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\begin{center}
\vskip 1cm{\LARGE\bf On Some Conjectures about Arithmetic \\
\vskip .1in
Partial Differential Equations}
\vskip 1cm
\large
Ram Krishna Pandey and Rohit Saxena \\
Department of Mathematics \\
Indian Institute of Technology Roorkee\\
Roorkee - 247667\\
India \\
\href{mailto:ramkpandey@gmail.com}{\tt ramkpandey@gmail.com} \\
\href{saxenarrohit@gmail.com}{\tt saxenarrohit@gmail.com} \\
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\def\tit{y=\frac{1}{2}}
\def\text{Remember the pred defining notation $y=mx$}
\def\eqn{$x_{p}'=ax^n$}
\def\anot{${\alpha}_{0}$}
\def\sm{\setminus}
\def\pr{p^R}
\def\tit{y=\frac{1}{2}}
\def\orig{x_p'=a}
\def\pl{p^L}
\def\q{\mathbb{Q}}
\def\z{\mathbb{Z}}
\def\nat{\mathbb{N}}
\def\io{i_0}
\def\e1{i+C$p^{i+1}$ = L}
\def\anot{${\alpha}_{0}$}
\def\nat{\mathbb{N}}
\def\z{\mathbb{Z}}
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\begin{abstract}
In this paper, we study the arithmetic partial differential equations $x_{p}'=ax^n$ and $x_{p}'=a$. We solve a conjecture of Haukkanen, Merikoski, and Tossavainen (HMT, in short) about the number of solutions (conjectured to be finite) of the equation $x_{p}'=ax^n$ and improve a theorem of HMT about finding the solutions of the same equation. Furthermore, we also improve another theorem of HMT about the solutions of the equation $x_{p}'=a$ and discuss one more conjecture of HMT about the number of solutions of $x_{p}'=a$.
\end{abstract}
\section{Introduction}\label{intro}
Let the symbols $\mathbb{Z}$, $\mathbb{Q}$, and $\mathbb{R}$ have their
usual meaning. We follow the notation used by Haukkanen, Merikoski, and
Tossavainen \cite{hauk} (HMT, in short), except for
$\mathbb{N}$, which here denotes the set of positive integers
$\{1,2, \ldots\}$. We use $\mathbb{P} = \{ 2,3,5, 7,\ldots \}$
for the set of all prime numbers.
Let $a \in \mathbb{Q} \setminus \{0\}$. Then there are unique $L \in
\mathbb{Z}$ and $M \in \mathbb{Q} \setminus \{0\}$ such that $a = M
p^L$ and $p \nmid M$. The {\it arithmetic partial derivative} of $a \in
\mathbb{Q} \setminus \{0\}$, denoted by $a_p'$, is defined by HMT
\cite{hauk} as follows:
\[a_p' = M L p^{L-1}.\]
A comprehensive list of references is given in \cite{hauk} for the readers about the history of the arithmetic derivatives and their several generalizations.
In this paper, we study the arithmetic partial differential equations
$x_{p}'=ax^n$, and $x_{p}'=a$. In Section~\ref{finite}, we resolve
Conjecture 29 of HMT \cite{hauk} about the finiteness of the number of
solutions of the equation $x_{p}'=ax^n$, and give an efficient algorithm
(Theorem \ref{thm2}) to find the solutions of the same equation
in Section \ref{finitesoln}. In Section \ref{sepecial}, we improve
\cite[Theorem 1]{hauk} concerning the solutions of the equation
$x_{p}'=a$, and give some necessary and sufficient conditions for
certain nontrivial solutions in Theorems \ref{thm3} and \ref{thm4},
respectively. Further, we discuss HMT's Conjecture 27 about the number
of solutions of $x_{p}'=a$ and, based on our findings, we hypothesize
that this conjecture is false.
\section{Number of solutions of $x_{p}'=ax^n$}\label{finite}
\begin{theorem} \label{finiteness}
The solution set of the equation $x_{p}'=ax^n$, $a\in\mathbb{Q} \setminus\{0\}$, $p\in\mathbb{P}$, $n\in\mathbb{Z} \setminus \{0,1\}$, is finite.
