In this paper we prove that if a generalized polynomial function f satisfies the condition f(x) f(y) = 0 for all solutions of the equation x ² + y ² = 1, then f is identically equal to 0.

In this paper we consider a generalized monomial or polynomial $$ f : \mathbb {R}\rightarrow \mathbb {R}$$f : R → R that satisfies the additional equation $$ f(x) f(y) = 0 $$f ( x ) f ( y ) = 0 for the pairs $$ (x,y) \in D \,$$( x , y ) ∈ D , where $$ D \subseteq {\mathbb {R}}^{2} $$D ⊆ R 2 is given by some algebraic condition. In the particular cases when f is a generalized polynomial and there exist non-constant regular polynomials p and q that fulfill $$\begin{aligned} D = \{\, (p(t),q(t)) \,\vert \, t \in \mathbb {R}\,\} \end{aligned}$$D = { ( p ( t ) , q ( t ) ) | t ∈ R } or f is a generalized monomial and there exists a positive rational m fulfilling $$\begin{aligned} D = \{\, (x,y) \in {\mathbb {R}}^{2} \,\vert \, x^2 - m y^2 = 1 \,\}, \end{aligned}$$D = { ( x , y ) ∈ R 2 | x 2 - m y 2 = 1 } , we prove that $$ f(x) = 0 $$f ( x ) = 0 for all $$ x \in \mathbb {R}\,$$x ∈ R .

In this paper we give necessary conditions for quadratic functions $$ f :\mathbb {R}\rightarrow \mathbb {R}$$f : R → R that satisfy the additional equation $$ y^2 f(x) = x^2 f(y) $$y 2 f ( x ) = x 2 f ( y ) under the condition $$ xy = 1 \,$$x y = 1 .

In this paper we consider a generalized polynomial f : ℝ → ℝ of degree two that satisfies the additional equation f ( x ) f ( y ) = 0 for the pairs ( x, y ) ∈ D , where D ⊆ ℝ ² is given by some algebraic condition. In the particular cases when there exists a positive rational m fulfilling D = { ( x , y ) ∈ ℝ 2 | x 2 - m y 2 = 1 } , D = \left\{ {\left( {x,y} \right) \in \mathbb{R}{^2}|{x^2} - m{y^2} = 1} \right\}, we prove that f ( x ) = 0 for all x ∈ ℝ.