ON A $\boldsymbol {k}$-ADDITIVE UNIQUENESS SET FOR MULTIPLICATIVE FUNCTIONS

Abstract

Let $k\geq 2$ be an integer. We prove that the 2-automatic sequence of odious numbers $\mathcal {O}$ is a k-additive uniqueness set for multiplicative functions: if a multiplicative function f satisfies a multivariate Cauchy’s functional equation $f(x_1+x_2+\cdots +x_k)=f(x_1)+f(x_2)+\cdots +f(x_k)$ for arbitrary $x_1,\ldots ,x_k\in \mathcal {O}$ , then f is the identity function $f(n)=n$ for all $n\in \mathbb {N}$ .

1 Introduction

An arithmetic function $f:\mathbb {N}\to \mathbb {C}$ is multiplicative if $f(1)=1$ and $f(mn)=f(m)f(n)$ whenever m and n are relatively prime. Let $\mathcal {M}$ denote the set of complex-valued multiplicative functions.

A set $E\subseteq \mathbb {N}$ is an additive uniqueness set of a set of arithmetic functions $\mathcal {F}$ if there is exactly one element $f\in \mathcal {F}$ that satisfies

$$ \begin{align*} f(m+n)=f(m)+f(n) \quad \text{for all} \ m,n\in E. \end{align*} $$

For example, $\mathbb {N}$ and $\{1\}\cup 2\mathbb {N}$ are trivially additive uniqueness sets of $\mathcal {M}$ .

This concept was introduced by Spiro [13] in 1992. She proved that the set of primes is an additive uniqueness set of $\mathcal {M}_0=\{f\in \mathcal {M} \mid f(p_0)\neq 0 \ \text {for some prime}\ p_0\}$ and asked whether other interesting sets were additive uniqueness sets for multiplicative functions. Spiro’s work has been extended in many directions.

Let $k\geq 2$ be a fixed integer. If there is only one function $f\in \mathcal {F}$ which satisfies ${f(x_1+x_2+\cdots +x_k)=f(x_1)+f(x_2)+\cdots +f(x_k)}$ for arbitrary $x_i\in E$ , $i\in \{1,2,\ldots ,k\}$ , then E is called a k-additive uniqueness set of $\mathcal {F}$ .

In 2010, Fang [5] proved that the set of primes is a 3-additive uniqueness set of $\mathcal {M}_0$ . In 2013, Dubickas and ${\breve{\mathrm{S}}}$ arka [4] generalised Fang’s result to sums of arbitrary primes.

In 1999, Chung and Phong [3] showed that the set of positive triangular numbers $T_n=\tfrac 12n(n+1)$ , $n\in \mathbb {N}$ , and the set of positive tetrahedral numbers ${\textit {Te}}_n=\tfrac 16n(n+1)(n+2)$ , $n\in \mathbb {N}$ , were new additive uniqueness sets for $\mathcal {M}$ . Park [11] extended their work to sums of k triangular numbers, $k\geq 3$ .

In 2018, Kim et al. [7] proved that the set of generalised pentagonal numbers $P_n=\tfrac 12n(3n-1)$ , $n\in \mathbb {Z}$ , is an additive uniqueness set for $\mathcal {M}$ . Recently, they showed that the set of positive pentagonal numbers and the set of positive hexagonal numbers $H_n=n(2n-1)$ , $n\in \mathbb {N}$ , are new additive uniqueness sets for the collection of multiplicative functions [8]. They also conjectured that among the sets of s-gonal numbers, only the sets of triangular, pentagonal and hexagonal numbers are additive uniqueness sets for $\mathcal {M}$ .

Park [9] proved that the set of nonzero squares is a k-additive uniqueness set of $\mathcal {M}$ for every $k\geq 3$ , although it is not a 2-additive uniqueness set [2]. In 2020, he showed that $\{p-1 \mid p \ \text { is a prime}\}$ is an additive uniqueness set for $\mathcal {M}$ [10].

Recently, the author [6] proved that the set of practical numbers is a k-additive uniqueness set of $\mathcal {M}$ for every $k\geq 2$ .

A set $S\subseteq \mathbb {N}$ is called an additive basis (respectively, an asymptotic additive basis) of order j for $\mathbb {N}$ if there is a constant j such that every natural number (respectively, every sufficiently large natural number) can be written as a sum of at most j members of S. For example, the classical Lagrange theorem asserts that the set of squares is an additive basis of order $4$ , and Gauss (1796) proved that the triangular numbers form an additive basis of order $3$ . The famous binary Goldbach conjecture is equivalent to the assertion that the set of primes is an asymptotic additive basis of order $3$ .

A set $S\subseteq \mathbb {N}$ is called k-automatic if there exists a deterministic finite automaton M that recognises the language of base k representations of elements of S [1].

