<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[12pt,reqno]{article}

\usepackage[usenames]{color}
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsthm}
\usepackage{amsfonts}
\usepackage{amscd}
\usepackage{graphicx}

\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}
\usepackage{latexsym}
\usepackage{epsf}
\usepackage{breakurl}

\setlength{\textwidth}{6.5in}
\setlength{\oddsidemargin}{.1in}
\setlength{\evensidemargin}{.1in}
\setlength{\topmargin}{-.1in}
\setlength{\textheight}{8.4in}

\newcommand{\seqnum}[1]{\href{https://oeis.org/#1}{\rm \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 the Minimal Number of Solutions\\
\vskip .03in
of the Equation $\phi(n+k) = M \phi(n),\ M=1,2$ \\
}
\vskip 1cm
\large
{Matteo Ferrari}\\
Dipartimento di Scienze Matematiche ``G.L.\ Lagrange''\\
Politecnico di Torino \\  Corso Duca degli Abruzzi 24 \\ 10138  Torino\\
Italy\\
\href{mailto: matteo.ferrari@polito.it}{\tt matteo.ferrari@polito.it}\\
\vskip 1cm
{Lorenzo Sillari}\\
Scuola Internazionale Superiore di Studi Avanzati (SISSA) \\ 
Via Bonomea 265\\ 34136 Trieste\\
Italy\\
\href{mailto: lsillari@sissa.it}{\tt lsillari@sissa.it}\\ 
\end{center}

\vskip .2 in

\begin{abstract} 
We fix a positive integer $k$ and look for solutions $n \in \mathbb{N}$ of the equations $\phi (n+k) = \phi (n)$ and $\phi(n+k) = 2 \phi (n)$. For $k \le 12 \cdot 10^{100}$, we prove that Fermat primes can be used to build five solutions for the first equation when $k$ is even, and five for the second one when $k$ is odd. Furthermore, for $k \le 4 \cdot 10^{58}$, we show that for the second equation there are at least three solutions when $k$ is even. Our work increases the previously known minimal number of solutions for both equations.
\end{abstract}

\section{Introduction}

Euler's phi function $\phi(n)$ counts the number of positive integers less than or equal to $n$ that are coprime with $n$. In 1956, Sierpi\'{n}ski \cite{Sierpinski1956} considered the equation
\begin{equation}\label{equation:M:1}
\phi (n+k) = \phi (n),
\end{equation}
and proved that, for every positive integer $k$, there exists at least one solution. This result was improved by Schinzel \cite{Schinzel1958}, who showed that \eqref{equation:M:1} has at least two solutions for every positive integer $k \le 8 \cdot 10^{47}$. The upper bound on $k$ was extended to $2 \cdot 10^{58}$ by Schinzel and Wakulicz \cite{SchinzelWakulicz1959} for all $k$, and to $1.38 \cdot 10^{26595411}$ by Holt \cite{Holt2003} for even $k$. All these results were obtained exhibiting explicit solutions, built proving (or assuming) the existence of prime numbers satisfying certain conditions. Graham, Holt and Pomerance \cite{GrahamHoltPomerance1999} showed that, for even $k$, if we take $j$ and $r$ such that $j$ and $j+k$ have the same prime factors, and the numbers $jr/g +1, (j+k)r/g  +1$, with $g= \gcd (j,j+k)$, are both primes not dividing $j$, then 
\[
n = j \biggl( \frac{j+k}{g}r +1 \biggr) 
\]
is a solution of \eqref{equation:M:1} \cite[Theorem 1]{GrahamHoltPomerance1999}. Despite the lack of general results, numerical evidence suggests that \eqref{equation:M:1} has infinitely many solutions for every fixed $k$. This can be proven for some special values of $k$ (cf.\ \cite{Sungjing2021}).

