1 Introduction

Since 1988, when Dhombres published his paper [4] concerning the alienation of functional equations, many results arising from considering a functional equation

$$\begin{aligned} E(f,g)=E_{1}(f)+E_{2}(g)=0 \end{aligned}$$

resulting from adding up two functional equations side by side, have appeared. We ask whether or not equation \(E(f,g)=0\) splits back to the system of these two functional equations, i.e.,

$$\begin{aligned} \left\{ \begin{array}{l} E_{1}(f)=0 \\ E_{2}(g)=0 \end{array} \right. . \end{aligned}$$
(1.1)

If the answer is“yes”, we say that the equations in (1.1) are strongly alien to each other. The problem of alienation was studied by many authors, we cite here just some of them [8, 11, 12, 14, 17, 18]. For a number of references on the subject, see the survey [9] by Ger and Sablik.

Let S be a semigroup that need not be abelian, \((H,+)\) an abelian group and \(\sigma \) be an endomorphism of S such that \(\sigma \circ \sigma (x)=x\) for any \(x\in S\). That the set S is a semigroup means that it is equipped with an associative composition rule \((x,y)\mapsto xy.\)

The aim of the present paper is to characterize the solutions of some functional equations resulting from summing up side by side the linear functional equations defining the additive maps

$$\begin{aligned} f(xy)=f(x)+f(y),\ x,y\in S, \end{aligned}$$
(1.2)

the quadratic equation which here means

$$\begin{aligned} f(xy)+f(x\sigma (y))=2f(x)+2f(y),\ x,y\in S, \end{aligned}$$
(1.3)

Drygas’ equation which here means

$$\begin{aligned} f(xy)+f(x\sigma (y))=2f(x)+f(y)+f(\sigma (y)),\ x,y\in S, \end{aligned}$$
(1.4)

and the following version of Jensen’s equation

$$\begin{aligned} f(xy)+f(x\sigma (y))=2f(x),\ x,y\in S. \end{aligned}$$
(1.5)

The monographs by Aczél and Dhombres [1] and by Stetkær [15] contain many references about the additive, Jensen’s, Drygas’ and the quadratic functional equations. Versions of equations similar to (1.3), (1.4) and (1.5) were studied on semigroups and monoids in [5, 6, 16].

Our main contributions to the theory of functional equations are the following: First we investigate, on the semigroup S, the alienation of ( 1.2) and (1.3), i.e., the solutions \(f,g:S\rightarrow H\) of the equation

$$\begin{aligned} f(xy)+g(xy)+g(x\sigma (y))=f(x)+f(y)+2g(x)+2g(y), \end{aligned}$$
(1.6)

which was studied on abelian semigroups by Fadli in [7].

As a consequence, we describe the solutions of the functional equation

$$\begin{aligned} f(xy)+g(xy)+g(xy^{-1})=f(x)+f(y)+2g(x)+2g(y),\ x,y\in G, \end{aligned}$$
(1.7)

which was treated, under some conditions on an abelian group G, by Ger [10] and Adam [2].

Second, we treat the solutions \(f,g:S\rightarrow H\) of the following equation

$$\begin{aligned} f(xy)+f(x\sigma (y))+g(xy)+g(x\sigma (y))=2f(x)+2g(x)+2g(y). \end{aligned}$$
(1.8)

This allows us to show that Jensen’s equation (1.5) and the quadratic equation (1.3) are strongly alien to each other on any semigroup.

Finally, in the last section, we characterize the solutions \(f,g:S\rightarrow H\) of the following equation

$$\begin{aligned} f(xy)+g(xy)+g(x\sigma (y))=f(x)+f(y)+2g(x)+g(y)+g(\sigma (y)), \end{aligned}$$
(1.9)

and this enables us to show in Corollary 4.2, with the additional condition that \((H,+)\) is divisible by 2 and 3, that the additive Cauchy equation (1.2) and Drygas’ functional equation (1.4) are alien in the sense of Dhombres on any semigroup. We will finish this section by finding, on 2-divisible abelian groups, the solutions of the functional equation

$$\begin{aligned} f(xy)+g(xy)+g(xy^{-1})=f(x)+f(y)+2g(x)+g(y)+g(y^{-1}), \end{aligned}$$
(1.10)

which was treated by the authors in [3].

