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\begin{center}
\vskip 1cm{\LARGE\bf 
Multiplicative Functions Additive on the \\
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Sums of Two Positive Triangular Numbers
}
\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}
Let $T_2$ be the set of sums of two positive triangular numbers. If a multiplicative function $f$ satisfies $f(a+b) = f(a) + f(b)$ for all $a,b \in T_2$, then $f$ is the identity function provided that $f(n) \ne 0$ for some $n \ne 1,3,5$. 
\end{abstract}


\section{Introduction}

A function $f:\mathbb{N} \to \mathbb{C}$ is called \emph{multiplicative} if $f(1) = 1$ and $f(mn) = f(m)\,f(n)$ for all relatively prime integers $m$ and $n$.

In 1992 Spiro showed that if a multiplicative function $f$ satisfies the condition
\[
	f(p+q) = f(p)+f(q)
\]
for all primes $p, q$ and $f(p_0) \ne 0$ for some prime $p_0$, then $f$ is the identity function \cite{Spiro}. 
She called this property \emph{additive uniqueness}. 

Since her paper, many mathematicians have studied the additive uniqueness of various sets. Fang, Dubickas, and \v{S}arka \cite{Fang, DS} obtained the same result with the extended condition
\[
	f(p_1+p_2+\dots+p_k) = f(p_1) + f(p_2) + \dots + f(p_k)
\]
with $k \ge 3$. 

Chung \cite{Chung} found all multiplicative functions satisfying
\[
	f(m^2+n^2) = f(m^2)+f(n^2)
\]
for all $m,n \in \mathbb{N}$. In this case, the function $f$ is not determined uniquely. Later, the author \cite{Park2018AequatMath} showed that if the condition is changed to
\[
	f(a_1^2 + a_2^2 + \dots + a_k^2)
	= f(a_1^2) + f(a_2^2) + \dots + f(a_k^2)
\]
with $k \ge 3$, then $f$ is the identity function. 

Let $T$ be the set of positive triangular numbers. That is,
\[
	T = \left\{ \frac{n(n+1)}{2} \,|\, n = 1,2,\ldots \right\}
	= \left\{ 1, 3, 6, 10, 15, 21, 28, 36, 45, \ldots \right\}.
\]
Chung and Phong \cite{CP} showed that $T$ is an additive uniqueness set for multiplicative functions. The author and his colleagues \cite{KKLP} showed that the set 
\[
	P = \left\{
		\frac{n(3n-1)}{2} \,|\, n \in \mathbb{Z}, n \ne 0
	\right\}
	= \{ 1, 2, 5, 7, 12, 15, 22, 26, 35, 40, \dots \}
\]
of positive generalized pentagonal numbers is an additive uniqueness set for multiplicative functions. Also, the condition for the additive uniqueness of the sets $T$ and $P$ can be extended as was done for the set of primes and the set of squares \cite{Park2021, H, Park2023}.

The author \cite{Park2022} found all multiplicative functions $f$ satisfying the condition
\[
	f(a^2+b^2+c^2+d^2) = f(a^2+b^2) + f(c^2+d^2)
\]
for all positive integers $a$, $b$, $c$, and $d$.

Now we consider the problem of replacing squares with triangular numbers in the previous condition. Let $T_2$ be the set of sums of two elements of $T$. That is,
\[
	T_2 = \left\{ 2, 4, 6, 7, 9, 11, 12, 13, 16, 18, 20, 21, 22, 24, 25, 27, 29, 30, \ldots \right\}.
\]
Then the following holds.

\begin{theorem}
If a multiplicative function $f$ satisfies the condition
\[
	f(a+b) = f(a)+f(b)
\]
for all $a,b \in T_2$, then $f$ is one of the following:
\begin{enumerate}
\item $f(n) = n$, the identity function,
\item $f(n) = 0$ for all $n \ne 1,3,5$ and $f(3)f(5) = 0$.
\end{enumerate}
\label{thm1}
\end{theorem}


\section{Proof}

We use induction. First, we compute some values of $f(n)$. Second, we represent $f(n)$ as a sum of $f(a)+f(b)$ with $a,b \in T_2$ for sufficiently large $n$.

\begin{lemma}
If $f(2) \ne 0$, then $f(n) = n$ for $n \le 21$.
\end{lemma}
\begin{proof}
It is enough to consider powers of primes. 
Note that $f(4) = f(2)+f(2) = 2f(2)$. Since
\begin{align*}
	f(2)\,f(3) 
	&= f(6) \\
	&= f(2+4) = f(2) + f(4) = 3f(2),
\end{align*}
we obtain that $f(3) = 3$.