In the same spirit of Sierpi\'{n}ski, Makowski \cite{Makowski1974} considered the equation
\begin{equation}\label{equation:M:2}
    \phi (n+k) = 2 \phi (n),
\end{equation}
finding one solution for all fixed $k$. More recently, the same equation was studied by Hasanalizade \cite{Hasanalizade2021}, who proved that \eqref{equation:M:2} has at least two solutions for all $k \le 4 \cdot 10^{58}$ and at least three solutions for some odd $k \le 4 \cdot 10^{58}$ \cite[Theorem 1]{Hasanalizade2021}. Hasanalizade also gave \cite[Lemma 1]{Hasanalizade2021} a modified version of \cite[Theorem 1]{GrahamHoltPomerance1999}, valid for odd $k$ multiple of $3$ and for equation \eqref{equation:M:2}.

In this paper we find new solutions for \eqref{equation:M:1} and \eqref{equation:M:2}. These solutions are obtained in two ways. First, we consider Fermat prime numbers and show that for each Fermat prime it is possible to build a solution for \eqref{equation:M:1}, when $k$ is even, and for \eqref{equation:M:2}, when $k$ is odd. Numerically, we show that these solutions can be actually built for $ k \le 2 \cdot 10^{100}$. Since only five Fermat primes are known, this method provides five different solutions. Second, we exhibit a new solution for equation \eqref{equation:M:2} for even $k$. Together with the results of Hasanalizade, this gives our main theorem.
\begin{theorem}\label{main}
Equation \eqref{equation:M:2} has at least three solutions for all $k \le 4 \cdot 10^{58}$.
\end{theorem}
We conclude the paper by providing several ways of building particular solutions when certain conditions are met.

\section{The equation \texorpdfstring{$\phi(n) = \phi(n+k)$}{}}

Let $F_m$ denote the $m$-th Fermat number
\[
F_m = 2^{2^m}+1.
\]
It is known that $F_m$ is prime for $m=0, \ldots ,4$, while it is composite for all $5 \le m \le 32$. It is not known whether or not there exist other values of $m$ such that $F_m$ is prime.

Let $a$ and $b$ be positive integers. If the prime factors of $a$ are contained in the prime factors of $b$ we write $a \mid ^* b$. If $a \mid ^* b$, then it is easy to see that
\[
\phi (ab) = a \phi(b).
\]
Moreover, $a$ has the same prime factors as $b$ if and only if $a \mid ^*  b $ and $b \mid ^*  a$. In this case, we have that
\[
a  \phi(b) = b  \phi (a).
\]
We prove the following result:

\begin{theorem}\label{even:M:1}
    Equation \eqref{equation:M:1} has at least six solutions for all even $k \le 12 \cdot 10^{100}$.
\end{theorem}
\begin{proof}
We build a solution for each $F_m$ prime, exploiting the elementary fact that $ (F_m-1) \mid ^* k$ for even $k$.

\medskip
\noindent
\textbf{Case 1:} $\gcd (F_m,k)=1$. We have that
\[
n = (F_m -1) k = 2^{2^m} k
\]
is a solution of \eqref{equation:M:1}. In fact 
\[
    \phi (n + k ) = \phi (F_m k ) = \phi(F_m) \phi(k) = (F_m -1) \phi (k), 
\]
and
\[
    \phi(n) = \phi ((F_m-1)k) = (F_m-1) \phi(k).
\]
Note that in this case we have no upper bound on the values of $k$.


\medskip\noindent
\textbf{Case 2:} $F_m \mid k$. Assume that there exists a positive integer $r$ such that $(F_m -1)r +1$ and $F_m r +1$ are both primes and do not divide $k$. Then 
\[
n = (F_m-1)(F_m r+1)k
\]
is a solution of \eqref{equation:M:1}. In fact
\[
    \phi(n+k) = \phi ( ( (F_m-1)F_m r + F_m ) k )
    = \phi ( F_m ((F_m-1)r+1 )k )
    = F_m (F_m-1) r \phi(k), 
\]
while
\[
    \phi(n) = \phi ( (F_m-1)(F_mr+1)k ) 
    = F_m (F_m-1) r \phi(k).
\]
We have that $(F_m-1)r+1$ and $F_mr+1$ are both prime, for the values of $m$ and $r$ given in table \ref{m:r}.

