1 Introduction

Let \( \mathbb {R}\,\), \( \mathbb {Q}\,\), and \( \mathbb {N}\) denote the set of all real numbers, rationals, and positive integers, respectively. We call a function \( f : \mathbb {R}\rightarrow \mathbb {R}\) additive if \( f(x+y) = f(x) + f(y) \) holds for all \( x,y \in \mathbb {R}\,\). The function f is called \(\mathbb {Q}\) -homogeneous if the equation \( f(qx) = q f(x) \) is fulfilled by every \( q \in \mathbb {Q}\) and \( x \in \mathbb {R}\,\). As it is also well-known [6, Theorem 5.2.1], if \( f : \mathbb {R}\rightarrow \mathbb {R}\) is additive, then f is \(\mathbb {Q}\)-homogeneous as well.

We define the following sets:

$$\begin{aligned} S_0&= \{ (x, y) \in \mathbb {R}^2 \,\vert \, xy = 1 \},\\ S_1&= \{ (x, y) \in \mathbb {R}^2 \,\vert \, x^2 - y^2 = 1 \}, \\ S_2&= \{ (x, y) \in \mathbb {R}^2 \,\vert \, x^2 + y^2 = 1 \}. \end{aligned}$$

Moreover, if p and q are regular, non-constant, real polynomials, while m is a positive real number, we shall also consider the sets

$$\begin{aligned} R_{p,q}= & {} \{\, (p(t),q(t)) \,\vert \, t \in \mathbb {R}\,\}, \\ S_{1,m}= & {} \{ (x, y) \in \mathbb {R}^2 \,\vert \, x^2 - m y^2 = 1 \}. \end{aligned}$$

Z. Kominek, L. Reich and J. Schwaiger [5] investigated additive real functions that satisfy the additional equation

$$\begin{aligned} f(x)f(y) = 0 \end{aligned}$$
(1)

for every \((x,y) \in D \), considering various subsets D of \( \mathbb {R}^2 \). In several cases, involving \( D = R_{p,q} \) and \( D = S_2 \), they obtained \( f(x) = 0 \) for every \( x \in \mathbb {R}\). Their result for \( D = S_2 \) was extended by Z. Boros and W. Fechner [1] to the situation when f is a generalized polynomial. On the other hand, P. Kutas [7] has recently established the existence of a non-zero additive function \( f : \mathbb {R}\rightarrow \mathbb {R}\) fulfilling (1) for all \( (x,y) \in S_0 \,\). The case of bounded f(x)f(y) on \( S_2 \) was investigated by these authors [2].

The purpose of the present paper is to involve the case \( D = S_{1,m} \) into this research for every positive rational m. We note that, in some sense, \( S_{1,1} = S_1 \) is on a half way from \( S_0 \) to \( S_2 \,\), as it is geometrically analogous to \( S_0 \) and algebraically analogous to \( S_2 \,\). Moreover, motivated by [1, Theorem 1], we wish to extend the investigation of the cases \( D = R_{p,q} \) and \( D = S_{1,m} \) for a generalized polynomial or monomial \( f \,\), respectively.

2 Concepts and lemmas

Let \( n \in \mathbb {N} \) . A function \( F : \mathbb {R}^{n} \rightarrow \mathbb {R} \) is called n-additive if F is additive in each of its variables. Clearly, an n-additive function is also \(\mathbb {Q}\)-homogeneous in each variable.

Given a function \(F:\mathbb {R}^{n}\rightarrow \mathbb {R}\), by the diagonalization (or trace) of F we understand the function \( f : \mathbb {R} \rightarrow \mathbb {R} \) arising from F by putting all the variables (from \(\mathbb {R}\)) equal:

$$\begin{aligned} f(x)=F(x, \ldots , x) \qquad \left( x\in \mathbb {R}\right) . \end{aligned}$$

If, in particular, f is a diagonalization of an n-additive function \( F : \mathbb {R}^{n} \rightarrow \mathbb {R} \,\), we say that f is a generalized monomial of degree \( n \,\). It is convenient to assume that generalized monomials of degree zero are precisely constant mappings. It is also clear that the set of all monomials of degree n is a real linear space with respect to the pointwise operations for any non-negative integer n.

If f is a finite sum of generalized monomials, then f is called a generalized polynomial.

For more information concerning these notions the reader is referred to the monograph by M. Kuczma [6, Chapter 15.9] as well as to the short introduction in [1].

In order to make use of the already mentioned \(\mathbb {Q}\)-homogeneity property of n-additive functions, in our arguments we shall repeatedly apply the following observation. If a regular real polynomial P(s) equals zero for infinitely many distinct values of the variable s, then it is identically zero, i.e., the coefficient of \( s^k \) equals zero for every non-negative integer k (up to the degree of P). Clearly, it follows from the fact that for a not identically zero polynomial P the equation \( P(s) = 0 \) is satisfied only by a finite number of distinct values of s. This idea is explicitly stated in [3, Lemma 1]. The application of this idea in the theory of functional equations goes back to the paper by Nishiyama and Horinouchi [8].