Notation The following notation will be used throughout the paper unless explicitly stated otherwise. Sis a semigroup that need not be abelian and \((H,+)\) is a uniquely 2 -divisible abelian group. For any complex-valued function f on S we use the notation

$$\begin{aligned} f_{e}=\dfrac{f+f\circ \sigma }{2}\ \text {and}\ f_{o}=\dfrac{f-f\circ \sigma }{2}. \end{aligned}$$

We say that f is \(\sigma \)-even if \(f\circ \sigma =f\), and \(\sigma \)-odd if \(f\circ \sigma =-f\). We denote by \(\mathcal {A}(S)\) the set of all additive maps from S to H, and put

$$\begin{aligned} \mathcal {A}^{\pm }(S):=\{a\in \mathcal {A}(S):a\circ \sigma =\pm a\}. \end{aligned}$$

By \(\mathcal {N}(S,H,\sigma )\) we mean the set of solutions \(\theta :S\rightarrow H\) of the homogeneous equation:

$$\begin{aligned} \theta (xy)-\theta (x\sigma (y))=0,\ x,y\in S. \end{aligned}$$
(1.11)

We say that \(Q:S\rightarrow H\) is a quadratic map if it satisfies the functional equation (1.3) and \(J:S\rightarrow H\)is a Jensen map if it solves the functional equation (1.5).

2 Alienation of the cauchy and the quadratic equations

In this section we determine, in terms of additive maps, quadratic maps and elements of the set \(\mathcal {N}(S,H,\sigma ),\) the solutions \( f,g:S\rightarrow H\) of the Pexider type functional equation (1.6), namely

$$\begin{aligned} f(xy)+g(xy)+g(x\sigma (y))=f(x)+f(y)+2g(x)+2g(y),\ x,y\in S. \end{aligned}$$

Theorem 2.1

The solutions \(f,g:S\rightarrow H\) of (1.6) are the functions of the forms

$$\begin{aligned} \left\{ \begin{array}{l} f(x)=a(x)+2\theta (x) \\ g(x)=Q(x)-\theta (x) \end{array} \right. ,\ x\in S, \end{aligned}$$

where \(\theta \in \mathcal {N}(S,H,\sigma )\), \(a\in \mathcal {A}^{-}(S)\) and \( Q:S\rightarrow H\) is a quadratic map.

Proof

Assume that \(f,g:S\rightarrow H\) is a solution of the equation (1.6). Let \(f_{e}\) and \(g_{e}\) denote the \(\sigma \)-even parts of f and g, respectively. Simple computations show that the pair \(f_{e},g_{e}:S \rightarrow H\) is a solution of (1.6), i.e., we have

$$\begin{aligned} f_{e}(xy)+g_{e}(xy)+g_{e}(x\sigma (y))=f_{e}(x)+f_{e}(y)+2g_{e}(x)+2g_{e}(y),\ x,y\in S. \end{aligned}$$
(2.1)

Replacing y by \(\sigma (y)\) in (2.1), we see that

$$\begin{aligned} f_{e}(x\sigma (y))+g_{e}(x\sigma (y))+g_{e}(xy)=f_{e}(x)+f_{e}(y)+2g_{e}(x)+2g_{e}(y), \end{aligned}$$
(2.2)

for all \(x,y\in S\). Subtracting (2.1) from (2.2), we get

$$\begin{aligned} f_{e}(xy)-f_{e}(x\sigma (y))=0,\ x,y\in S. \end{aligned}$$
(2.3)

This implies that \(f_{e}\) is a solution of the homogeneous equation, i.e., there exists \(\theta ^{\prime }\in \mathcal {N}(S,H,\sigma )\) satisfying \( \theta ^{\prime }\circ \sigma =\theta ^{\prime }\) such that \(f_{e}=2\theta ^{\prime }\). Going back to (2.1), we infer that

$$\begin{aligned} (g_{e}+\theta ^{\prime })(xy)+(g_{e}+\theta ^{\prime })(x\sigma (y))=2(g_{e}+\theta ^{\prime })(x)+2(g_{e}+\theta ^{\prime })(y), \end{aligned}$$
(2.4)

for all \(x,y\in S\). This means that there exists a quadratic map \( Q:S\rightarrow H\) such that \(g_{e}=Q-\theta ^{\prime }\). Hence, it remains to determine \(f_{o}\) and \(g_{o}\) (the \(\sigma \)-odd parts of f and g, respectively). According to (1.6) and with simple computations, we get that

$$\begin{aligned} f_{o}(xy)+g_{o}(xy)+g_{o}(x\sigma (y))=f_{o}(x)+f_{o}(y)+2g_{o}(x)+2g_{o}(y),\ x,y\in S. \end{aligned}$$
(2.5)