\begin{table}[h]
\centering
\begin{tabular}{|c|c|}
			\hline
			$m$ &amp; $r$   \\
			\hline
	 0 &amp; $10^{100} + 9760$\\
     1 &amp; $10^{100} + 60128$\\
     2 &amp; $10^{100} + 150326$\\
     3 &amp; $10^{100} + 51326$\\
     4 &amp; $10^{100} + 14786$\\
     \hline
\end{tabular}
\caption{Values of $m$ and $r$.}
\label{m:r}
\end{table}

The numbers $(F_m-1)r+1$ and $F_mr+1$ are certainly prime with $k$, taking $ k \le 12 \cdot 10^{100}$. Indeed, since $2F_m \mid k$, taking $k &lt; F_m ((F_m-1)r+1)$, neither $(F_m-1)r+1$ nor $F_mr +1$ divide $k$.
We point out that different choices of $m$ and $r$ provide different solutions. To these we must add the previously known solutions. In particular, Sierpi\'{n}ski's solution might coincide with one of our solutions for some $k$, while Schinzel's solution differs from the ones we provided. Indeed, Schinzel found a solution building a sequence of odd primes $q_1,q_2,\ldots,q_\ell$, such that
$2q_i - 1$ is prime for $i= 1,\ldots,\ell$, and $2q_i - 1 \ne q_j$ 
for $1 \leq i, j \leq \ell$. Then, for even $k &lt; q_1q_2 \cdots q_\ell$, there exists a prime $q_j$ in the sequence such that both $q_j$ and $2q_j-1$ do not divide $k$. Schinzel's solution $n=(2q_j-1)k$ differs from our solutions, since the former is an odd multiple of $k$, while the latter is an even multiple of $k$. This brings the minimal number of known solutions to six for all even $k \le 2 \cdot 10^{100}$.
\end{proof}

\begin{remark}\label{rem:Dickson}
    It is a well-known conjecture by Dickson that for any fixed $a, b$ there exist infinitely many positive integers $r$ such that both $ar+1$ and $br+1$ are prime. Following Graham, Holt and Pomerance \cite{GrahamHoltPomerance1999}, we write $\mathcal{P}(a,b)$ if such a property holds. Assuming $\mathcal{P}(F_m,F_m-1)$ for $m =0, \ldots ,4$, we can remove the upper bound on $k$, obtaining that \eqref{equation:M:1} has at least six solutions for all even $k$.
\end{remark}


\begin{remark}\label{rem:infinite}
    Assume that $F_m$ divides $k$, for some fixed $m$. Then, for any $r$ such that $(F_m-1)r+1$ and $F_mr+1$ are both prime and do not divide $k$, we find a solution $n$ depending on $r$. Different values of $r$ yield different solutions. As a consequence, if $\mathcal{P}(F_m,F_m-1)$ is true, then equation \eqref{equation:M:1} has an infinite number of solutions for all even $k$ such that $F_m \mid k$.
\end{remark}


\section{The equation \texorpdfstring{$\phi(n+k) = 2 \phi(n)$}{}}

Our main results concern the number of solutions of \eqref{equation:M:2}. As for equation \eqref{equation:M:1}, there is substantial difference in the solutions that can be found for even $k$ and odd $k$. For equation \eqref{equation:M:1}, it has already been observed \cite[Table 1]{GrahamHoltPomerance1999} that solutions abound for even $k$, while they are scarce for odd $k$. After an extensive numerical investigation, we find that the same phenomenon holds for equation \eqref{equation:M:2}, with the parity of $k$ inverted. Indeed, we have an analogous version of Theorem \ref{even:M:1} (with a similar proof), that holds for odd $k$ instead of even $k$.

\begin{theorem}\label{odd:M:2}
    Equation \eqref{equation:M:2} has at least five solutions for all odd $k \le 6 \cdot 10^{100}$.
\end{theorem}
\begin{proof}
We build a solution for each $F_m$ prime, noting that $\gcd (F_m-1,k)=1$, and that
\[
F_m -1 = 2 \phi (F_m -1).
\]
\textbf{Case 1: }$\gcd (F_m,k)=1$. We have that
\[
n = (F_m -1) k = 2^{2^m} k
\]
is a solution of \eqref{equation:M:2}. Again, in this case we have no upper bound on the values of $k$ for which we find solutions.