We shall also need to verify the following statements.

Lemma 2.1

If \( f : \mathbb {R}\rightarrow \mathbb {R}\) is a generalized polynomial, \( I \subseteq \mathbb {R}\) is a non-degenerated interval and \( f(x) = 0 \) for every \( x \in I \,\), then \( f(x) = 0 \) for all \( x \in \mathbb {R}\,\).

Proof

Due to our assumptions, there exist a positive integer n and k-additive mappings \( A_k : {\mathbb {R}}^k \rightarrow \mathbb {R}\) \( (k = 1,\dots ,n) \) such that

$$\begin{aligned} f(x) = \sum _{k=0}^{n} A_{k}^{*}(x) \end{aligned}$$
(2)

for every \( x \in \mathbb {R}\,\), where \( A_{0}^{*}(x) = A_0 \in \mathbb {R}\) and

$$\begin{aligned} A_{k}^{*}(x) = A_k (x,x,\dots ,x) \quad (x \in \mathbb {R}\,,\ k = 1,2,\dots ,n). \end{aligned}$$

According to the hypothesis, \( I \subseteq \mathbb {R}\) is an interval with positive length. From the density of \(\mathbb {Q}\) in \(\mathbb {R}\) we can see that for any real number \( x \ne 0 \) there exist infinitely many \( r \in \mathbb {Q}\) such that \( rx \in I \) and thus

$$\begin{aligned} 0 = f(rx) = \sum _{k=0}^{n} A_{k}^{*}(rx) = \sum _{k=0}^{n} r^k A_{k}^{*}(x) \,. \end{aligned}$$

We have just obtained a polynomial of degree (at most) n with infinitely many rational zeroes. This implies that the polynomial is identically zero, hence \( A_{k}^{*}(x) = 0 \) for every \( k \in \{\, 0 ,\,, 1 ,\,, \dots ,\, n \,\} \), which yields \( f(x) = 0 \,\). In particular, we have \( 0 = A_{0}^{*}(x) = A_0 = f(0) \,\). Therefore, f vanishes on \( \mathbb {R}\,\). \(\square \)

Lemma 2.2

If \( f : \mathbb {R}\rightarrow \mathbb {R}\) is a generalized monomial and \( p : \mathbb {R}\rightarrow \mathbb {R}\) is a regular real polynomial, then \( \ f \circ p \ \) is a generalized polynomial.

Proof

Let j be a positive integer such that f is a monomial of degree j, i.e., f is the diagonalization of the j-additive mapping \( A : \mathbb {R}^{j} \rightarrow \mathbb {R}\,\). Moreover, let n be a non-negative integer and \( a_k \in \mathbb {R}\) \((k=0,1,\dots .n)\) such that

$$\begin{aligned} p(x) = \sum _{k=0}^{n} a_k x^k \end{aligned}$$

for every \( x \in \mathbb {R}\,\). Then

$$\begin{aligned} f(p(x))= & {} f \left( \sum _{k=0}^{n} a_k x^k \right) = A \left( \sum _{k_1 = 0}^{n} a_{k_1} x^{k_1}, \dots , \sum _{k_j = 0}^{n} a_{k_j} x^{k_j} \right) \\ {}= & {} \sum _{k_1 = 0}^{n} \dots \sum _{k_j = 0}^{n} A \left( a_{k_1} x^{k_1} , \dots , a_{k_j} x^{k_j} \right) . \end{aligned}$$

For any fixed non-negative integers \( k_l \in \{\, 0 ,\,, 1 ,\,, \dots ,\, n \,\} \) \( (l=1,2,\dots ,j) \), let \( s = \sum \limits _{l=1}^{j} k_l \) and

$$\begin{aligned} G(t_{1},\dots , t_{s}) = A \left( a_{k_1} t_1 \dots t_{k_1} \,,\, a_{k_2} t_{k_1 + 1} \dots t_{k_1 + k_2} \,,\, \dots \,,\, a_{k_j} t_{s - k_j + 1} \dots t_{s} \right) , \end{aligned}$$

where any empty product equals 1 (i.e., for \( k_l = 0 \) we have only \( a_{k_l} \) in the l-th entry of A). Due to the distributivity of multiplication of real numbers and the j-additivity of \( A \,\), G is s-additive and

$$\begin{aligned} A \left( a_{k_1} x^{k_1} \,,\, a_{k_2} x^{k_2} \,,\, \dots \,,\, a_{k_j} x^{k_j} \right) = G(x,x,\dots ,x) \,. \end{aligned}$$

Thus \( \ f \circ p \ \) is a finite sum of generalized monomials, hence it is a generalized polynomial. \(\square \)

3 Main results

Now we can establish our main theorems. The first one involves two non-constant regular polynomials with possibly different degrees.