\end{theorem}
\begin{proof}
If we look at the equation, we observe that an obvious solution to the equation is at $x=0$, provided $n>0$, for each prime number $p$. We ignore this solution as a trivial solution and consider only non-zero solutions for the equation. Express $x$ as $x={\beta}p^{\alpha}$, $p \nmid \beta$, $\alpha \in \mathbb{Z} \setminus \{0\}$, $p$ being a prime number. Then $x_{p}' = \beta \alpha p^{\alpha -1}$. As we have $x_{p}'=ax^n$, we get
$\beta \alpha p^{\alpha -1} = a \beta^n p^{n\alpha}$, which implies
\[\left(\frac{{\beta}^{n-1}a}{\alpha}\right)(p^{(n-1)\alpha +1}) = 1.\]
Write $a = Mp^L$, $M\in \mathbb{Q} \setminus \{0\}$, $p\nmid M$, $L \in \mathbb{Z}$, and $\alpha = {\alpha}_{0} p^R$, $\alpha_{0} \in \mathbb{Z}$, $R \in \mathbb{N}$, $p \nmid \alpha_{0}$. Then, we get
\[\left(\frac{Mp^L {\beta}^{n-1}}{ {\alpha}_{0} p^R}\right)\left(p^{(n-1)\alpha +1}\right)= 1\]
or
\[\left(\frac{M{\beta}^{n-1}}{ {\alpha}_{0}}\right)\left(p^{(n-1)\alpha +1+L-R}\right) = 1.\]
Since $p \nmid \left(\frac{M{\beta}^{n-1}}{ {\alpha}_{0}}\right)$, we have
\begin{eqnarray}\label{relation}
(n-1)\alpha +1+L-R &=& 0,
\end{eqnarray}
and
\begin{eqnarray}\label{beta}
\frac{M{\beta}^{n-1}}{ {\alpha}_{0}} &=& 1.
\end{eqnarray}
Substituting $\alpha = {\alpha}_{0} p^R$ in (\ref{relation}), we get
\begin{eqnarray}\label{finalrel}
(n-1)\alpha_0 p^R +1+L-R &=& 0.
\end{eqnarray}
Equation~(\ref{beta}) plays an important role in determining the solution
set and proving its finiteness. We first concentrate on the term
$\alpha=\alpha_0 p^R$ in the solution $x=\beta p^{\alpha}$, and prove
that only a finite number of values of $R$ are possible for which
$\alpha$ forms the solution $x$ of the equation. Then, through
equation (\ref{beta}), we conclude that the number of
corresponding values of $\beta$ is also finite, as $M$ is a constant.
We consider two separate cases for $R=0$, and for $R \neq 0$.
\bigskip
\noindent \textbf {Case 1:} ($R=0$). From (\ref{finalrel}) we have
that $(n-1)\alpha_0 +1+L = 0,$ which implies
\[\alpha_0 = -\left(\frac{1+L}{n-1}\right).\]
As $\alpha_0$ is an integer, we get $(n-1)|(1+L)$.
We remark here that if $(n-1)\nmid (1+L)$, then we do not get any
solution in this case.
\bigskip
\noindent \textbf {Case 2:} ($R\neq0$). We rewrite equation (\ref{finalrel}) as
\begin{eqnarray}\label{1.4}
(n-1)\alpha_0 &=& \frac{R-1-L}{p^R}.
\end{eqnarray}
Since $n, \alpha_0 \in \mathbb{Z}$, we have $(n-1)\alpha_0 \in \mathbb{Z}$ . Moreover, as $R\neq0$, so $R\in\mathbb{N}$ . We further divide this case into the following two subcases.
\bigskip
\noindent \textbf{Case 2.1:} ($R = 1+L$). From equation (\ref{1.4}), we get $(n-1)\alpha_0 = 0.$ Since $n\neq1$, hence $\alpha_0=0$ implies that $\alpha=0$. Thus, the only possible value of $\alpha$ is $0$.
\bigskip
\noindent \textbf{Case 2.2:} ($R \neq 1+L$). Clearly, if $R$ is not bounded, then there exists an $R_0 \in \mathbb{N}$ such that the right-hand side expression of (\ref{1.4}) becomes a fraction for $R \geq R_0$, which is not possible.
Hence, $R$ can attain only a finite number of values. So, a necessary condition on $R$ for a solution is $(n-1) | \left(\frac{R-1-L}{p^R}\right)$.
We get a value of $\alpha_0 = \frac{R-1-L}{(n-1)p^R}$ corresponding to every value of $R$, which satisfies the above condition. We thus obtain finite number of pairs $($\anot, $R)$ giving finite number of values of $\alpha = \alpha_0 p^R$ at which the solution is possible.
So far, we have analyzed all possible values of $R$ and have come to the conclusion that only finite number of values of $R$ are possible which may form the solution $x={\beta}p^{\alpha}$ with $\alpha = {\alpha}_{0} p^R$. Now, we need to prove that the corresponding values of $\beta$ also form a finite set.
Clearly, by (\ref{beta}), we can write $\beta = {\left(\frac{\alpha_{0}}{M}\right)}^{\frac{1}{n-1}}.$ Hence, we conclude that for a given value of \anot, at most two values of $\beta$ are possible. As $\beta \in \mathbb{Q}$, the
quantity ${\left(\frac{\alpha_{0}}{M}\right)}^{\frac{1}{n-1}}$ must be a rational number of the form $\left(\frac{E}{F} \right)$, $F\neq 0$, $E,F \in \mathbb{Z}$. So this acts as a filtering condition on $\alpha_0$ to further qualify for the solution set. So, we get a final condition on $\alpha$ to be satisfied so that $\alpha_0$ and the corresponding value of $R$ can give us a solution of the equation.