\medskip\noindent
\textbf{Case 2: } $F_m \mid k$. Assume that there exists a positive integer $r$ such that $(F_m -1)r +1$ and $F_m r +1$ are both primes and do not divide $k$. Then 
\[
n = (F_m-1)(F_m r+1)k
\]
is a solution of \eqref{equation:M:1}. Reasoning similarly to Theorem \ref{even:M:1}, we have five solutions for all odd $k \le 6 \cdot 10^{100}$. Indeed, since $F_m \mid k$, taking $k &lt; F_m ((F_m-1)r+1)$, neither $(F_m-1)r+1$ nor $F_mr +1$ divide $k$.
\end{proof}

\begin{remark}
Remarks \ref{rem:Dickson} and \ref{rem:infinite} are also valid in this case, replacing even $k$ with odd $k$. In particular for $m=0$, Remark \ref{rem:infinite} applied to equation \eqref{equation:M:2} has, as a consequence, the second part of \cite[Lemma 1]{Hasanalizade2021}.
\end{remark}

When $k$ is even, equation \eqref{equation:M:2} has at least two solutions. The solution $n=k$ was noticed by Makowski \cite{Makowski1974}. Another solution was found by Hasanalizade \cite{Hasanalizade2021}, building a sequence of primes \cite[\seqnum{A001259}]{oeis}. Here we recall the construction. Take a sequence of primes $3=p_1 &lt; p_2 &lt; \cdots &lt; p_m$ satisfying, for all $i = 2, \ldots, m$, the following conditions:
\begin{itemize}
\item $(p_i-2) \mid \prod\limits_{j \le i-1} p_j$,
\item $(p_i-1) \mid ^*  2 \prod\limits_{j \le i-1} p_j$,
\end{itemize}
and such that $\prod_{j} p_j$ does not divide $k$. Let $p_\ell$ be the smallest prime of the sequence such that $\gcd (p_\ell,k)=1$. Then $n = p_\ell/(p_\ell-2)k$ is a solution of equation \eqref{equation:M:2}. Building the explicit sequence of primes $\{3, 5, 7, 17, 19, 37, 97, 113, 257, 401, 487, 631, 971, 1297, 1801, 19457, 22051$, $28817, 65537, 157303, 160001\}$, we have one solution different from $n=k$ for all even $k \le 4 \cdot 10^{58}$.

Now we find a new third solution for even $k$.

