\documentclass[12pt,reqno]{article} \usepackage[usenames]{color} \usepackage{amssymb} \usepackage{graphicx} \usepackage{amscd} \usepackage[colorlinks=true, linkcolor=webgreen, filecolor=webbrown, citecolor=webgreen]{hyperref} \definecolor{webgreen}{rgb}{0,.5,0} \definecolor{webbrown}{rgb}{.6,0,0} \usepackage{color} \usepackage{fullpage} \usepackage{float} \usepackage{psfig} \usepackage{graphics,amsmath,amssymb} \usepackage{amsthm} \usepackage{amsfonts} \usepackage{latexsym} \usepackage{epsf} \setlength{\textwidth}{6.5in} \setlength{\oddsidemargin}{.1in} \setlength{\evensidemargin}{.1in} \setlength{\topmargin}{-.1in} \setlength{\textheight}{8.4in} \newcommand{\seqnum}[1]{\href{http://oeis.org/#1}{\underline{#1}}} \begin{document} \begin{center} \epsfxsize=4in \leavevmode\epsffile{logo129.eps} \end{center} \theoremstyle{plain} \newtheorem{theorem}{Theorem} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{conjecture}[theorem]{Conjecture} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \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} \\ \end{center} \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}} \vskip .2 in \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 a possibility of 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 of 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 Return to \htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}. \vskip .1in \end{document} .