Then we can obtain $f(7) = 7$, since
\begin{align*}
	f(2)\,f(7) 
	&= f(14) \\
	&= f(2+12) = f(2) + f(3)\,f(4) = 7\,f(2).
\end{align*}
Also, we obtain $f(2) = 2$ and $f(4) = 4$, since
\begin{align*}
	f(2)\,f(7) 
	&= f(14) \\
	&= f(7) + f(7) = 2\,f(7).
\end{align*}

Then $f(5) = 5$ from $f(2)\,f(5) = f(2)\,f(3) + f(4)$ and
\begin{align*}
	f(8) &= f(4) + f(4) = 8,
	&f(9) &= f(2) + f(7) = 9,\\
	f(11) &= f(4) + f(7) = 11,
	&f(13) &= f(4) + f(9) = 13,\\
	f(16) &= f(7) + f(9) = 16,
	&f(17) &= f(4) + f(13) = 17,\\
	f(19) &= f(7) + f(3)\,f(4) = 19.
\end{align*}
Hence $f(n) = n$ for $n \le 21$.
\end{proof}

\begin{lemma}\label{lem:f(2)=0,f(3)!=0}
If $f(2) = 0$ and $f(3) \ne 0$, then $f(n) = 0$ for all $4 \le n \le 21$.
\end{lemma}
\begin{proof}
Cleary, $f(4) = f(2) + f(2) = 0$ and $f(8) = f(4) + f(4) = 0$. We obtain $f(7) = 0$, since 
\[
	f(7) + f(7) = f(14) = f(2)\,f(7) = 0.
\]
Then $f(11) = f(9) = 0$, since
\[
	f(4) + f(7) = f(11) = f(2) + f(9).
\]
Also, $f(5) = f(13) = 0$, since
\[
	f(3)\,f(5) = f(15) = f(2)\,f(3) + f(9) = f(2) + f(13).
\]
We obtain $f(16) = 0$, since
\[
	f(16) = f(7) + f(9).
\]
Then $f(17) = f(19) = 0$, since
\[
	f(17) = f(4) + f(13)\quad\text{and}\quad
	f(19) = f(7) + f(3)\,f(4).
\]
Hence $f(n) = 0$ for all $4 \le n \le 21$.
\end{proof}

\begin{lemma}\label{lem:f(2)=0,f(5)!=0}
If $f(2) = 0$ and $f(5) \ne 0$, then $f(n) = 0$ for all $2 \le n \le 21$ except $n=5$.
\end{lemma}

\begin{proof}
The proof is almost identical to the proof of Lemma~\ref{lem:f(2)=0,f(3)!=0}.

We obtain $f(2) = f(4) = f(8) = 0$, since
\[
	f(4) = f(2) + f(2)\quad\text{and}\quad 
	f(8) = f(4) + f(4).
\]
Also, $f(7) = f(9) = f(11) = 0$, since
\[
	f(7) + f(7) = f(2)\,f(7)\quad\text{and}\quad
	f(4) + f(7) = f(11) = f(2) + f(9).
\]

Then $f(3) = f(13) = 0$, since
\[
	f(3)\,f(5) = f(15) = f(2)\,f(3) + f(9) = f(2) + f(13).
\]
We obtain $f(16) = f(17) = f(19) = 0$, since
\[
	f(16) = f(7)+f(9),\quad
	f(17) = f(4)+f(13)\quad\text{and}\quad
	f(19) = f(7)+f(3)\,f(4).
\]

Hence $f(n) = 0$ for all $2 \le n \le 21$ and $n \ne 5$.
\end{proof}


\begin{lemma}
If $f(2) = f(3) = f(5) = 0$, then $f(n) = 0$ for all $2 \le n \le 21$.
\end{lemma}

\begin{proof}
We can derive $f(13) = 0$ from $f(3)\,f(5) = f(2) + f(13)$.

