{\alpha-2}=\alpha(\gamma+\alpha-1)p^{-2}x. 
\end{eqnarray*}
Since
$$
\alpha (\gamma+\alpha-1)p^{-2}\ge p(1+p-1)p^{-2}=1,
$$
the theorem follows.
\end{proof}

\section{Examples}
\label{equationex}

\begin{example}
\label{sec2}
$x'_p=a$, where $p\nmid a\in\mathbb{Q}\setminus\{0\}$.

\medskip
\noindent
In~(\ref{acp}), $\tilde{a}=a$ and $\alpha=0$. The only solution of~(\ref{pdifeq})
is $x=p$ by~(\ref{c1b}). Therefore, the only solution is
$$
x=ap.
$$
\end{example}

\begin{example}
$x'_p=p\,\,(\alpha=1,n=0,r=1)$.

\medskip
\noindent
There is no~$m$ satisfying~(\ref{m1}).
If $p\ne 2$, then $r\ne p-1$. So, by~(\ref{x2}),
$$
x=\frac{p^2}{2}.
$$
If $p=2$, then $r=p-1$, and there is no solution.
\end{example}

\begin{example}
$x'_p=p^{-1}\,\,(\alpha=n=-1,r=p-1)$.

\medskip
\noindent
The condition (\ref{mccnd}) implies that
$-1-m=hp^{(m+1)p-1}$. Writing $h=-k$, where $k\in\mathbb{Z}_+$, this reads
$$
1+m=kp^{(m+1)p-1}.
$$
But
$$
kp^{(m+1)p-1}\ge 2^{(m+1)\cdot 2-1}=
2^{2m+1}&gt;2m+1\ge m+1
$$
for all $m\in\mathbb{N}$, meaning that there is no solution of type~(\ref{x1}). Neither does
(\ref{x2}) apply because $r=p-1$. Thus, there is no solution.
\end{example}

\begin{example}
\label{pp}
$x'_p=p^p\,\,(\alpha=p,n=1,r=0)$.

\medskip
\noindent
The only $m$ satisfying~(\ref{m1}) is
$m=0$. Then $h=1$ in~(\ref{mccnd}) and
$$
x=p^p
$$
by~(\ref{x1}). Since $r\ne p-1$, also
$$
x=\frac{p^{p+1}}{p+1}
$$
applies by~(\ref{x2}).
\end{example}

\begin{example}
$x'_p=p^{-p}\,\,(\alpha=-p,n=-1,r=0)$.

\medskip
\noindent
By~(\ref{mccnd}),
$$
-1-m=hp^{-mp}.
$$
If $m=0$, then $h=-1$. If $m&gt;0$, the right-hand side is not an integer
(since $p\nmid h$), while the left-hand side is an integer. Therefore, only $m=0$ and $h=-1$
satisfy~(\ref{mccnd}). Further, (\ref{x1}) gives
$$
x=-p^{-p}.
$$
Because $r\ne p-1$, we have another solution
$$
x=\frac{p^{1-p}}{1-p}
$$
by~(\ref{x2}).
\end{example}

\begin{example}
$x'_2=2^{26}\,\,(p=2,\alpha=26,n=13,r=0)$.

\medskip
\noindent
The condition (\ref{mccnd}) is now equivalent to
$$
13-m=2^{2m}h.
$$
If $m=0$, then $h=13$. If $m=1$, then $h=3$. If $m\ge 2$, then
$2^{2m}&gt;13-m$; hence $2^{2m}\nmid (13-m)$ for
$2\le m\le 12$, implying that a suitable $h$ does not exist.
So, by~(\ref{x1}),
$$
x=\frac{2^{26}}{13}\quad{\rm and}\quad x=\frac{2^{24}}{3}.
$$
Again $r\ne p-1$, and (\ref{x2}) yields the third solution
$$
x=\frac{2^{27}}{27}.
$$
\end{example}

\begin{example}
$x'_2=2^{-22}\,\,(p=2,\alpha=-22,n=-11,r=0)$.