Theorem 3.1

Let p and q be polynomials of degrees at least one. If the generalized polynomial \( f : \mathbb {R}\rightarrow \mathbb {R}\) satisfies the equation

$$\begin{aligned} f(p(x)) f(q(x)) = 0 \end{aligned}$$
(3)

for every \( x \in \mathbb {R}\,\), then \( f(x) = 0 \) identically.

Proof

Since generalized polynomials are obtained as finite sums of generalized monomials, Lemma 2.2 implies that both \( f \circ p \) and \( f \circ q \) are generalized polynomials. Now we can use a result [4, Theorem 2] by Halter-Koch, Reich and Schwaiger claiming that the set of generalized polynomials is an integral domain. In particular, if the product of two generalized polynomials is identically zero, then one of those generalized polynomials has to be identically zero as well. Therefore the functional equation (3) implies that either \( f(p(x)) = 0 \) identically or \( f(q(x)) = 0 \) identically. Due to our assumptions the ranges \( p(\mathbb {R}) \) and \( q(\mathbb {R}) \) are unbounded intervals, hence f vanishes on an unbounded interval. Applying Lemma 2.1 we obtain that \( f(x) = 0 \) for all \( x \in \mathbb {R}\,\). \(\square \)

Our second theorem involves particular hyperbolas. The major tool in our arguments is obtained by a family of linear transformations that leave such a hyperbola invariant.

Lemma 3.2

Let m denote a positive real number. Suppose that \( x \,,\ y \,,\ \alpha \) and \( \beta \) are real numbers such that

$$\begin{aligned} x^2 - m y^2 = 1 \quad \text{ and } \quad \alpha ^2 - m \beta ^2 = 1 \,. \end{aligned}$$

Then we have

$$\begin{aligned} (\alpha x + \beta m y)^2 - m (\beta x + \alpha y)^2 = 1 \end{aligned}$$

as well.

Proof

It is obtained by a straightforward calculation. \(\square \)

Remark 3.3

The geometric interpretation of this observation is that, for any \( (\alpha ,\beta ) \in S_{1,m} \,\), the linear transformation on \( {\mathbb {R}}^2 \) given by the matrix

$$\begin{aligned} \left( \begin{array}{cc} \alpha &{} m \beta \\ \beta &{} \alpha \end{array} \right) \end{aligned}$$

leaves the hyperbola \( S_{1,m} \) invariant.

Theorem 3.4

Let m denote a positive rational. Suppose that \( f :\mathbb {R}\rightarrow \mathbb {R}\) is a generalized monomial and \( f(x)f(y) = 0 \) for all solutions of the equation \( x^2 - m y^2 = 1 \,\). Then f is identically equal to zero.

Proof

Given a generalized monomial f, we can associate a positive integer k and a k-additive and symmetric functional \( A :\mathbb {R}^k \rightarrow \mathbb {R}\) with f in such a way that

$$\begin{aligned} f(x) = A (x, \ldots , x) \end{aligned}$$
(4)

for all \( x \in \mathbb {R}\,\).

Now, let \( x \in \mathbb {R}\) such that \( x \ge 1 \,\). Then there exists \( 0 \le y \in \mathbb {R}\) such that \( x^2 - m y^2 = 1 \,\). If \( \alpha , \beta \) are rational numbers such that \( \alpha ^2 - m \beta ^2 = 1 \,\), Lemma 3.2 and our assumptions on f imply

$$\begin{aligned} f (\alpha x + \beta m y) f (\beta x + \alpha y) = 0 \,. \end{aligned}$$
(5)

Next, let us denote

$$\begin{aligned} a_{l} = A (\underbrace{x, \ldots , x}_{l}, \underbrace{y, \ldots ,y}_{k-l}) \end{aligned}$$

for \( l = 0, 1, \ldots k \,\). With this notation we can calculate that

$$\begin{aligned} f(\alpha x + \beta m y) = A (\alpha x + \beta m y \,,\, \ldots \,,\, \alpha x + \beta m y) = \sum _{l=0}^{k} {k \atopwithdelims ()l} (\alpha )^l (\beta m)^{k-l} a_{l} \end{aligned}$$

and

$$\begin{aligned} f(\beta x + \alpha y) = A (\beta x + \alpha y \,,\, \ldots \,,\, \beta x + \alpha y) = \sum _{l=0}^{k} {k \atopwithdelims ()l} \alpha ^{k-l} \beta ^{l} a_{l} \,. \end{aligned}$$

Due to Eq. (5), for every pair of rationals \((\alpha , \beta )\) fulfilling \( \alpha ^2 - m \beta ^2 = 1 \,\), at least one of the foregoing expressions is equal to zero.