Replacing x by \(\sigma (x)\) in (2.5), we obtain

$$\begin{aligned} f_{o}(x\sigma (y))+g_{o}(xy)+g_{o}(x\sigma (y))=f_{o}(x)-f_{o}(y)+2g_{o}(x)-2g_{o}(y), \end{aligned}$$
(2.6)

for all \(x,y\in S\). Subtracting (2.5) from (2.6), we get

$$\begin{aligned} f_{o}(xy)-f_{o}(x\sigma (y))=2f_{o}(y)+4g_{o}(y),\ x,y\in S. \end{aligned}$$
(2.7)

Furthermore, if we add (2.5) and (2.6), we see that

$$\begin{aligned} (f_{o}+2g_{o})(xy)+(f_{o}+2g_{o})(x\sigma (y))=2(f_{o}+2g_{o})(x),\ x,y\in S. \end{aligned}$$
(2.8)

Thus, there exists a Jensen’s map \(J:S\rightarrow H\) such that \( f_{o}+2g_{o}=J\) and so (2.7) reduces to

$$\begin{aligned} f_{o}(xy)-f_{o}(x\sigma (y))=2J(y),\ x,y\in S. \end{aligned}$$
(2.9)

Subtracting the two identities obtained from (2.9) and replacing y by yz and \(y\sigma (z)\) respectively, we find that

$$\begin{aligned}&f_{o}(xyz)-f_{o}(xy\sigma (z))+f_{o}(x\sigma (y)z)-f_{o}(x\sigma (y)\sigma (z)) \\&\quad =2J(yz)-2J(y\sigma (z)). \end{aligned}$$

Applying (2.9) twice to the left hand side of the last equation, we obtain

$$\begin{aligned} 4J(z)=2J(yz)-2J(y\sigma (z)),\ y,z\in S, \end{aligned}$$

which is equivalent to

$$\begin{aligned} J(yz)-J(y\sigma (z))=2J(z), \end{aligned}$$
(2.10)

because H is uniquely 2-divisible. From (2.8) and (2.10), we infer that J is a \(\sigma \)-odd additive function. Then there exists \(a\in \mathcal {A}^{-}(S)\) such that \(f_{o}+2g_{o}=a\). In view of (2.7), we obtain that

$$\begin{aligned} f_{o}(xy)-f_{o}(x\sigma (y))=2a(y),\ x,y\in S, \end{aligned}$$

which is equivalent to

$$\begin{aligned} f_{o}(xy)-f_{o}(x\sigma (y))=a(xy)-a(x\sigma (y)),\ x,y\in S. \end{aligned}$$

Hence, there exists \(\theta ^{\prime \prime }\in \mathcal {N}(S,H,\sigma )\) satisfying \(\theta ^{\prime \prime }\circ \sigma =-\theta ^{\prime \prime }\) such that \(f_{o}=a+2\theta ^{\prime \prime }\), which yields that \( g_{o}=-\theta ^{\prime \prime }\) (because H is uniquely 2-divisible). Therefore

$$\begin{aligned} f=a+2\theta \ \text {and}\ g=Q-\theta , \end{aligned}$$

where \(\theta =\theta ^{\prime }+\theta ^{\prime \prime }\in \mathcal {N} (S,H,\sigma )\), \(a\in \mathcal {A}^{-}(S)\) and \(Q:S\rightarrow H\) is a quadratic map.

The other direction of the proof can be trivially shown. \(\square \)

As a direct consequence of Theorem 2.1 we have the following result.