This proves that there exist only a finite number of values of $\beta$ corresponding to every value of $\alpha_0$ or $\alpha$, which themselves have finite possible values for the solution set. Hence, $x={\beta}p^{\alpha}$ has only finitely many solutions.
\end{proof}
\section{Solutions of $x_{p}' = ax^n$ }\label{finitesoln}
In this section, we find all solutions of the equation $x_{p}'=ax^n$, $a\in\mathbb{Q} \setminus \{0\}$, $p\in\mathbb{P}$, $n\in\mathbb{Z} \setminus \{0,1\}$. The derivation of the solutions following the notation of Section \ref{finite} is given below.
Let us recall equation (\ref{finalrel}) and consider again two separate cases for $R=0$, and $R \neq 0$.
\bigskip
\noindent \textbf {Case 1:} ($R=0$). We get a solution if $(n-1)\mid (1+L)$, by the argument used in Theorem \ref{finiteness}.
\bigskip
\noindent \textbf {Case 2:} ($R \neq 0$).
The basic approach for the derivation is to consider the cases for the values of $\alpha_0$ such that either $(n-1)\alpha_0 > 0$ or $(n-1)\alpha_0 < 0$ or $(n-1)\alpha_0 = 0$, where $n$ is a constant and the sign of $\alpha_0$ depends upon the sign of $(n-1)$. The upper and lower bounds for the possible values of $R$ have been derived in all the cases through which we can get corresponding $\beta$ and can form the solution. The necessary condition to be satisfied by $R$ is that on substituting it in equation (\ref{finalrel}), $\alpha_0$ must come out to be an integer. If not, then that value is ignored and we proceed to a next value in the range. This condition acts as a filtering condition for the values of $R$.
From equation (\ref{finalrel}), it is clear that $1+L-R \equiv 0 \pmod p$. Since $1+L$ is a constant, we have $1+L \equiv 0 \pmod p$ implies that $R \equiv 0 \pmod p$, and $1+L \not \equiv 0 \pmod p$ implies that $R \not \equiv 0 \pmod p$. We can further reduce the solution ranges derived for each cases by examining the above two cases. So, we discuss below each subcase one by one.
\bigskip
\noindent \textbf {Case 2.1:} ($(n-1)\alpha_0 > 0$).
Clearly, $(n-1)$\anot$\pr >0$ for all $R$ and $p$. We have $\pr > R$ for all $R\in \nat$. Clearly, $(n-1)\alpha_0 \in \mathbb{Z}$. So, $(n-1)\alpha_0p^R - R > 0$ for all $R$. By (\ref{finalrel}),
\begin{eqnarray} \label{3.1}
(n-1)\alpha_0p^R - R &=& -1-L.
\end{eqnarray}
We get $-1-L > 0$ or $L < -1.$ At $L \geq -1$, this case does not give any solution. Now, there are two possibilities.
\bigskip
\noindent \textbf {Case 2.1.1:} ($(n-1)\alpha_0 = 1$).
Clearly, $(n-1)\alpha_0 = 1$ implies that $n=2$, and $\alpha_0 = 1$, as $n \in \mathbb{Z} \setminus\{0,1\}$. Introducing a new variable $K=-1-L$ and combining it with the equation (\ref{3.1}), we get $R+K=p^R,$ which implies
\begin{eqnarray}\label{3.2}
R= \log_p (R+K).
\end{eqnarray}
This equation gives us a relation, which also gives a filtering condition on $R$ that
\begin{eqnarray}\label{2.3}
R+K\equiv 0 \pmod p.
\end{eqnarray}
We can rewrite equation (\ref{3.2}) in the following two ways:
\begin{eqnarray}\label{3.4}
R &=& \log_p R + \log_p (1+K/R).
\end{eqnarray}
\begin{eqnarray}\label{3.5}
R &=& \log_p K + \log_p (1+R/K).
\end{eqnarray}
Now, we consider three cases for the values of $R$ and examine in each case the possibility and range for the solution.
\bigskip
\noindent \textbf{Case 2.1.1.1:} ($R > K$). $R > K$ implies that $\log_p (1+K/R)$ $<1$. So, $R = \log_p R +\log_p (1+K/R)$ $\Rightarrow$ $R<\log_p R+1$, which implies that $p^{R-1} < R.$ Clearly, this does not hold for any values of $p$ and $R$. So, we cannot get any solution in this case.