\begin{theorem}\label{even:M:2}
Equation \eqref{equation:M:2} has at least three solutions for all even $k \le 4 \cdot 10^{58}$.
\end{theorem}
\begin{proof}
Assume that there exists a sequence of prime numbers $2 = p_1 &lt; p_2 &lt; \cdots &lt; p_m$ and of positive integers $a_2 &lt; \cdots &lt; a_m$ such that, for all $i = 2, \ldots , m$, the following conditions hold:
\begin{itemize}
    \item $p_i = 2a_i +1$,
    \item $a_i \mid ^* \prod\limits_{j \le i-1} p_j$,
    \item $(a_i +1) \mid \prod\limits_{j \le i-1} p_j$,
\end{itemize}
and such that $\prod_j p_j$ does not divide $k$. Let $p_\ell$ be the smallest prime such that $\gcd (p_\ell,k)=1$. Then $n = a_\ell k /(a_\ell+1)$ is a solution of \eqref{equation:M:2}. In fact
\[
\phi(n+k) = \phi\biggl( \frac{2a_\ell +1}{a_\ell +1}k \biggr) = 2a_\ell \phi\biggl(\frac{k}{a_\ell +1}\biggr),
\]
and
\[
2  \phi(n) = 2  \phi\biggl( \frac{a_\ell} {a_\ell +1}k \biggr) = 2a_\ell  \phi\biggl(\frac{k}{a_\ell +1}\biggr).
\]
A sequence of primes \cite[\seqnum{A358717}]{oeis}, up to $10^8$, satisfying the above assumptions is given by 
\[
\{2,3,5,11,19,37,73,109,1459,2179,2917,4357,8713\}.
\]
The product of the prime numbers in the above sequence is of the order of $6 \cdot 10^{26}$, and $\prod_j p_j$ does not divide $k$, for $k &lt; \prod_j p_j$. To improve the upper bound on $k$, we slightly modify our argument. We proceed as follows: 
\begin{itemize}
\item If $k$ is not divisible by $2 \cdot 3 \cdot 5 \cdot 11 \cdot 19$, then we simply apply our argument with no changes.
\item If $2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 19 \mid  k$, we take the sequence starting with $2, 3, 5, 7$ and then built using the rules as explained before. The solution to \eqref{equation:M:2} will be again $n=a_\ell k/(a_\ell +1)$, where $\ell$ is the smallest index such that $\gcd (p_\ell ,k)=1$. The sequence \cite[\seqnum{A358718}]{oeis} continues as
\begin{align*}
\{&amp;2, 3, 5, 7, 11, 13, 19, 29, 37, 41, 43, 59, 73, 83, 109, 113, 131, 163, 173, 181,\\ 
&amp;227, 257, 331, 347, 353, 379, 419, 491, 523, 571, 601, 653, 661, 677, \ldots, 12011\}.
\end{align*}
In this case, since the sequence contains the number $7$, it becomes dense enough in the primes to give a large upper bound. We find that such upper bound can be taken to be of the order of $2 \cdot 10^{310}$.
\item If $2 \cdot 3 \cdot 5 \cdot 11 \cdot 19 \mid k, \gcd (7,k)=1$, and $\gcd (13,k)=1$, then $n = 36k/55$ is a solution of \eqref{equation:M:2}.
\item If $2 \cdot 3 \cdot 5 \cdot 11 \cdot 13 \cdot 19 \mid  k, \gcd (7, k)=1$, and $\gcd (23,k)=1$, then $n = 66k/95$ is a solution of \eqref{equation:M:2}.
\item If $2 \cdot 3 \cdot 5 \cdot 11 \cdot 13 \cdot 19 \cdot 23 \mid  k$ and $\gcd (7,k)=1$, we find, proceeding as before, the following sequence \cite[\seqnum{A358719}]{oeis}:
\begin{align*}
\{&amp;2, 3, 5, 11, 13, 23, 19, 37, 73, 109, 131, 229, 457, 571, 1459, 1481, 2179, 2621,\\
&amp;2917, 2963, 4357, 8713, 49921, 1318901, 3391489, 6782977, 13565953\},
\end{align*}
that gives the upper bound of $2\cdot 10^{83}$.
\end{itemize}
We conclude the proof by observing that our solution differs from the previously known solutions, since in our case $n&lt;k$. Indeed, Hasanalizade's solutions \cite{Hasanalizade2021} satisfy  $n \ge k$.
\end{proof}

\begin{remark}
Our trick can be repeated as many times as desired to improve the upper bound on $k$, as long as we can find particular solutions.
\end{remark}

As a consequence we obtain our main theorem.
\begin{proof}[Proof of Theorem \ref{main}]
Theorem \ref{odd:M:2} proves that for all odd $k \le 2 \cdot 10^{100}$ there are at least five solutions to equation \eqref{equation:M:2}, while Theorem \ref{even:M:2} proves that for all even $ k \le 4 \cdot 10^{58}$ we have at least three solutions. Overall, we get three solutions for all $k \le 4 \cdot 10^{58}$.
\end{proof}

\begin{remark}
We recall that while the upper bound on $k$ for our new solution is fairly high, increasing the upper bound of $4 \cdot 10^{58}$ for Hasanalizade's solution would require a lot of effort, as pointed out by Holt \cite{Holt2003}.
\end{remark}

\section{Special solutions}

We open the section with a modified version of the first part of \cite[Lemma 1]{Hasanalizade2021}. Our result removes the assumption $3 \mid  k$.

