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\begin{center}
\vskip 1cm{\Large\bf 
The 3-Additive Uniqueness of 
Generalized Pentagonal \\
\vskip .1in
Numbers for Multiplicative Functions
}
\vskip 1cm
\large
Poo-Sung Park\\
Department of Mathematics Education \\
Kyungnam University \\
Changwon 517670 \\
South Korea \\
\href{mailto: pspark@kyungnam.ac.kr}{\tt pspark@kyungnam.ac.kr}
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\begin{abstract}
The set $\mathcal{P}$ of all nonzero generalized pentagonal numbers is $k$-additively unique for multiplicative functions for fixed $k \geq 2$. That is, if a multiplicative function $f$ satisfies the condition
\[
	f(a_1 + a_2 + \dots + a_k) = f(a_1) + f(a_2) + \dots + f(a_k)
\]
for arbitrary $a_1, a_2, \dots, a_k \in \mathcal{P}$, then $f$ is the identity function. The known proof for $k=3$ uses the generalized Riemann hypothesis (GRH). Here, we give a proof bypassing GRH.
\end{abstract}

\section{Introduction}

In 1992 Spiro \cite{Spiro} coined the term \emph{additive uniqueness} to describe a curious property of a set $E \subset \mathbb{N}$ for a set $S$ of arithmetic functions. If a function $f \in S$ satisfying the condition
\[
	f(x+y) = f(x) + f(y) \text{ for all }a,b \in E
\]
is uniquely determined, she called $E$ an additive uniqueness set for $S$.

Many mathematicians have been studying and generalizing her seminal work. The concept of \emph{$k$-additive uniqueness} is a generalization. A set $E$ is called a $k$-additive uniqueness set for $S$, if a function $f \in S$ satisfying the condition
\[
	f(x_1+x_2+\dots+x_k) = f(x_1) + f(x_2) + \dots + f(x_k)\text{ for all }x_i \in E
\]
is uniquely determined.

In 2018, the author and colleagues \cite{KKLP2018} proved that the set
\[
	\mathcal{P} = \biggl\{ P_m = \frac{m(3m-1)}{2} \,\bigg|\, m \in \mathbb{Z}, m\ne 0 \biggr\}
	= \{ 1, 2, 5, 7, 12, 15, \ldots \}
\]
of all nonzero generalized pentagonal numbers is an additive uniqueness set for multiplicative functions. Here, a multiplicative function refers to an arithmetic function $f: \mathbb{N} \to \mathbb{C}$ satisfying $f(1) = 1$ and $f(ab) = f(a)f(b)$ for all $a,b$ with $\gcd(a,b) = 1$. In addition, they \cite{KKLP2021} extended this result to the set 
\[
	\biggl\{ P_m = \frac{m(3m-1)}{2} \,\bigg|\, m \in \mathbb{Z}, m \geq 1 \biggr\}
	= \{ 1, 5, 12, 22, 35, 51, \ldots \}
\]
of nonzero ordinary pentagonal numbers.

In 2022, Hasanalizade \cite{H} extended the author's finding for generalized pentagonal numbers to $k$-additive uniqueness with $k \geq 3$. However, his proof for the case of $k=3$ relies on the proposition that every integer $&gt;2$ can be written as a sum of three nonzero generalized pentagonal numbers \cite[Theorem 4.3]{KLO}, which requires the generalized Riemann hypothesis (GRH). In this article, we present an elementary proof that does not rely on GRH.

\section{Main result}

\begin{theorem}
Let $f$ be a multiplicative function and let $\mathcal{P}$ be the set of nonzero generalized pentagonal numbers. If
\[
	f(a+b+c) = f(a) + f(b) + f(c)
\]
for every $a, b, c \in \mathcal{P}$, then $f$ is the identity function.
\end{theorem}

\begin{proof}
We have $f(1)=1$ and thus $f(3) = 3$. Note that 
\begin{align*}
	f(5) &amp;= f(1)+f(2)+f(2) = 1+2\,f(2),\\
	f(7) &amp;= f(1) + f(1) + f(5) = 3+2f\,f(2).
\end{align*}
Then $f(2) = 2$ by
\begin{align*}
	f(9) 
	&amp;= f(2) + f(2) + f(5) = 1 + 4\,f(2) \\
	&amp;= f(1) + f(1) + f(7) = 5 + 2\,f(2).
\end{align*}

So, clearly, $f(n) = n$ for $n \leq 9$.

We use induction. Assume that $f(n) = n$ for all $n &lt; N$.

If $N=ab$ with $a, b \geq 2$ and $\gcd(a,b) = 1$, then $f(N) = f(a)\,f(b) = ab$ by the induction hypothesis. So it suffices to consider the case when $N$ is a power of a prime.

Consider the case of $N = 2^r$. We have that $2^r = 3m \pm 1$ for some $m$ according to the parity of $r$.