\bigskip
\noindent \textbf {Case 2.1.1.2:} ($R = K$). Substituting $R=K$ in (\ref{3.4}) or in (\ref{3.5}), we get $R=\log_p K + \log_p 2$ or $R=\log_p (2R)$, which implies
$\pr = 2R.$ This relation is possible only for $p=2$ and $R=1$. So, for $R=K$, we can expect a solution only if $p=2$ and $R=1$. In this case, $K+R \equiv 0 \pmod p$ is always satisfied. So, this case may yield a solution when $p=2$.
\bigskip
\noindent \textbf {Case 2.1.1.3:} ($R < K$). Clearly, $R < K$ implies that $\log_p (1+R/K) < 1$. So, $R= \log_p K +\log_p (1+R/K)\Rightarrow R<\log_p K+1$, which gives $R \in \{1, 2, \ldots, \lceil\log_p K\rceil\}$.
So, the feasible values of $R$ at which we may get the solution must lie in the set $\{1, 2, \ldots, \lceil\log_p K\rceil\}$. Further, $R+K \equiv 0 \pmod p$ must be satisfied. So, we take only those values of $R$ which are in the set $\{1, 2, \ldots, \lceil\log_p K\rceil\}$ and satisfy $R+K \equiv 0 \pmod p.$
\bigskip
\noindent \textbf {Case 2.1.2:} ($(n-1)\alpha_0 \neq 1$). Rewrite equation (\ref{3.1}) as $(n-1)\alpha_0p^R = R+K.$ Clearly, $(n-1)$\anot$ > 1$ implies that $R+K > \pr$, which implies
$R <\log_p (R+K).$
We can rewrite the above inequality in the following two ways:
\begin{eqnarray}\label{3.6}
R &<& \log_p R + \log_p (1+K/R),
\end{eqnarray}
\begin{eqnarray}\label{3.7}
R &<& \log_p K + \log_p (1+R/K).
\end{eqnarray}
Again proceeding in the same way as in the last case, we take the following three cases:
\bigskip
\noindent \textbf {Case 2.1.2.1:} ($R > K$). Clearly, $R>K$ implies that $\log_p (1+K/R)$ $<1$. So, $R < \log_p R +\log_p (1+K/R)$ $\Rightarrow$ $R<\log_p R+1$, which implies that $p^{R-1} < R$, which is not possible. So, we do not get any solution in this case.
\bigskip
\noindent \textbf {Case 2.1.2.2:} ($R < K$). Clearly, $R < K$ implies that $\log_p (1+R/K)$ $<1$. So, $R< \log_p K +\log_p (1+R/K)\Rightarrow R<\log_p K+1$. That is, $R \in \{1, 2, \ldots, \lceil\log_p K\rceil\}$. Further, $R+K \equiv 0 \pmod p$ must be satisfied. So, we only take those values of $R$ which are in the set $\{1, 2, \ldots, \lceil\log_p K\rceil\}$ and satisfy $R+K \equiv 0 \pmod p.$
\bigskip
\noindent \textbf {Case 2.1.2.3:} ($R = K$).
Substituting $R=K$ in (\ref{3.6}) or in (\ref{3.7}), we get $R<\log_p K + \log_p 2$ or $R<\log_p (2R)$, which implies $\pr < 2R.$ This inequality cannot be satisfied for any values of $R$ and $p$ in their respective domains.
Thus, we see that if $(n-1)\alpha_0>0$, then we get solutions only for $R0$, because $(n-1)$\anot $<0$.
This implies $K<0$ or $L>-1$. As $(n-1)$\anot $<0$, we have $(n-1)$\anot$\pr<0$. So, we get $R+K<0$ or $L>R-1$. Thus, we get two conditions: $L>-1$, and $R<1+L$ for the feasibility of this case.
By introducing two new variables $F$ and $W$, both of them are positive and such that $(n-1)$\anot$ = -F$, and $K=-W$, we rewrite equation (\ref{3.8}) as
\begin{eqnarray}
R + Fp^R &=& W,
\end{eqnarray}
where all of $W, F$, and $R$ are greater than zero.
Clearly, since $W > Fp^R$, we have $W>p^R$ or $R<\log_p W$ or $R<\log_p (-K)$ or equivalently, $R<\log_p (1+L).$
Thus we get an upper bound for the possible values of $R$, in the given case
$R \in \{1, 2, \ldots, \lceil\log_p K\rceil\}$. Further, $R+K \equiv 0 \pmod p$ must be satisfied. So, we take only those values of $R$ which are in the set $\{1, 2, \ldots, \lceil\log_p K\rceil\}$ and satisfy $R+K \equiv 0 \pmod p.$
\bigskip
\noindent \textbf {Case 2.3:} ($(n-1)\alpha_0 = 0$). Clearly, we have \anot $=0$ as $n \neq 1$. So $\alpha$ = 0 in this case.
