Multiplicative Functions Additive on the Sums of Two Positive Triangular Numbers

Poo-Sung Park Department of Mathematics Education Kyungnam University Changwon 517670 South Korea

                     pspark@kyungnam.ac.kr

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 ∈ T_{2}, then f is the identity function provided that f(n) ≠ 0 for some n ≠ 1, 3, 5.

1 Introduction

A function f : ℕ → ℂ is called 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}) ≠ 0 for some prime p_{0}, then f is the identity function [11]. She called this property additive uniqueness.

Since her paper, many mathematicians have studied the additive uniqueness of various sets. Fang, Dubickas, and Šarka [4, 3] obtained the same result with the extended condition

f(p_{1} + p_{2} + ⋯ + p_{k}) = f(p_{1}) + f(p_{2}) + ⋯ + f(p_{k})

with k ≥ 3.

Chung [1] found all multiplicative functions satisfying

f(m^{2} + n^{2}) = f(m^{2}) + f(n^{2})

for all m, n ∈ ℕ. In this case, the function f is not determined uniquely. Later, the author [7] showed that if the condition is changed to

f(a_{1}^{2} + a_{2}^{2} + ⋯ + a_{k}^{2}) = f(a_{1}^{2}) + f(a_{2}^{2}) + ⋯ + f(a_{k}^{2})

with k ≥ 3, then f is the identity function.

Let T be the set of positive triangular numbers. That is,

T = \{\frac{n(n + 1)}{2} | n = 1, 2, …\} = \{1, 3, 6, 10, 15, 21, 28, 36, 45, …\} .

Chung and Phong [2] showed that T is an additive uniqueness set for multiplicative functions. The author and his colleagues [6] showed that the set

P = \{\frac{n(3n − 1)}{2} | n ∈ ℤ, n ≠ 0\} = \{1, 2, 5, 7, 12, 15, 22, 26, 35, 40, … \}

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 [8, 5, 10].

The author [9] 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} = {2, 4, 6, 7, 9, 11, 12, 13, 16, 18, 20, 21, 22, 24, 25, 27, 29, 30, …} .

Then the following holds.

Theorem 1. If a multiplicative function f satisfies the condition

f(a + b) = f(a) + f(b)

for all a, b ∈ T_{2}, then f is one of the following:

1. f(n) = n, the identity function,

2. f(n) = 0 for all n ≠ 1, 3, 5 and f(3)f(5) = 0.

2 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 ∈ T_{2} for sufficiently large n.

Lemma 2. If f(2) ≠ 0, then f(n) = n for n ≤ 21.

Proof. It is enough to consider powers of primes. Note that f(4) = f(2) + f(2) = 2f(2). Since

f(2) f(3) = f(6)
= f(2 + 4) = f(2) + f(4) = 3f(2),

we obtain that f(3) = 3.

Then we can obtain f(7) = 7, since

f(2) f(7) = f(14)
= f(2 + 12) = f(2) + f(3) f(4) = 7 f(2).

Also, we obtain f(2) = 2 and f(4) = 4, since

f(2) f(7) = f(14)
= f(7) + f(7) = 2 f(7).

Then f(5) = 5 from f(2) f(5) = f(2) f(3) + f(4) and

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.

Hence f(n) = n for n ≤ 21.

Lemma 3. If f(2) = 0 and f(3) ≠ 0, then f(n) = 0 for all 4 ≤ n ≤ 21.

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) and f(19) = f(7) + f(3) f(4).

Hence f(n) = 0 for all 4 ≤ n ≤ 21.

Lemma 4. If f(2) = 0 and f(5) ≠ 0, then f(n) = 0 for all 2 ≤ n ≤ 21 except n = 5.

Proof. The proof is almost identical to the proof of Lemma 3.

We obtain f(2) = f(4) = f(8) = 0, since

f(4) = f(2) + f(2) and f(8) = f(4) + f(4).

Also, f(7) = f(9) = f(11) = 0, since

f(7) + f(7) = f(2) f(7) and 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), f(17) = f(4) + f(13) and f(19) = f(7) + f(3) f(4).

Hence f(n) = 0 for all 2 ≤ n ≤ 21 and n ≠ 5.

Lemma 5. If f(2) = f(3) = f(5) = 0, then f(n) = 0 for all 2 ≤ n ≤ 21.

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 ≤ n ≤ 21 by the same methods as the proofs of Lemmas 3 and 4.

Now we prove Theorem 1. Note that 21 ∈ 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 positive triangular numbers. Because, if n − 21 = a + b + c for some triangular numbers a ≥ b ≥ c ≥ 0, then

n = \begin{cases}a + b + c + 21 & if a ≥ b ≥ c ≥ 1, \\ a + b + 10 + 11 & if a ≥ b ≥ 1 and c = 0, \\ a + 1 + 10 + 10 & if a ≥ 1 and b = c = 0. \end{cases}

Hence, Theorem 1 can be proved by induction according to the values of f(2), f(3), and f(5).

Since 3 ∉ T_{2}, information about f(3) is always from f(3) f(n) with 3 ∤ n. Thus, when f(2) = 0 and f(3) ≠ 0, we can set f(3) to an arbitrary number. So can f(5), if f(2) = 0 and f(5) ≠ 0.

3 Acknowledgment

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

References

[1] P. V. Chung, Multiplicative functions satisfying the equation f(m^{2} + n^{2}) = f(m^{2}) + f(n^{2}), Math. Slovaca 46 (1996), 165–171.

[2] P. V. Chung and B. M. Phong, Additive uniqueness sets for multiplicative functions, Publ. Math. Debrecen 55 (1999), 237–243.

[3] A. Dubickas and P. Šarka, On multiplicative functions which are additive on sums of primes, Aequationes Math. 86 (2013), 81–89.

[4] J.-H. Fang, A characterization of the identity function with equation f(p + q + r) = f(p) + f(q) + f(r), Combinatorica 31 (2011), 697–701.

[5] E. Hasanalizade, Multiplicative functions k-additive on generalized pentagonal numbers, Integers 22 (2022), #A43.

[6] B. Kim, J. Y. Kim, C. G. Lee, and P.-S. Park, Multiplicative functions additive on generalized pentagonal numbers, C. R. Math. Acad. Sci. Paris 356 (2018), 125–128.

[7] P.-S. Park, On k-additive uniqueness of the set of squares for multiplicative functions, Aequationes Math. 92 (2018), 487–495.

[8] P.-S. Park, Multiplicative functions which are additive on triangular numbers, Bull. Korean Math. Soc. 58 (2021), 603–608.

[9] P.-S. Park, Multiplicative functions which are additive on sums of two nonzero squares, Integers 22, (2022), #A87.

[10] P.-S. Park, The 3-additive uniqueness of generalized pentagonal numbers for multiplicative functions, J. Integer Sequences 26 (2023), Article 23.5.7.

[11] C. A. Spiro, Additive uniqueness sets for arithmetic functions, J. Number Theory 42 (1992), 232–246.

2020 Mathematics Subject Classification: Primary 11N64. Keywords: additive uniqueness, multiplicative function, sum of two positive triangular numbers.

(Concerned with sequence \underline{A051533}.) Received December 31 2024; revised versions received August 14 2025; September 1 2025. Published in Journal of Integer Sequences, September 8 2025. Return to Journal of Integer Sequences home page.