\medskip
\noindent
Now (\ref{mccnd}) reads
$$
-11-m=2^{2m}h.
$$
If $m=0$, then $h=-11$. If $m=1$, then $h=-3$. Proceeding as above, we see
that no $m\ge 2$ works, and we obtain three solutions
$$
x=-\frac{2^{-22}}{11},\quad
x=-\frac{2^{-24}}{3},\quad x=-\frac{2^{-21}}{21}.
$$
\end{example}

\begin{example}
\label{sec2end}(Due to Jori M\"antysalo.)
$x'_2=2^{2058}\,\,(p=2,\alpha=2058,n=1029,r=0)$.

\medskip
\noindent
Computer experiments suggest $m=0$, $m=1$, and $m=5$. Then, respectively,
$h=1029$, $h=257$, and $h=1$, giving us three solutions of type~(\ref{x1}).
Also, (\ref{x2}) applies. All in all, we have four solutions
$$
x=\frac{2^{2058}}{1029},\quad x=\frac{2^{2056}}{257},\quad
x=2^{2048},\quad x=\frac{2^{2059}}{2059}.
$$
\end{example}

\begin{example}
$x'_p=-px^n$, $n\in\mathbb{Z}\setminus\{0\}$.

\medskip
\noindent
If $n=1$, then, by Theorem~\ref{axthm}, $x=cp^{-p^2}$ where $p\nmid c\in\mathbb{Q}$.
To obtain nontrivial solutions in the case of $n\ne 1$, we see as in the proof of Corollary~\ref{axncor2} that
necessarily $n=2$, $n=3$, or $n=-1$; then $\tilde{\xi}=-2$ (for $p\ne 2$),
$\tilde{\xi}=-1$, and $\tilde{\xi}=1$, respectively. Further, $\xi=\tilde{\xi}$.
Again (\ref{t5.5}) holds, but now (ii) in Theorem~\ref{axnthm}
is violated for $n=-1$. Thus, the nontrivial solution is, for $n=3$,
$$
x=\Big(\frac{-1}{-p^2}\Big)^\frac{1}{2}=\frac{1}{p}
$$
and, for $n=2$, $p\ne 2$,
$$
x=\frac{-2}{-p^2}=\frac{2}{p^2}.
$$
\end{example}

\begin{example}
$x'_p=p^{-1}x^n$, $n\in\mathbb{Z}\setminus\{0\}$.

\medskip
\noindent
If $n=1$, then, again by Theorem~\ref{axthm}, 
$$
x=cp,
$$
where $p\nmid c\in\mathbb{Q}$. If $n\ne 1$, we have, by
applying~(\ref{t5.4a}) and using
the notation of the proof of Corollary~\ref{axncor1},
$$
\gamma=-1+(n-1)\xi+1=(n-1)\tilde{\xi}p^\gamma.
$$
As in that proof, we see that no $\gamma\in\mathbb{N}$ works.
Consequently, there is only the trivial solution for $n\ge 2$ and
no solution for $n\le -1$.
\end{example}

\begin{example}
$x'_p=p^{1-(n-1)p^2}x^n$,
$n\in\mathbb{Z}\setminus\{0,1\}$.

\medskip
\noindent
By~(\ref{t5.4a}),
$$
\gamma=1-(n-1)p^2+(n-1)\xi+1=
2-(n-1)p^2+(n-1)\tilde{\xi}p^\gamma,
$$
i.e.,
$$
\tilde{\xi}=\frac{\gamma+(n-1)p^2-2}{(n-1)p^\gamma}.
$$
We proceed as in the proof of Corollary~\ref{axncor2}.