Now that we have the final ranges for the values of $R$ in each case, so, we can find the possible values of $\alpha = \alpha_0 p^R$. First, we find the value of $\alpha_0$ corresponding to each $R$. We accept only those values of $R$ which are inside the range and giving an integral value of $\alpha_0$, otherwise, reject it. This way, we get the possible values of $\alpha_0$ and $R$, which are then used to find corresponding $\alpha$. Then, substituting the value of $\alpha_0$ in equation (\ref{beta}), we can find the corresponding value of $\beta$. If $\beta$ comes out to be rational, this means solution exists for the given $\alpha_0$ and $x = \beta p^{\alpha}$ is the solution of the equation $x_p ' = ax^n$. Otherwise, we test the next value of $\alpha_0$. This is how the algorithm works.
We summarize above discussion in the following:
\begin{theorem}\label{thm2}
The equation $x_p' = ax^n$, where $p \in \mathbb{P}$, $a \in \mathbb{Q} \setminus \{0\}$ with $a = Mp^L$, $M \in \mathbb{Q} \setminus \{0\}$, $p \nmid M$, $L \in \mathbb{Z}$ has a nontrivial solution $(0 \neq) x = \beta p^{\alpha}$, $p \nmid \beta$, $\alpha \in \mathbb{Z} \setminus \{0\}$ with $\alpha = \alpha_0 p^R$, $\alpha_0 \in \mathbb{Z}$, $R \in \mathbb{N}$, $p \nmid \alpha_0$ if and only if any one of the following conditions hold
\begin{enumerate}
\item $(n-1) \mid (1+L), \alpha = - \frac{1+L}{n-1}$, and $\beta = (\frac{\alpha}{M})^{\frac{1}{n-1}} \in \mathbb{Q}$.
\item $(-2 \neq) L < -1$, $R \in \{1, 2, \ldots, \lceil\log_p (-1-L)\rceil\}$ with $R-1-L \equiv 0 \pmod p$ such that $\alpha_0 = \frac{R-1-L}{(n-1)p^R} \in \mathbb{Z}$, and $\beta = (\frac{\alpha}{M})^{\frac{1}{n-1}} \in \mathbb{Q}$.
\item $L = -2$, $p = 2$, $R=1$, $\alpha_0(n-1)=1$, $\beta = (\frac{\alpha}{M})^{\frac{1}{n-1}} \in \mathbb{Q}$.
\item $L > -1$, $R \in \{1, 2, \ldots, \lceil\log_p (1+L)\rceil\}$ with $R-1-L \equiv 0 \pmod p$ such that $\alpha_0 = \frac{R-1-L}{(n-1)p^R} \in \mathbb{Z}$, and $\beta = (\frac{\alpha}{M})^{\frac{1}{n-1}} \in \mathbb{Q}$.
\end{enumerate}
Furthermore, all solutions are found in this way.
\end{theorem}
\section{Solutions of $x_p' = a$}\label{sepecial}
In this section, we discuss the solutions of $\orig$. Let us express $a$ in the form $M\pl$ with $p\nmid M$, $M\in \q$, $L\in \z$. We improve Theorem 1 of \cite{hauk} and give a better bound for the solution range.
An ``alternate step" approach has been introduced to reach the solution even faster.
Let $y=x/M$, so that ${y_p'=\pl}$. Following Theorem 1 of \cite{hauk}, we start with the sets $I_0$ and $I$ exactly same as in Theorem 1 of \cite{hauk} depending on whether $L>0$, $L<0$ or $L=0$. We then improve the set $I_0$ and hence improve the set $I$, which is the candidate for the solutions.
\bigskip
\noindent \textbf{Case 1:} ($L>0$). Let $I_0 = \{0, 1, 2, \ldots, L-1\}$, and $I= \{ i \in I_0 : p^{i+1} || (L-i)\}.$ Then Theorem 1 of \cite{hauk} implies that $y = \frac{p^{L+1}}{L-i}$ is a solution of the equation ${y_p'=p^L}$ for each $i\in I$. Besides this, there is one more possibility of a solution at when $p \nmid (L+1)$, giving $y = \frac{p^{L+1}}{L+1}$ as a solution. We concentrate only on positive values of $i$ in $I_0$. We can test separately for the possibilities at $i=0$ and at when $p\nmid(L+1)$.
We first derive a necessary condition for the existence of at least one solution of the equation for the positive values of $i$. Suppose that there exists a solution at $i$ and $p^{i+1} || (L-i)$. Let us write $L-i = Cp^{i+1}$, where $p\nmid C$, $C \in \nat$. Then
\begin{eqnarray}\label{3.25}
i+Cp^{i+1} &=& L.
\end{eqnarray}
Since $C\geq1$, and $i>0$, we have $L > Cp^{i+1} > p^{i+1}$. This implies that $i+1 < \log_p L$ or $i < \log_p L -1$.