If $N = 2^{2s+1}$, then $N = 3m-1$ for some $m \in \mathbb{N}$. Note that 
\begin{align*}
	&amp;P_{m} + P_{-2m+1} + P_{-2m+1}\\
	&amp;= m \cdot \frac{3m-1}{2} + (2m-1)(3m-1) + (2m-1)(3m-1) \\
	&amp;= \frac{3m-1}{2} \cdot (9m-4)
\end{align*}
and
\begin{align*}
	&amp;P_{2m-1} + P_{-2m+1} + P_{-2m+1}\\
	&amp;= (2m-1)(3m-2) + (2m-1)(3m-1) + (2m-1)(3m-1) \\
	&amp;= (2m-1)(9m-4).
\end{align*}
Since each factor $&lt; 3m-1$ is fixed by $f$ by the induction hypothesis, letting $x = f(3m-1)$ and $y = f(9m-4)$, we have that
\begin{align*}
	m \cdot \frac{3m-1}{2} + 2(2m-1)\,x 
	&amp;= \frac{3m-1}{2}\,y \\
	(2m-1)(3m-2) + 2(2m-1)\,x
	&amp;= (2m-1)\,y.
\end{align*}
Thus $x = 3m-1$ and $y = 9m-4$. So $f(2^{2s+1}) = 2^{2s+1}$.

If we change $m$ to $-m$, then we can show that $f$ fixes $9m+4$ and $N = 2^{2s} = 3m+1$ in the same way.

Now suppose $N = 3^r$. Then
\begin{align*}
	P_{3^{r-1}} + P_{3^{r-1}} + P_{3^{r-1}} 
	&amp;=  3^{r-1} \cdot \frac{3^r-1}{2} + 3^{r-1} \cdot \frac{3^r-1}{2} + 3^{r-1} \cdot \frac{3^r-1}{2} 
	&amp;= 3^r \cdot \frac{3^r-1}{2},
\end{align*}
and thus $f(3^r) = 3^r$.

Finally, let $N = p^r$ with odd prime $p \ne 3$. Assume $N = 6m-1$ for some $m$.
Note that
\begin{align*}
	P_{m} + P_{-m} + P_{2m} 
	&amp;= \frac{m(3m-1)}{2} + \frac{m(3m+1)}{2} + m(6m-1)  \\
	&amp;= m(9m-1)
\end{align*}
and
\begin{align*}
	P_{-m} + P_{2m} + P_{2m} 
	&amp;= \frac{m(3m+1)}{2} + m(6m-1) + m(6m-1)  \\
	&amp;= \frac{3m(9m-1)}{2}.
\end{align*}
The first expression leads to an equation
\[
	3m + x = y
\]
where $x = f(6m-1)$ and $y = f(9m-1)$.

For the second expression, there  are two cases according to $m$. If $9m-1$ is even, then $m$ is odd and
\[
	f\biggl( \frac{3m(9m-1)}{2} \biggr) = f(3m)\,f\biggl(\frac{9m-1}{2}\biggr) = 3m \cdot \frac{9m-1}{2},
\]
since $3m$ and $\frac{9m-1}{2}$ are smaller than $N = 6m-1$. If $9m-1$ is odd, then $m$ is even and
\[
	f\biggl( \frac{3m(9m-1)}{2} \biggr) = f\biggl(\frac{3m}{2}\biggr)\,f(9m-1) = \frac{3m}{2} \cdot f(9m-1) = \frac{3m}{2} \cdot y.
\]
In both cases, we conclude that $x = f(6m-1) = 6m-1$ and $y = f(9m-1) = 9m-1$.

Similarly, we can show that $f(p^r) = p^r$ by changing $m$ to $-m$, when $p^r = 6m+1$. The proof is complete.
\end{proof}

\section{Acknowledgment}

This work was supported by the Ministry of Education of the Republic of Korea and the National Research Foundation of Korea (NRF-2021R1A2C1092930).


\begin{thebibliography}{9}

\bibitem{H} E. Hasanalizade,
Multiplicative functions $k$-additive on generalized pentagonal numbers,
{\it Integers} {\bf 22} (2022), \#A43.

\bibitem{KKLP2018} B. Kim, J. Y. Kim, C. G. Lee, and P.-S. Park, 
Multiplicative functions additive on generalized pentagonal numbers, 
{\it C. R. Math. Acad. Sci. Paris} {\bf 356} (2018), 125--128.

\bibitem{KKLP2021} B. Kim, J. Y. Kim, C. G. Lee, and P.-S. Park, 
Multiplicative functions additive on polygonal numbers, 
{\it Aequationes Math.} {\bf 95} (2021), 601--621.

\bibitem{KLO} D. Kim, J. Lee and B.-H. Oh, 
A sum of three nonzero triangular numbers, 
{\it Int. J. Number Theory} {\bf 17} (2021), 2279--2300.

\bibitem{Spiro} C. A. Spiro, 
Additive uniqueness sets for arithmetic functions, 
{\it J. Number Theory} {\bf 42} (1992), 232--246.

\end{thebibliography}

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\noindent 2020 {\it Mathematics Subject Classification}: Primary 11N64.

\noindent \emph{Keywords:} $k$-additive uniqueness, multiplicative function, generalized pentagonal number.

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\noindent (Concerned with sequence
\seqnum{A001318}.)

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\vspace*{+.1in}
\noindent
Received April 14 2023;
revised version received June 9 2023.
Published in {\it Journal of Integer Sequences}, June 9 2023.

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