Arithmetic functions commutable with sums of squares

TL;DR

This paper characterizes functions from natural numbers to complex numbers that preserve the sum of squares structure for three or more variables, revealing specific functional forms that satisfy this property.

Contribution

It provides a complete characterization of functions that commute with sums of squares for multiple variables, extending understanding of functional equations related to quadratic forms.

Findings

Identifies all functions satisfying the sum of squares functional equation for k ≥ 3

Shows the structure of such functions is highly constrained

Provides explicit forms of functions that preserve sum of squares

Abstract

In this note, we characterize all functions $f : \mathbb{N} \rightarrow \mathbb{C}$ such that $f(x_1^2+ \cdots + x_k^2)=f(x_1)^2+ \cdots + f(x_k)^2$, where $k \geq 3$ and $x_1, \cdots, x_k$ are positive integers.

Equations

$$A_{n+2}+A_{n-2}+A_1=f(2n^2+9)=2A_n+A_3(n\ge 3)$$

$$A_{n+1}+A_{n-1}+2A_2+(k-4)A_1=f(2n^2+k+6)=2A_n+(k-3)A_1+A_3(n\ge 3)$$

$$C=(3f(1{)}^2{)}^2=9A^2$$

$$C+2D=f(3{)}^2+2f(4{)}^2=f(41)=f(6{)}^2+f(1{)}^2+f(2{)}^2=(2A+B{)}^2+A+B$$

$$2A+(3B{)}^2=2f(1{)}^2+f(12{)}^2=f(146)=f(11{)}^2+f(3{)}^2+f(4{)}^2=(2A+C{)}^2+C+D$$

$$\displaystyle\begin{multlined}(A+2C)^{2}+C+A=f(19)^{2}+f(3)^{2}+f(1)^{2}=f(371)\\ =f(17)^{2}+f(9)^{2}+f(1)^{2}=(2B+C)^{2}+(A+2B)^{2}+A\end{multlined}(A+2C)^{2}+C+A=f(19)^{2}+f(3)^{2}+f(1)^{2}=f(371)\\ =f(17)^{2}+f(9)^{2}+f(1)^{2}=(2B+C)^{2}+(A+2B)^{2}+A$$

$$E+2A=f(5{)}^2+2f(1{)}^2=f(27)=3f(3{)}^2=3C$$

$$E+2B=f(5{)}^2+2F(2{)}^2=f(33)=f(1{)}^2+2f(4{)}^2=A+2D$$

$$\displaystyle\begin{multlined}(2A+C)^{2}+A+B=f(11)^{2}+f(1)^{2}+f(2)^{2}=f(126)\\ =f(9)^{2}+f(6)^{2}+f(3)^{2}=(A+2B)^{2}+(2A+B)^{2}+C\end{multlined}(2A+C)^{2}+A+B=f(11)^{2}+f(1)^{2}+f(2)^{2}=f(126)\\ =f(9)^{2}+f(6)^{2}+f(3)^{2}=(A+2B)^{2}+(2A+B)^{2}+C$$

$$\displaystyle\begin{multlined}(A+3B)^{2}+(2A+2B)^{2}+A+B=f(13)^{2}+f(10)^{2}+f(1)^{2}+f(2)^{2}\\ =f(274)=f(16)^{2}+f(4)^{2}+2f(1)^{2}=(4B)^{2}+(4A)^{2}+2A\end{multlined}(A+3B)^{2}+(2A+2B)^{2}+A+B=f(13)^{2}+f(10)^{2}+f(1)^{2}+f(2)^{2}\\ =f(274)=f(16)^{2}+f(4)^{2}+2f(1)^{2}=(4B)^{2}+(4A)^{2}+2A$$

$$\displaystyle\begin{multlined}(4B)^{2}+3B=f(16)^{2}+3f(2)^{2}=f(268)\\ =f(13)^{2}+2f(7)^{2}+f(1)^{2}=(A+3B)^{2}+2(3A+B)^{2}+A\end{multlined}(4B)^{2}+3B=f(16)^{2}+3f(2)^{2}=f(268)\\ =f(13)^{2}+2f(7)^{2}+f(1)^{2}=(A+3B)^{2}+2(3A+B)^{2}+A$$

$$(3A+B{)}^2+3A=f(7{)}^2+3f(1{)}^2=f(52)=3f(4{)}^2+f(2{)}^2=48A^2+B$$

$$E+3A=f(5{)}^2+3f(1{)}^2=f(28)=3f(3{)}^2+f(1{)}^2=3C+A$$

$$E+3B=f(5{)}^2+3f(2{)}^2=f(37)=2f(4{)}^2+f(1{)}^2+f(2{)}^2=32A^2+A+B$$

$$2(A-B{)}^2+A+2B=3C$$

$$5A+3C=8B$$

Full text

Arithmetic functions commutable with sums of squares

Jungin Lee

Abstract. In this note, we characterize all functions $f:\mathbb{N}\rightarrow\mathbb{C}$ such that $f(x_{1}^{2}+\cdots+x_{k}^{2})=f(x_{1})^{2}+\cdots+f(x_{k})^{2}$, where $k\geq 3$ and $x_{1},\cdots,x_{k}$ are positive integers.