So, we get a new upper limit for the value of $i$ in $I_0$, which is $\lceil\log_p L\rceil - 2$. So, now we replace $I_0$ by a much smaller set $\{1,2, \ldots, \lceil\log_p L\rceil - 2\}$. The new upper bound is of logarithmic order of $L$ and thus it will be much easier to work with. Also a necessary condition for the existence of a solution for given $p$ and $L$ is
\begin{center}
$L \geq 1+p^2$,
\end{center}
which follows from equation (\ref{3.25}). If $L<1+p^2$, then we do not get any solution for $i > 0$. So, we can have at most two solutions for the given equation: one for $i=0$, and another in the case $p\nmid (L+1)$.
Beginning with $i=1$, we start testing whether it is included in the
set $I$. Let $i=i_0$ be some value of $i$ that satisfies the condition
for inclusion in the set $I$. Now, we derive the condition for the
possibility of getting an alternate solution for a value of $i$, higher
than that of $i_0$ and the step size from the initial value $i_0$ at
which we can get another $i$, so that we do not have to traverse each
and every value of $i \in I$ till $\lceil\log_p L\rceil - 2$.
Sometimes, even $\log_p L$ may be large. In such cases, the step size
method described below helps in reducing the work greatly.
Since we get a solution at $i=i_0$, we have
\begin{eqnarray}\label{3.26}
i_0 + C_0p^{i_0+1} &=& L,
\end{eqnarray}
where $p\nmid C_0$. Let the alternate solution exist at $i=i_1$. So, we have
\begin{eqnarray}\label{3.27}
i_1 + C_1p^{i_1+1} &=& L,
\end{eqnarray}
where $p \nmid C_1, i_1 > i_0$.
From equation (\ref{3.26}), we have
\begin{eqnarray} \label{3.28}
C_0< \frac{L}{p^{i_0+1}}.
\end{eqnarray}
From equations (\ref{3.26}) and (\ref{3.27}), we get
\begin{center}
$i_0 + C_0 p^{i_0+1} = i_1 + C_1 p^{i_1+1}$
\end{center}
\begin{eqnarray}\label{3.29}
\Rightarrow
i_1 = i_0 + p^{i_0+1}(C_0 - C_1 p^{i_1-i_0}).
\end{eqnarray}
We have $p \mid C_1 p^{i_1-i_0}$, $p\nmid C_0$. Hence, $p\nmid (C_0 - C_1 p^{i_1-i_0})$. Let $K=C_0 - C_1 p^{i_1-i_0}$. Then
\begin{eqnarray}\label{3.30}
i_1 = i_0 + Kp^{i_0+1}, ~~p\nmid K.
\end{eqnarray}
So, we conclude that the candidate of $i\in I$ for the alternate solution is in the form of (\ref{3.30}). The step size is $Kp^{i_0+1}$, where $K>0$ and not divisible by $p$.
From equation (\ref{3.30}), $i_1 - i_0 = Kp^{i_0+1}$. Now, since $i_1>i_0$, $i_1-i_0>0$, we have $K>0$ or $C_0-C_1p^{i_1-i_0} > 0$ . This gives
$C_1<\frac{C_0}{p^{i_1-i_0}}.$ Hence, $C_1\geq1$ $\Rightarrow$ $\frac{C_0}{p^{i_1-i_0}}>1$, which implies
\begin{eqnarray}\label{3.31}
i_1-i_0<\log_p C_0.
\end{eqnarray}
Combining (\ref{3.28}), (\ref{3.30}), and (\ref{3.31}), we get
\begin{eqnarray}\label{3.32}
K < \frac{1}{p^{i_0+1}} \log_p \left(\frac{L}{p^{i_0+1}}\right).
\end{eqnarray}
We get an upper bound for the number of steps in terms of $Kp^{i_0+1}$, within which we can expect an alternate solution of the equation, once we get an initial solution. Starting from $K=1$, we traverse till the upper bound in (\ref{3.32}). As $p\nmid K$, we also exclude all those values which are divisible by $p$. Here, we introduce a new set called \textit {Alternate Step Range Set} or {\it ASR}, in short, containing the possible values of $K$ for a given $i_0$. Let $U = \frac{1}{p^{i_0+1}} \log_p \left(\frac{L}{p^{i_0+1}}\right)$. Then
\[ASR = \{ 1,2,\ldots, \lfloor U \rfloor\} \setminus \{p, 2p, \ldots\}.\]
By iterating through the ASR set, we can get the alternate solution of the equation within very few steps. Once we reach the alternate solution, say at $i=i_1$, we repeat the same steps and form the ASR range using $i=i_1$, which will then be used to get next higher value of $i$. We stop this process when we do not get an alternate solution. Under computational limits, this method is highly efficient in reaching all the solutions.
We now derive a necessary condition for the existence of an alternate solution, given that a solution exists at $i=i_0$. If an alternate solution exists, the minimum value of $K$ must be $1$. So, we get
\begin{center}
$1< \frac{1}{p^{i_0+1}} \log_p \left(\frac{L}{p^{i_0+1}}\right)$.