\begin{lemma}\label{lemma:modified}
Let $k$ be an odd positive integer. Suppose that $j$ is a positive integer such that $j$ and $2j +k$ have the same prime factors. Take $g= \gcd (j, 2j+k)$ and consider a positive integer $r$ such that $2jr/g +1$ and $(2j+k)r/g+1$ are both prime and coprime with $k$. Then
\[
n = 2j \biggl( \frac{2j+k}{g}r+1 \biggr)
\]
is a solution of equation \eqref{equation:M:2}.
\end{lemma}
\begin{proof}
We have that 
\[
\phi(n+k) = \phi\biggl(2j\biggl(\frac{2j+k}{g}r+1\biggr) + k\biggr) 
    = \phi \biggl( (2j+k) \biggl(\frac{2j}{g}r + 1\biggr) \biggr) 
    =2j  \phi(2j+k) \frac{r}{g},
    \]
    and
    \[
2  \phi(n)  = 2\phi \biggl(2j\biggl(\frac{2j+k}{g}r+1\biggr)\biggr) 
    = 2  \phi(2j) \phi\biggl(\frac{2j+k}{g}r+1\biggr) \\
    = 2  \phi(2j)(2j+k) \frac{r}{g}.
\]
We conclude the result using that
\[
2j  \phi(2j+k) = 2j  \phi(2(2j+k)) = \phi(4j(2j+k)) =  2  \phi(2j) (2j+k).
\]
\end{proof}
Motivated by explicit computations, we would like to find more general formulas that provide new families of solutions to \eqref{equation:M:2}. In practice, we always find at least four solutions in the range $n \le 10^6$, for $k \le 10^4$. The only value of $k$ for which we find exactly $4$ solutions is $k=6$. In that case, the solutions are
\[
n \in \{4,6,7,10\}.
\] 
A computer search yielded no other solutions $n \le 10^8$ for $k=6$. We notice that $n=6$ is the solution given by Makowski and $n=10$ is the solution from Hasanalizade, while $n=4$ is obtained via our method with $a_\ell=2$. There is no known family of solutions providing $n=7$. In an attempt to find such a family, we prove the following result. 
\begin{proposition}\label{prop:special}
Assume that there exists a prime $p$ such that $2p-1$ is prime, $\gcd (p,k)=1, \gcd (2p-1,k)=1$, and $(p-1) \mid  k$. Then
\[
n= \frac{p}{p-1}k
\]
is a solution of \eqref{equation:M:2}.
\end{proposition}
\begin{proof}
It is immediate to check that
    \[
 \phi(n+k) = \phi\biggl( k\frac{p}{p-1} +k \biggr) = \phi\biggl( k\frac{2p-1}{p-1} \biggr) = (2p-2) \phi\biggl(\frac{k}{p-1} \biggr), 
 \]
 and
 \[
  2  \phi(n) = 2\phi\biggl( k\frac{p}{p-1} \biggr) = 2(p-1)  \phi\biggl(\frac{k}{p-1} \biggr).
 \]
\end{proof}
We can take $p=7$ and obtain the solution $n =7$, for $k=6$. Unfortunately, this cannot be generalized to all even $k$, since the condition $(p-1) \mid  k$ is satisfied only for a finite number of $p$ and it might happen that none of them satisfies the other requirements (e.g., $k=10$). 

We observe that for all odd $k$, thanks to Proposition \ref{prop:special}, we can take $p=2$ and recover the solution $n=2k$ given by Makowski.

Finally, we conclude the paper  by giving one more way of building solutions in another special case.

\begin{proposition}
Let $k$ be an even positive integer. Assume that there exists $m$ such that $m \mid k, \gcd (m+2,k)=1, \gcd (m+4,k)=1$, and $\phi (m+2)=\phi(m+4)$. Then
\[
n = \frac{m+4}{m}k
\]
is a solution of \eqref{equation:M:2}.
\end{proposition}
\begin{proof}
Again, it is immediate to check that

 \[
 \phi(n+k) = \phi\biggl( k\frac{m+4}{m} +k \biggr) = 2  \phi\biggl( k\frac{m+2}{m} \biggr) = 2  \phi(m+2)\phi\biggl(\frac{k}{m} \biggr), 
 \]
 and
 \[
  2  \phi(n) = 2   \phi\biggl( k\frac{m+4}{m} \biggr) = 2 \phi(m+4)\phi\biggl(\frac{k}{m} \biggr) .
 \]
\end{proof}
 
