Remarks on the plurality function (Q1895251)

This paper deals with the conditional functional equation \[ f(x)f(y) \neq\underline 0 \Rightarrow f(x+y)=f(x)f(y) \tag \(*\) \] where \(f : \mathbb{R}^n_+ \to \mathbb{R}^n_+\) \((\mathbb{R}^n_+=[0,+\infty)^n \setminus\{\underline 0\})\), \(\underline 0=(0, \ldots, 0) \in \mathbb{R}^n\) and sum and product are componentwise. The solutions of \((*)\) are called plurality functions. In a previous paper [Aequationes Math. 47, No. 2-3, 175-190 (1994; Zbl 0804.39004)] the author proved that if \(f\) is a solution of \((*)\) and there exists a rational number \(r \neq 1\) such that \(f(rx)=f(x)\), then \(f\) assumes values in the set \(O(n)=\{(x_1, \ldots, x_n) \in \mathbb{R}^n_+\): \(x_\nu=0\) or \(x_\nu=1\), \(\nu=1, \ldots, n\}\). The main result of the present paper shows that the statement above remains true if \(r \neq 1\) is an algebraic number. Furthermore some remarks about plurality functions are presented.