1 Introduction

In 1996, Chung [2] classified multiplicative functions satisfying $f(m^{2}+n^{2})=f(m)^{2}+f(n)^{2}$ for all $m,n\in\mathbb{N}$. Ba$\check{s}$i$\acute{c}$ [1] generalized this result to arbitrary arithmetic functions. Recently, Park ([3], [4]) proved that for every integer $k\geq 3$, a multiplicative function $f$ satisfies $f(x_{1}^{2}+\cdots+x_{k}^{2})=f(x_{1})^{2}+\cdots+f(x_{k})^{2}$ for all $x_{1},\cdots,x_{k}\in\mathbb{N}$ is an identity function. We generalize Park’s result to arithmetic functions, as Ba$\check{s}$i$\acute{c}$ generalized Chung’s result.

2 Results

Theorem**.**

*Let $k\geq 3$ be an integer. If a function $f:\mathbb{N}\rightarrow\mathbb{C}$ satisfies $f(x_{1}^{2}+\cdots+x_{k}^{2})=f(x_{1})^{2}+\cdots+f(x_{k})^{2}$ for every $x_{1},\cdots,x_{k}\in\mathbb{N}$, then one of the following holds:

(1) $f\equiv 0$

(2) $f(n)=\left\{\begin{matrix}n\,\,(\exists x_{1},\cdots,x_{k}\in\mathbb{N}\,\,\,n=x_{1}^{2}+\cdots+x_{k}^{2})\\ \pm n\,\,(otherwise)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\end{matrix}\right.$

(3) $f(n)=\left\{\begin{matrix}\frac{1}{k}\,\,(\exists x_{1},\cdots,x_{k}\in\mathbb{N}\,\,\,n=x_{1}^{2}+\cdots+x_{k}^{2})\\ \pm\frac{1}{k}\,\,(otherwise)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\end{matrix}\right.$

(In (2) and (3), each $\pm$ sign is independent.)*

Proof.

Denote $f(n)^{2}$ by $A_{n}$. If $k=3$, then

[TABLE]

so it is enough to show that (1), (2) or (3) holds for $n\leq 4$. If $k\geq 4$, then

[TABLE]

so it is enough to show that (1), (2) or (3) holds for $n\leq 3$. For convenience, we denote $A_{1},\cdots,A_{5}$ by $A,B,C,D$ and $E$.

Case I. $k=3$

[TABLE]

From the equations (3a), (3b), (3g) and (3h), we can obtain $(B-4A)(B+8A-1)=0$.

(i) $B=4A$ : If we substitute $B=4A$ and $C=9A^{2}$ to the equation (3f), we obtain $27A^{2}(A-1)(9A+5)=0$. If $A=0$, $B=C=0$ and $D=0$ by (3b). If $A=1$, $B=4,C=9$ and $D=16$ by (3b). If $A=-\frac{5}{9}$, $B=-\frac{20}{9},C=\frac{25}{9}$ and this contradicts to (3k).

(ii) $B=1-8A$ : If we substitute $B=1-8A$ and $C=9A^{2}$ to the equations (3f) and (3k), we obtain $(9A-1)(27A^{3}+39A^{2}-52A+8)=0$ and $(9A-1)(9A^{3}+5A^{2}-29A+4)=0$, respectively. Thus $A=\frac{1}{9}$, $B=C=\frac{1}{9}$, and $D=\frac{1}{9}$ by (3b).

Case II. $k=4$

[TABLE]

From the equation (4c), we can obtain $(A-B)(11A-3B+1)=0$.

(i) $B=A$ : If we substitute $B=A$ to the equation (4f), we obtain $32A^{2}=2A$, so $A$ is [math] or $\frac{1}{16}$. By the equations (4h) and (4i), $(A,B,C)=(0,0,0),(\frac{1}{16},\frac{1}{16},\frac{1}{16})$.

(ii) $B=\frac{11A+1}{3}$ : If we substitute $B=\frac{11A+1}{3}$ to the equations (4f) and (4g), we obtain $(A-1)(160A+14)=0$ and $(A-1)(16A-1)=0$, respectively. Thus $A=1$, $B=4$ and $C=9$ by (4h) and (4i).

Case III. $k\geq 5$

[TABLE]

From $(kA)^{2}+((k-2)A+2B)^{2}+2B+(k-4)A=f(k)^{2}+f(k+6)^{2}+2f(2)^{2}+(k-4)f(1)^{2}=f(2k^{2}+13k+40)=2f(k+3)^{2}+3f(3)^{2}+(k-5)f(1)^{2}=2((k-1)A+B)^{2}+3C+(k-5)A$, we obtain (5a). From $(D+4A)+(A+3C)=f(20)+f(28)=5B+(D+3B)$, we obtain (5b). From the equations (5a) and (5b), we obtain $2(A-B)(A-B+3)=0$.

(i) $B=A$ : By the equation (5b), $C=A$ and by the equation (2), $A_{n}=A$ for every $n\in\mathbb{N}$. Then $A=A_{k}=(kf(1)^{2})^{2}=k^{2}A^{2}$, so $A$ is [math] or $\frac{1}{k^{2}}$.

(ii) $B=A+3$ : By the equation (5b), $C=A+8$ and by the equation (2), $A_{n}-n^{2}=A-1$ for every $n\in\mathbb{N}$. Then $((k-1)A+(A+3))^{2}-(k+3)^{2}=A_{k+3}-(k+3)^{2}=A_{k}-k^{2}=(kA)^{2}-k^{2}$, so $A=1$. ∎