\medskip

{\em Case~1.} $\gamma=0$. Again, $\tilde\xi\in\mathbb{Z}$ if and
only if $n-1=\pm 1$ or $n-1=\pm 2$; hence necessarily
$n=2$, $n=3$, or $n=-1$. We have $\tilde{\xi}=p^2-2$ (for $p\ne 2$) and
$\tilde{\xi}=p^2\mp 1$, respectively.
Also $\xi=\tilde{\xi}$. But $\delta=\nu_q(p^2\mp 1)$ is odd for at least one
$q\in\mathbb{P}\setminus\{p\}$. (If not, then $p^2\mp 1=b^2$ for
some $b\in\mathbb{Z}$, so $(b-p)(b+p)=\mp 1$, implying a contradiction.)
Then
$$
\frac{\nu_q(\xi)-\nu_q(a)}{n-1}=
\frac{\nu_q(p^2\pm 1)-\nu_q(p^{1-(n-1)p^2})}{\pm 2}=\pm\frac{\delta}{2}
\notin\mathbb{Z},
$$
violating~(\ref{t5.5}). So, only the subcase $n=2$, $p\ne 2$, $\xi=p^2-2$ remains, which
yields
$$
x=\Big(\frac{\xi}{ap}\Big)^\frac{1}{n-1}=
\frac{p^2-2}{p^{1-(2-1)p^2}p}=(p^2-2)p^{p^2-2}.
$$

\medskip

{\em Case~2.} $\gamma=1$. Then
$$
\tilde{\xi}=\frac{1+(n-1)p^2-2}{(n-1)p}=p-\frac{1}{(n-1)p}
\notin\mathbb{Z}.
$$

\medskip

{\em Case~3.} $\gamma=2$. Then
$$
\tilde{\xi}=\frac{2+(n-1)p^2-2}{(n-1)p^2}=1;
$$
so $\xi=p^2$. Since (\ref{t5.5}) is satisfied, we obtain
$$
x=\Big(\frac{p^2}{p^{1-(n-1)p^2}p}\Big)^\frac{1}{n-1}=p^{p^2}.
$$

\medskip

{\em Case~4.} $\gamma\ge 3$. Then
\begin{eqnarray*}
|\tilde{\xi}|=\frac{|\gamma+(n-1)p^2-2|}{|n-1|p^\gamma}\le
\frac{|n-1|p^2+\gamma}{|n-1|p^\gamma}\le
\frac{p^2+\gamma}{p^\gamma}=
\\
\frac{1}{p^{\gamma-2}}+\frac{\gamma}{p^\gamma}\le
\frac{1}{p}+\frac{3}{p^3}\le\frac{1}{2}+\frac{3}{8}&lt;1.
\end{eqnarray*}

Thus, $x=p^{p^2}$ is a nontrivial solution for all $n\ne 0,1$. Also
$(p^2-2)p^{p^2-2}$ is such a solution for $n=2$, $p\ne 2$. These are the only nontrivial solutions.
\end{example}

\begin{example}
$x''_{pp}=px$.

\medskip
\noindent
By~(\ref{xpp}), an equivalent condition is that $x=0$ or
$$
\tilde{\xi}p^\gamma(\tilde{\xi}p^\gamma+\gamma-1)=p^3.
$$
As in the proof of Theorem~\ref{xppthm}, this reduces to
$$
p^\gamma[p^\gamma\pm(\gamma-1)]=\pm p^3.
$$
It is easy to see that no $\gamma\in\mathbb{N}$ applies. So, there are no
other solutions than the trivial one.
\end{example}

\begin{example}
$x''_{pp}=0$.