A number is odious if the number of ones in its base 2 representation is odd. The set of odious numbers is 2-automatic. Let $\mathcal {O}$ be the set of odious numbers, that is,

$$ \begin{align*} \mathcal{O}=\{1,2,4,7,8,11,13,14,16,19,21,22,25,26,28,31,\ldots\}. \end{align*} $$

Using automata theory, Rajasekaran et al. [12] proved the following result.

Theorem 1.1 (Rajasekaran et al., 2020).

A natural number is the sum of exactly two odious numbers if and only if it is not of the form $2\cdot 4^i-1$ for $i\ge 0$ .

The next theorem, also from Rajasekaran et al. [12], shows that the set of odious numbers is an asymptotic additive basis of order 3.

Theorem 1.2 (Rajasekaran et al., 2020).

Every natural number $N>15$ is the sum of three distinct odious numbers.

We prove the following theorem showing that the set of odious numbers is an additive uniqueness set of $\mathcal {M}$ .

Theorem 1.3. Fix $k\geq 2$ . The set $\mathcal {O}$ of odious numbers is a k-additive uniqueness set of $\mathcal {M}$ : if a multiplicative function f satisfies

$$ \begin{align*} f(x_1+x_2+\cdots+x_k)=f(x_1)+f(x_2)+\cdots+f(x_k) \end{align*} $$

for arbitrary $x_1,\ldots ,x_k\in \mathcal {O}$ , then f is the identity function.

It would be interesting to see whether a result similar to Theorem 1.3 holds for other classes of automatic sets.

2 Proof of Theorem 1.3

The proof consists of four parts.

Case I: $k=2$ . It is easy to show by induction that $f(2^k)=2^k$ for all $k\in \mathbb {N}$ , because $f(2)=f(1+1)=2$ and $f(2^{k+1})=f(2\cdot 2^k)=2f(2^k)$ . Suppose that N is an integer such that $f(n)=n$ for all $n\le N$ . We show that $f(N+1)=N+1$ . If $N+1\ne 2\cdot 4^i-1$ for $i\ge 1$ , then by Theorem 1.1 there are two distinct odious numbers $x,y$ such that $N+1=x+y$ and $x,y\le N$ . Thus, $f(N+1)=f(x)+f(y)$ so that $f(N+1)=N+1$ . If $N+1=2\cdot 4^i-1$ for some $i\ge 1$ , then

$$ \begin{align*} 2^{2i+1}=f(2^{2i+1})=f(2\cdot4^i-1+1)=f(N+1)+1, \end{align*} $$

since $2\cdot 4^i-1=2^{2i+1}-1=\underbrace {11\ldots 1}_{2i+1}{}_2\in \mathcal {O}$ . Therefore, $f(N+1)=N+1$ . Note that in this case we do not use the multiplicativity of f.

Case II: $k=3$ . Clearly, $f(3)=3$ and $f(10)=f(2)f(5)=f(2)[2f(2)+1]$ . On the other hand, $f(10)=f(4+4+2)=2f(4)+f(2)$ and $f(4)=f(2+1+1)=f(2)+2$ . Hence, $f^2(2)-f(2)-2=0$ with two solutions $f(2)=-1$ and $f(2)=2$ . The first solution yields $f(4)=1$ , which leads to the contradiction

$$ \begin{align*} f(6)& =f(4+1+1)=f(4)+2=3 \\ & =3f(2)=-3. \end{align*} $$

Therefore, we conclude that $f(2)=2$ . From this, it is easy to check that $f(n)=n$ for $1\leq n\leq 15$ . Assume that $f(n)=n$ for all $n\leq N$ . We have $N\geq 15$ . We show that $f(N+1)=N+1$ . By Theorem 1.2, there exist distinct odious numbers $x,y$ and z such that $N+1=x+y+z$ where $x,y,z<N$ . Hence, the assumption $f(n)=n$ for all $n\leq N$ yields $f(N+1)=f(x+y+z)=f(x)+f(y)+f(z)=x+y+z=N+1$ .

Case III: $k=4$ . By Theorem 1.2 and straightforward calculations, every integer $\ge 4$ can be written as a sum of four odious numbers.

Note that $f(4)=4$ , $f(6)=f(2)f(3)=f(2+2+1+1)=2f(2)+2$ and $f(12)=4f(3)=f(4+4+2+2)=8+2f(2)$ . For convenience, let $a=f(2)$ , $b=f(3)$ . This gives the system of equations

$$ \begin{align*} \begin{cases} ab=2a+2\\ 2b=a+4. \end{cases} \end{align*} $$

We obtain the two solutions $f(2)=-2$ , $f(3)=1$ and $f(2)=2$ , $f(3)=3$ . The first solution yields $f(5)=f(2+1+1+1)=1$ , which leads to the contradiction

$$ \begin{align*} f(10)&=f(4+4+1+1)=10\\ &=f(2)f(5)=-2. \end{align*} $$

Thus, we can conclude that $f(2)=2$ , $f(3)=3$ . So, $f(n)=n$ for $n\leq 4$ , and f must be the identity function by induction.