\end{center}
From the above relation we get a necessary condition for the existence of an alternate solution for $i$ greater than the given initial value $i_0$ as
\begin{eqnarray}\label{3.33}
L > p^{\left(p^{i_0+1}+i_0+1\right)}.
\end{eqnarray}
Thus, we get a new condition for the existence of the alternate solution for $i_0\in \nat$. If inequality (\ref{3.33}) is not satisfied, this means that there exists no solution for $i>i_0$. Moreover, $i_0\geq1$, so putting $i_0$ in (\ref{3.33}), we conclude that if $L \leq p^{(p^2+2)}$, then we cannot have more than one solution for the positive values of $i$. In such a situation, we can get at most three solutions of the partial differential equation, one in this range and the other two for $i=0$, and for $p\nmid (L+1)$.
\bigskip
\noindent \textbf{Case 2:} ($L < 0$). Let $I_0 = \mathbb{N} \cup \{0\}$, and $I$ is same as in Case 1. One can test separately at $i=0$, and for $p\nmid(L+1)$. So, we take only the positive values of $i$. Let $L=-Q$. Then $p^{i+1} || (L-i)$ or $p^{i+1} || (Q+i)$. Write
\begin{eqnarray}\label{3.34}
Q+i &=& Cp^{i+1}, \hspace{0.1cm} p\nmid C, ~~C>0.
\end{eqnarray}
We now derive the condition for the existence of a solution for this range. Rewrite equation (\ref{3.34}) as
\begin{eqnarray}
\frac{Q}{p^{i+1}} + \frac{i}{p^{i+1}} &=& C.
\end{eqnarray}
Since $0<\frac{i}{p^{i+1}} <1$, we have $\frac{Q}{p^{i+1}} > C-1$. This implies that $\frac{Q}{C-1}> p^{i+1}$.
Here, $(C-1)$ is in the denominator, so, one can test separately at $C=1$ and for the rest of the cases, we assume $C>1$. At $C>1$, $\frac{Q}{C-1} < Q$. So, we get $p^{i+1} < Q$, which gives
\begin{eqnarray}
i< \log_p Q-1.
\end{eqnarray}
Thus, we get an upper bound on the value of $i$, which is $\lceil\log_p Q\rceil -2$. So, the infinite set $I_0$ has now been reduced to $I_0 = \{ 1,2,\ldots, \lceil\log_p Q\rceil -2$\}.
Also, $Q>(C-1)p^{i+1}$, so for the existence of a solution at $C>1$, $Q> p^{i+1}$. The minimum value of $i$ may be $1$, so a necessary condition for the existence of a solution is $Q>p^2$ or $L<-p^2$.
Now, we examine the range where an alternate solution is possible and also derive the possibility of an alternate solution.
Let there exists a solution at $i=i_1$. Here, we consider $i_1$ to be the highest value of $i$ at which solution is possible and consider the alternate solution at some smaller value of $i$, unlike the previous case, where we considered alternate solution for the higher value of $i$ and started with a smaller value of $i$. So, let an alternate solution exist at $i=i_2$. Hence, we have the following two equations.
\begin{eqnarray*}
Q+i_1 &=& C_1 p^{i_1+1}, ~~p\nmid C_1,
\end{eqnarray*}
and
\begin{eqnarray*}
Q+i_2=C_2 p^{i_2+1}, ~~p\nmid C_2.
\end{eqnarray*}
Hence,
\begin{eqnarray*}
i_1 - i_2 &=& p^{i_2+1}(p^{i_1-i_2}C_1-C_2).
\end{eqnarray*}
Put $K$ $=$ $p^{i_1-i_2}C_1-C_2$ with $K\geq1$. We get $i_1 - i_2=K p^{i_2+1}$, $p\nmid K$, which implies $K p^{i_2+1}1$. Considering
the inequality (\ref{3.43}), we put $i_2=1$, as this would be the
minimum value of $i_2$, in case it exists. So, we get $1<\log_p
{i_1}-1$ or $i_1 > p^2$.
So if $i \leq p^2$, we terminate the process as there will not be any
alternate solution at a smaller value of $i$.
Now, we derive a necessary condition for the existence of at least two solutions for $C>1$. Considering inequality (\ref{3.43}), we put $i_2=1$, as this would be the minimum value of $i_2$, in case it exists, and for $i_1$, we substitute $i_1 < \log_p Q-1$. We get
\begin{align*}
& 1 < \log_p i_1 -1
\Rightarrow & 2 < \log_p (\log_p Q -1)
\Rightarrow & p^2+1 < \log_p Q
\Rightarrow & Q >p^{(p^2+1)}.
\end{align*}
This gives a necessary condition to have at least two solutions for $C > 1$.