\medskip
\noindent
By~(\ref{xpp}), this is equivalent to $x=0$ or $\tilde{\xi}=0$ (accepted
here) or $\tilde{\xi}p^\gamma+\gamma-1=0$. It is easy to see that
$\gamma=0$ and $\gamma=1$ apply; the first case gives us $\tilde{\xi}=1$
and the second one $\tilde{\xi}=0$.
So, $x=0$ or $\xi=0$ or $\xi=p$, which implies that $x=cp^d$,
where $p\nmid c$ and $d\in\{0,1\}$.
\end{example}

\section{Notes on solution sets}
\label{density}
Examples \ref{sec2}--\ref{sec2end} and further computer experiments
encourage us to state the following conjecture.

\begin{conjecture}
Let $p\in\mathbb{P}$ and $a\in\mathbb{Q}\setminus\{0\}$. The equation
$$
x'_p=a
$$
has at most four solutions if $p=2$, and at most two solutions otherwise.
\end{conjecture}
\noindent
What about the number of solutions of $x'_p=ax^n$? The case $n=1$ is easy.

\begin{theorem}
Let $p\in\mathbb{P}$ and $a\in\mathbb{Q}$. The number of
nontrivial solutions of the equation
$$
x'_p=ax
$$
is either zero or infinite.
\end{theorem}
\begin{proof}
Apply Theorem~\ref{axthm}.
\end{proof}
\noindent
The case $n\ge 2$ is difficult. Because we have not done thorough computer experiments, we only record
 the following speculation.
\begin{conjecture}
Let $p\in\mathbb{P}$, $a\in\mathbb{Q}\setminus\{0\}$,
and $n\in\mathbb{Z}\setminus\{0,1\}$. The equation
$$
x'_p=ax^n
$$
has only finitely many solutions.
\end{conjecture}

Our last topic concerns the density of infinite solution sets. We begin by verifying two auxiliary results.
\begin{lemma}
\label{rs}
Let $r,s\in\mathbb{P}$, $r\ne s$. The set
\begin{eqnarray}
\label{t}
T=\{\pm r^ms^n: m,n\in\mathbb{Z}\}
\end{eqnarray}
is dense in~$\mathbb{Q}$.
\end{lemma}
\begin{proof}
An equivalent claim is that the set
$$
\{r^ms^n: m,n\in\mathbb{Z}\}
$$
is dense in~$\mathbb{Q}_+$. By taking logarithms of the elements
of this set, we get another equivalent claim stating that the set
$$
\{m\ln{r}+n\ln{s}: m,n\in\mathbb{Z}\}
$$
is dense in~$\mathbb{R}$. This, is turn, is equivalent to the density
of
$$
T'=\Big\{m+n\,\frac{\ln{s}}{\ln{r}}:m,n\in\mathbb{Z}\Big\}.
$$
Because $\ln{s}/\ln{r}$ is irrational, $T'$ is dense in~$\mathbb{R}$
by Dirichlet's theorem \cite[Theorem~1.23]{Br}.
\end{proof}

\begin{lemma}
\label{spqthm}
Let $p,q\in\mathbb{P}$, $p\ne q$. The sets
$$
S_p=\{c\in\mathbb{Q}\setminus\{0\}:p\nmid c\}
$$
and
$$
S_{pq}=\{c\in\mathbb{Q}\setminus\{0\}:p,q\nmid c\}
$$
are dense in~$\mathbb{Q}$.
\end{lemma}
\begin{proof}
For $r,s\in\mathbb{P}\setminus\{p,q\}$, $r\ne s$, let $T$ be as
in~(\ref{t}). Since $T\subset S_{pq}\subset S_p$, the claim
follows from Lemma~\ref{rs}.
\end{proof}

\begin{theorem}
Let $p,q\in\mathbb{P}$, $p\ne q$, and $a\in\mathbb{Q}$, $ap\in\mathbb{Z}$.
The solution sets of the equations
$$
x'_p=ax,\quad x'_p=x'_q,\quad x''_{pq}=x''_{qp},\quad x''_{pp}=x, \quad
and \quad x''_{pp}+x''_{qq}=x
$$
are dense in~$\mathbb{Q}$.
\end{theorem}
\begin{proof}
Apply Theorems~\ref{axthm}, \ref{xpxqthm}, \ref{xpqthm}, \ref{xppthm}, \ref{xppxqqthm}
and Lemma~\ref{spqthm}.
\end{proof}

In fact, we can say more about the second and third equations.