Case IV: $k\geq 5$ . In this case we follow closely Park’s argument in [11]. It is clear that the sum of k odious numbers can represent k but cannot represent any number from 1 to $k-1$ . Since sums of four odious numbers represent all integers $\geq 4$ as in Case III, the sum

(2.1) $$ \begin{align} \underbrace{1+\cdots+1}_{k-4\ \text{times}}+x+y+z+w, \end{align} $$

where $x,y,z,w\in \mathcal {O}$ , can represent all integers $\geq k$ .

Let $k\geq 5$ . Note that

$$ \begin{align*} (k-2)+8&=(k-2)\cdot1+4+4\\ &=(k-2)\cdot1+7+1,\\ (k-3)+18&=(k-3)\cdot1+14+2+2\\ &=(k-3)\cdot1+7+7+4,\\ (k-4)+33&=(k-4)\cdot1+28+2+2+1\\ &=(k-4)\cdot1+14+14+4+1. \end{align*} $$

Let $a=f(2)$ , $b=f(4)$ , $c=f(7)$ . The above equalities give rise to the system of equations

$$ \begin{align*} \begin{cases} 2b=c+1\\ ac+2a=2c+b\\ bc+2a=2ac+b. \end{cases} \end{align*} $$

The solutions are

$$ \begin{gather*} f(2)=\tfrac{1}{4},\ f(4)=\tfrac{1}{2},\ f(7)=0\\ f(2)=f(4)=f(7)=1\\ f(2)=2,\ f(4)=4,\ f(7)=7. \end{gather*} $$

Observe that $f(k+1)=k-1+f(2)$ , $f(k+4)=k-2+f(4)+f(2)$ and $f(k+6)= k-4+f(4)+3f(2)$ .

If $\text {gcd}(4,k+1)=1$ , the equalities

$$ \begin{align*} f(4(k+1))=f(\underbrace{4+\cdots4}_{k-3\ \text{times}}+7+7+2)&=f(4)(k-3)+2f(7)+f(2)\\ &=f(4)f(k+1)=f(4)(k-1+f(2)) \end{align*} $$

exclude the first set of solutions $f(2)=\tfrac 14$ , $f(4)=\tfrac 12$ , $f(7)=0$ .

If $4\nmid k+1$ but $2 \mid k+1$ , then $\text {gcd}(4,k+4)=1$ , and the equalities

$$ \begin{align*} f(4(k+4))=f(\underbrace{4+\cdots4}_{k-3\ \text{times}}+14+7+7)&=f(4)(k-3)+f(2)f(7)+2f(7)\\ &=f(4)f(k+4)=f(4)(k-2+f(4)+f(2)) \end{align*} $$

exclude the first set of solutions.

Finally, if $4 \mid k+1$ , then $\text {gcd}(4,k+6)=1$ , and we consider

$$ \begin{align*} f(4(k+6))=f(\underbrace{4+\cdots4}_{k-3\ \text{times}}+28+7+1)&=f(4)(k-3)+f(4)f(7)+f(7)+1\\ &=f(4)f(k+6)=f(4)(k-4+f(4)+3f(2)), \end{align*} $$

which excludes the first set of solutions.

Now consider the second solution set $f(2)=f(4)=f(7)=1$ . Arrange the odious numbers into an increasing sequence, and let $x_n$ denote the nth term. Then, $f(x_1)=f(x_2)=f(x_3)=f(x_4)=1$ . As seen in Case III, every $x_n$ with $n\geq 3$ can be written as a sum of four odious numbers. From the equality

(2.2) $$ \begin{align} (k-5)+1+2+2+2+x_e =(k-5)+7+x_a+x_b+x_c+x_d \end{align} $$

we infer that $f(x_n)=1$ for all $n\geq 5$ inductively. But for sufficiently large n, $x_n$ can be represented as a sum of k odious numbers by (2.1), so $f(x_n)=k$ , which is a contradiction.

Hence, we conclude that $f(2)=2$ , $f(4)=4$ and $f(7)=7$ . Moreover, (2.2) yields $f(x_n)=x_n$ for every $n\geq 1$ .

If N is a sum of k odious numbers, then $f(N)=N$ . Otherwise, choose an integer $M\geq k$ such that $\text {gcd}(M,N)=1$ . Then, M and $MN$ can be represented as sums of k odious numbers by (2.1). By the multiplicativity of f,

$$\begin{align*}Mf(N)=f(M)f(N)=f(MN)=MN.\end{align*}$$

Therefore, $f(N)=N$ , and this completes the proof.