\bigskip
\noindent \textbf{Case 3:} ($L=0$). Clearly, $y_p'=1$ $\Rightarrow$ $y=p$ is the only solution.
Now, we have the values of $i$ for which we have the solution for $y_p=p^L$. We can get the corresponding solution of the equation $x_p'=a$, by multiplying $M$ to the solution obtained through the above methods, since $y=\frac{x}{M}$ $\Rightarrow$ $x=My$, we have $x_p' = M y_p'$.
Now we restate the improved version of \cite[Theorem 1]{hauk} and give
another theorem (using the notation used in the discussion) about the
nature of solutions of $x_p' = p^L$, which is the outcome of the above
discussion.
\begin{theorem}\label{thm3}
Let $p \in \mathbb{P}$ and $L \in \mathbb{Z}$. Further, let $I_0 = \{1,2, \ldots, \lceil \log_p L\rceil - 2\}$ for $L>0$, $I_0 = \{1,2, \ldots, \lceil \log_p (-L)\rceil - 2\}$ for $L<0$, and $I_0 = \emptyset$ for $L=0$. Let also $I = \{i \in I_0 : p^{i+1} || (L-i)\}$. Then $x = \frac{p^{L+1}}{L-i}$ is a solution of $x_p' = p^L$ for each $i \in I$. If $p \nmid (L+1)$, then also $x = \frac{p^{L+1}}{L+1}$ is a solution. All solutions are obtained in this way. 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}
\begin{theorem}\label{thm4}
\begin{enumerate}
\item Let $L > 0$.
\begin{itemize}
\item [(i)] A necessary and sufficient condition for the existence of a solution of $x_p' = p^L$ in the case $i > 0$, where $i \in I$, is $L \geq 1+p^2$.
\item [(ii)] A necessary and sufficient condition for the existence of at least two solutions of $x_p' = p^L$ in the case $i > 0$, where $i \in I$, is $L > p^{p^{i_0+1}+i_0+1}$ provided the first solution is obtained at $i_0 \in I$.
\end{itemize}
\item Let $L < 0$.
\begin{itemize}
\item [(i)] A necessary and sufficient condition for the existence of a solution of $x_p' = p^L$ in the case $i > 0$, where $i \in I$, is $-L > p^2$.
\item [(ii)] A necessary and sufficient condition for the existence of at least two solutions of $x_p' = p^L$ in the case $i > 0$, where $i \in I$, is $-L > p^{p^{2}+1}$.
\end{itemize}
\end{enumerate}
\end{theorem}
In the remark given below, we discuss about the possibilities of the number of solutions of $x_p' = a$. Through this discussion, we have a strong belief
that Conjecture 27 of \cite{hauk} is false.
\begin{remark}
The maximum number of possible solutions may be greater than four, as
is evident from the algorithm that on increasing the value of $L$, we
have a higher range with more number of testing steps in the
alternating sequence range. Two solutions are possible at $i=0$, and
when $p\nmid (L+1)$. Then, for the positive values of $i$, we have
derived the minimum positive value or maximum negative value for $L$,
so as to have at least one solution and an alternate solution.
The possibility of two solutions exists for any value of $L$, except at
$L=0$, where only one solution is possible. At negative values of $L$,
we have one more case, namely, $C=1$. So, for negative $L$,
we already get the possibility of the existence of three solutions. We
concentrate on the positive values of $i$ for further possibilities.
For $p=2$, the minimum value of $L$ must be $5$ in order for
three or more solutions to exist. If $L$ is negative, its maximum
value must be $-5$, in order for three or more
solutions to exist. Further, $L>2^{(2^2 +2)}=64$, for the possible
existence of at least one alternate solution, given that $L>0$, which
will also form the fourth solution. New solutions are possible, if we
further increase the value of $L$.
Similarly, for $p > 2$, we can easily test for first three solutions,
but for the alternate solution, the minimum value of $L$ is $3^{11}$
for $p=3$, and $5^{26}$ for $p=5$, and so on. Due to such a high value,
it is difficult to investigate for further solutions at $p>2$, but it
is quite possible to get more than three solutions if we increase the
limit drastically beyond the given values.
\end{remark}
\section{Acknowledgment} We are thankful to the anonymous referee for
his (her) useful comments.
\begin{thebibliography}{9}
\bibitem {hauk} P. Haukkanen, J. K. Merikoski, and T. Tossavainen, On arithmetic partial differential equations, {\it J. Integer Seq.} {\bf 19} (2016),
\href{https://cs.uwaterloo.ca/journals/JIS/VOL19/Tossavainen/tossa6.pdf} {Article 16.8.6}.
\end{thebibliography}
\bigskip
\hrule
\bigskip
\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 January 31 2017;
revised version received February 23 2017.
Published in {\it Journal of Integer Sequences}, March 26 2017.
\bigskip
\hrule
\bigskip
\noindent
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\htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}.
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