\section{Acknowledgments}
We thank the anonymous referee for suggesting an argument that improved the upper bound on $k$ in Theorems \ref{even:M:1} and \ref{odd:M:2}. We also thank the referee for very helpful comments that improved the exposition of the paper. The first author is partially supported by MIUR grant \emph{``Dipartimenti di Eccellenza 2018-2022''}, CUP E11G18000350001. The second author is partially supported by GNSAGA of INdAM. The numerical computations supporting our paper have been performed using Magma V2.27-5, and are available upon request to the first author.

\begin{thebibliography}{9}
    
    \bibitem{GrahamHoltPomerance1999}
        {S. W. Graham and J. J. Holt and C. Pomerance,}
        {On the solutions to $\phi(n) = \phi(n + k)$, in \textit{Number Theory in Progress: Proceedings of the International Conference on Number Theory organized by the Stefan Banach International Mathematical Center in Honor of the 60th Birthday of Andrzej Schinzel, Zakopane, Poland, June 30--July 9, 1997,} De Gruyter, 2012, pp. 867--882.}

    \bibitem{Hasanalizade2021}
    {E. Hasanalizade,}
        {On the equation {$\phi(n+k)=2\phi(n)$},}
        {\textit{Integers} \textbf{21} (2021), Paper No. A68.}

    \bibitem{Holt2003}
    { J. J. Holt,}
    {The minimal number of solutions to $\phi(n) = \phi(n + k)$,}
    {\textit{Math. Comp.} \textbf{72} (2003), 2059--2061.}

    \bibitem{Makowski1974}
    {A. Makowski,}
    {On the equation {$\phi(n+k)=2\phi(n)$},}
    {\textit{Elem. Math.} \textbf{29} (1974), 13.}
    
    \bibitem{Schinzel1958}
    {A. Schinzel,}
    {Sur l'\'{e}quation {$\phi(x+k)=\phi(x)$},}
    {\textit{Acta Arith.} \textbf{4} (1958), 181--184.}
    
    \bibitem{SchinzelWakulicz1959}
    {A. Schinzel and A. Wakulicz,}
    {Sur l'\'{e}quation {$\phi(x+k)=\phi(x)$} \uppercase{II},}
    {\textit{Acta Arith.} \textbf{5} (1959), 425--426.}

    \bibitem{Sierpinski1956}
    {W. Sierpi\'{n}ski,}
        {Sur une propri\'{e}t\'{e} de la fonction {$\phi(n)$},}
        {\textit{Publ. Math. Debrecen} \textbf{4} (1956), 184--185.}

    \bibitem{oeis}
    {N. J. A.~Sloane et al.,} 
        {The On-line Encyclopedia of Integer Sequences, 2020.} 
        {Available at \url{https://oeis.org/}.}
        
    \bibitem{Sungjing2021}
    {K. Sungjin,}
        {On the equations $\phi(n) = \phi(n + k)$ and $\phi(p-1) = \phi(q-1)$,}
        {\textit{Int. J. Number Theory} \textbf{17} (2021), 1287--1305.}
\end{thebibliography}



\bigskip
\hrule
\bigskip

\noindent 2020 {\it Mathematics Subject Classification}: Primary 11N25. Secondary 11Y99.

\noindent \emph{Keywords:} Euler's phi function.
\bigskip
\hrule
\bigskip

\noindent 
(Concerned with sequences
\seqnum{A001259},
\seqnum{A358717},
\seqnum{A358718}, and
\seqnum{A358719}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received September 7 2022;
revised versions received December 27 2022; January 14 2023.
Published in {\it Journal of Integer Sequences}, January 15 2023.

\bigskip
\hrule
\bigskip

\noindent
Return to
\htmladdnormallink{Journal of Integer Sequences home page}{https://cs.uwaterloo.ca/journals/JIS/}.
\vskip .1in


\end{document}

                                                                                

</pre></body></html>