It is obvious that $f(n) = 0$ for other $4 \le n \le 21$ by the same methods as the proofs of Lemmas~\ref{lem:f(2)=0,f(3)!=0} and \ref{lem:f(2)=0,f(5)!=0}. 
\end{proof}

Now we prove Theorem~\ref{thm1}. Note that $21 \in T$ and 
\[
	21 = 6 + 15 = 1 + 10 + 10.
\]
We know that every positive number can be represented as a sum of three triangular numbers including $0$ by Fermat's polygonal number theorem or Gauss's Eureka theorem. So, if $n > 21$, then $n$ is a sum of four \emph{positive} triangular numbers. Because, if $n-21 = a+b+c$ for some triangular numbers $a \ge b \ge c \ge 0$, then
\[
	n = \begin{cases}
	a + b + c + 21 & \text{if } a \ge b \ge c \ge 1,\\
	a + b + 10 + 11 & \text{if } a \ge b \ge 1 \text{ and } c = 0,\\
	a + 1 + 10 + 10 & \text{if } a \ge 1 \text{ and } b = c = 0.
	\end{cases}
\]

Hence, Theorem~\ref{thm1} can be proved by induction according to the values of $f(2)$, $f(3)$, and $f(5)$.

Since $3 \notin T_2$, information about $f(3)$ is always from $f(3)\,f(n)$ with $3 \nmid n$. Thus, when $f(2) = 0$ and $f(3) \ne 0$, we can set $f(3)$ to an arbitrary number. So can $f(5)$, if $f(2) = 0$ and $f(5) \ne 0$.


\section{Acknowledgment}

This work was supported by the National Research Foundation of Korea (NRF) grant funded by the Korean government (MSIT) (NRF-2021R1A2C1092930).


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\bibitem{Chung} P. V. Chung,
Multiplicative functions satisfying the equation $f(m^2 +n^2) = f(m^2)+f(n^2)$,
{\it Math. Slovaca} {\bf 46} (1996), 165--171.

\bibitem{CP} P. V. Chung and B. M. Phong,
Additive uniqueness sets for multiplicative functions,
{\it Publ. Math. Debrecen} {\bf 55} (1999), 237--243.

\bibitem{DS} A. Dubickas and P. \v{S}arka,
On multiplicative functions which are additive on sums of primes, 
{\it Aequationes Math.} {\bf 86} (2013), 81--89. 

\bibitem{Fang} J.-H. Fang,
A characterization of the identity function with equation
$f(p+q+r)=f(p)+f(q)+f(r)$,
{\it Combinatorica} {\bf 31} (2011), 697--701.

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

\bibitem{KKLP} 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{Park2018AequatMath} P.-S. Park,
On $k$-additive uniqueness of the set of squares for multiplicative functions,
{\it Aequationes Math.} {\bf 92} (2018),
487--495.

\bibitem{Park2021} P.-S. Park, 
Multiplicative functions which are additive on triangular numbers, 
{\it Bull. Korean Math. Soc.} {\bf 58} (2021), 
603--608.

\bibitem{Park2022} P.-S. Park,
Multiplicative functions which are additive on sums of two nonzero squares,
{\it Integers} {\bf 22}, (2022), \#A87.

\bibitem{Park2023} P.-S. Park,
The 3-additive uniqueness of generalized pentagonal numbers for multiplicative functions,
{\it J. Integer Sequences} \textbf{26} (2023),  
\href{https://cs.uwaterloo.ca/journals/JIS/VOL26/Park/park14.html}{Article 23.5.7}.

\bibitem{Spiro} C. A. Spiro, 
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\end{thebibliography}


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

\noindent \emph{Keywords:} additive uniqueness, multiplicative function, sum of two positive triangular numbers.

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\noindent (Concerned with sequence
\seqnum{A051533}.)

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\vspace*{+.1in}
\noindent
Received December 31 2024;
revised versions received August 14 2025; September 1 2025.
Published in {\it Journal of Integer Sequences}, September 8 2025.

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