\begin{theorem}
Let $p,q\in\mathbb{P}$, $p\ne q$.
The solution sets of the equations
$$
x'_p=x'_q=0,\quad x''_{pq}=x''_{qp}=0,\quad x''_{pp}=0,\quad
and\quad x''_{pp}+x''_{qq}=0
$$
are dense in~$\mathbb{Q}$.
\end{theorem}
\begin{proof}
The set~$S_{pq}$, which is dense by Lemma~\ref{spqthm}, is
contained in the solution set of each equation.
\end{proof}
\noindent
This theorem raises the following 

\begin{conjecture}
Let $p,q\in\mathbb{P}$, $p\ne q$, and $a\in\mathbb{Q}$.
The solution sets of the equations
$$
x''_{pq}=x''_{qp}=a,\quad x''_{pp}=a,\quad
and\quad x''_{pp}+x''_{qq}=a
$$
are dense in~$\mathbb{Q}$.
\end{conjecture}

\section{Concluding remarks}
\label{remarks}

Ufnarovski and \AA hlander~\cite{UA} studied certain arithmetic differential equations.
For example, they proved \cite[Corollary~3]{UA} that if the equation
$$
x'=a\quad(x\in\mathbb{Z}_+,\,a\in\mathbb{Z}_+\setminus\{1\})
$$
has a solution, then it has infinitely many solutions.
The equation $x'_p=a$
($x\in\mathbb{Q},\,a\in\mathbb{Q}\setminus\{0\}$), in contrast to that, has always only finitely
many solutions, as we saw in the proof of Theorem~\ref{athm}.
For another example,
they showed \cite[Theorem~6]{UA} that
$$
x'=x\quad(x\in\mathbb{Z}_+)
$$
if and only if $x=p^p$ for some $p\in\mathbb{P}$.
A special case of Theorem~\ref{axthm} gives a somewhat parallel result,
stating that $x_p=x$ ($x\in\mathbb{Z}_+$) if and only $x=cp^p$ where
$p\nmid c\in\mathbb{Z}_+$.

Kovi\v c~\cite{Ko} studied certain arithmetic partial differential equations.
We pursued this topic a few steps further, yet there is still much work to do
with both arithmetic derivative and arithmetic partial derivatives. Connections
between different derivations are of special interest. For
example, the {\em logarithmic derivative} of $a\in\mathbb{Q}\setminus\{0\}$
defined \cite[p.~13]{UA} by
$$
\mathrm{ld}\,a=\frac{a'}{a}
$$
can also be expressed as
$$
\mathrm{ld}\,a=\sum_{p\in\mathbb{P}}\frac{a'_p}{a}.
$$

\section{Acknowledgment}

We thank the referee for the useful comments provided.

\begin{thebibliography}{9}

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

\bibitem{Br} A.~Browder, \emph{Mathematical Analysis: An Introduction},
Springer, 1996.

\bibitem{HMMT} P.~Haukkanen, M.~Mattila, J.~K.~Merikoski, and T.~Tossavainen,
Can the arithmetic derivative be defined on a non-unique factorization domain?
\emph{J. Integer Seq.}~\textbf{16} (2013),
\href{https://cs.uwaterloo.ca/journals/JIS/VOL16/Tossavainen/tossavainen4.html}{Article~13.1.2}.

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

\bibitem{MP} R.~K.~Mistri and R.~K.~Pandey, Derivative of an ideal in a number
ring, \emph{Integers}~\textbf{14} (2014), \# A24.

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

\end{thebibliography}

\bigskip
\hrule
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\noindent 2010 {\it Mathematics Subject Classification:}
Primary 11A25; Secondary 11A51.

\noindent {\it Keywords:} 
arithmetic derivative, partial derivative.

\bigskip
\hrule
\bigskip

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

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received March 21 2016;
revised versions received September 12 2016; September 14 2016.
Published in {\it Journal of Integer Sequences}, October 11 2016.

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\noindent
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