What is more, we can find infinitely many distinct pairs \((\alpha _j , \beta _j)\) such that \( \alpha _j^2 - m \beta _j^2 = 1 \) and both \( \alpha _j \) and \( \beta _j \) are rationals, so let us take

$$\begin{aligned} \alpha _j = \frac{m j^2 + 1}{m j^2 - 1} \quad \textrm{and} \quad \beta _j = \frac{2j}{m j^2 - 1} \end{aligned}$$
(6)

for \( j \in \mathbb {N}\) such that \( m j^2 \ne 1 \,\).

Thus, for every \( j \in N_m \doteq \mathbb {N}\setminus \{ 1/\sqrt{m} \} \), we have either

$$\begin{aligned} 0 = \sum _{l=0}^{k} {k \atopwithdelims ()l} \left( \frac{m j^2 + 1}{m j^2 - 1} \right) ^{l} \left( \frac{2mj}{m j^2 - 1} \right) ^{k-l} a_{l} \end{aligned}$$

or

$$\begin{aligned} 0 = \sum _{l=0}^{k} {k \atopwithdelims ()l} \left( \frac{m j^2 + 1}{m j^2 - 1} \right) ^{k-l} \left( \frac{2j}{m j^2 - 1} \right) ^{l} a_{l} \,. \end{aligned}$$

Multiplying both equations by \( (m j^2 - 1)^k \) and introducing the functions

$$\begin{aligned} P (j)&= \sum _{l=0}^{k} {k \atopwithdelims ()l} \left( m j^2 + 1 \right) ^l \left( 2mj \right) ^{k-l} a_{l} \,,\\ \tilde{P} (j)&= \sum _{l=0}^{k} {k \atopwithdelims ()l} \left( m j^2 + 1 \right) ^{k-l} \left( 2j \right) ^{l} a_{l} \,, \end{aligned}$$

we have \( P (j) = 0 \) or \( \tilde{P} (j) = 0 \) for each integer \( j \in N_m \,\). Hence either P or \( \tilde{P} \) has infinitely many zeros. On the other hand, both P and \( \tilde{P} \) are polynomials of degree not greater than \( 2k \,\). Therefore, one of them has to be identically equal to \( 0 \,\). So either

$$\begin{aligned} 0 = P (0) = a_{k} = A_k (x \,,\, x \,,\, \ldots \,,\, x) = f(x) \end{aligned}$$

or

$$\begin{aligned} 0 = \tilde{P} \left( \frac{i}{\sqrt{m}} \right) = \left( \frac{2i}{\sqrt{m}} \right) ^{k} a_{k} = \left( \frac{2i}{\sqrt{m}} \right) ^{k} A_k (x \,,\, x \,,\, \ldots \,,\, x) = \left( \frac{2i}{\sqrt{m}} \right) ^{k} f(x) \,, \end{aligned}$$

i.e., \( f(x) = 0 \) (here i denotes the imaginary unit as polynomials with real coefficients can be considered as polynomials over the complex number field as well).

We have thus proved \( f(x) = 0 \) for every real number \( x \ge 1 \,\). Hence, applying Lemma 2.1, we obtain that \( f(x) = 0 \) for all \( x \in \mathbb {R}\,\). \(\square \)

Corollary 3.5

Let a and b denote positive real numbers such that \( \frac{a^2}{b^2} \) is rational. Suppose that \( f :\mathbb {R}\rightarrow \mathbb {R}\) is a generalized monomial and \( f(x)f(y) = 0 \) for all solutions of the equation \( \frac{x^2}{a^2} -\frac{y^2}{b^2} = 1 \,\). Then f is identically equal to zero.

Proof

Let u and w be real numbers fulfilling the condition \( u^2 - \frac{a^2}{b^2} w^2 = 1 \,\). Moreover, let \( g(t) = f(at) \) for all \( t \in \mathbb {R}\,\). Clearly, then g is a generalized monomial as well. For \( x = au \) and \( y = aw \) we have

$$\begin{aligned} \frac{x^2}{a^2} - \frac{y^2}{b^2} = u^2 - \frac{a^2}{b^2} w^2 = 1 \,, \end{aligned}$$

hence our assumption yields \( g(u) g(w) = f(au) f(aw) = f(x) f(y) = 0 \,\). Therefore g satisfies the assumptions in Theorem 3.4 with \( m = \frac{a^2}{b^2} \,\), hence g is identically equal to zero, which yields \( f(x) = g (x/a) = 0 \) for every \( x \in \mathbb {R}\) as well. \(\square \)