Corollary 2.2

([7]) Let \((S,+)\) be an abelian semigroup. The solutions \( f,g:S\rightarrow H \) of Eq. (1.6) are the functions of the forms

$$\begin{aligned} \left\{ \begin{array}{l} f(x)=A(x)-2\theta (x) \\ g(x)=B(x,x)+\theta (x) \end{array} \right. ,\ x\in S, \end{aligned}$$

where \(\theta \in \mathcal {N}(S,H,\sigma )\), \(A\in \mathcal {A}(S)\), and \( B:S\times S\rightarrow H\) is a symmetric bi-additive map such that \( B(x,\sigma (y))=-B(x,y)\).

Proof

Applying Theorem 2.1 and [13, Theorem 3] we get that

$$\begin{aligned} \left\{ \begin{array}{l} f(x)=a(x)+2\Theta (x) \\ g(x)=B(x,x)+\phi (x)-\Theta (x) \end{array} \right. ,\ x\in S, \end{aligned}$$
(2.11)

where \(\Theta \in \mathcal {N}(S,H,\sigma )\), \(a\in \mathcal {A}^{-}(S)\), \( \phi \in \mathcal {A}^{+}(S)\) and \(B:S\times S\rightarrow H\) is a symmetric bi-additive map such that \(B(x,\sigma (y))=-B(x,y)\). From (2.11), we can deduce easily the claimed result with \(A:=a+2\phi \in \mathcal {A}(S)\) and \( \theta :=\phi -\Theta \in \mathcal {N}(S,H,\sigma )\). Conversely, the formulas above define solutions of (1.6) on an abelian semigroup \( (S,+) \). \(\square \)

Remark 2.3

Suppose that S is a 2-divisible group, e its identity element and \(\sigma \) is its group inversion. Then any solution of (1.11 ) is a constant function. Indeed, for \(y=x\) in (1.11), we get that \( \theta (x^{2})=\theta (e)\). Hence, \(\theta \equiv \theta (e)\) (because S is 2-divisible).

As a consequence of Corollary 2.2 on uniquely 2-divisible abelian groups, we have the following statement due to Adam [2] about the alienation phenomenon of additivity and quadraticity up to a constant.

Corollary 2.4

([2]) Let S and H be uniquely 2-divisible abelian groups. Then, the pair of functions \(f,g:S\rightarrow H\) is a solution of Eq. (1.7) if and only if

$$\begin{aligned} \left\{ \begin{array}{l} f(x)=A(x)-2c \\ g(x)=B(x,x)+c \end{array} \right. ,\ x\in S, \end{aligned}$$
(2.12)

where \(c\in H\) is a constant, \(A\in \mathcal {A}(S),\) and \(B:S\times S\rightarrow H\) is a symmetric bi-additive map.

Proof

This corollary follows easily from Corollary 2.2 and Remark 2.3. \(\square \)

Remark 2.5

In [10], Ger showed, with the weaker conditions that S is a 2-divisible group (that need not be abelian) and \((H,+)\) is an abelian group (that need not be uniquely 2-divisible), that the pair \(f,g:S\rightarrow H\) is a solution of (1.7) if and only if

$$\begin{aligned} \left\{ \begin{array}{l} f(x)=A(x)-2c \\ g(x)=Q(x)+c \end{array} \right. ,\ x\in S, \end{aligned}$$

where \(c\in H\) is a constant, \(A\in \mathcal {A}(S)\) and \(Q:S\rightarrow H\) is a quadratic map.

3 Alienation of the jensen and the quadratic equations

In this section we determine the solution \(f,g:S\rightarrow H\) of the Eq. ( 1.8), namely

$$\begin{aligned} f(xy)+f(x\sigma (y))+g(xy)+g(x\sigma (y))=2f(x)+2g(x)+2g(y),\ x,y\in S. \end{aligned}$$

Our main contribution here is to show that Jensen’s and quadratic functional equations are strongly alien to each other on any semigroup.

Theorem 3.1

The pair \(f,g:S\rightarrow H\) satisfies the equation (1.8) if and only if

$$\begin{aligned} \left\{ \begin{array}{c} f(xy)+f(x\sigma (y))=2f(x) \\ g(xy)+g(x\sigma (y))=2g(x)+2g(y) \end{array} \right. \ \text {for all}\ x,y\in S. \end{aligned}$$
(3.1)

Proof

Let \(f,g:S\rightarrow H\) be a solution of (1.8). Replacing y by \( \sigma (y)\) in (1.8), we obtain

$$\begin{aligned} (f+g)(xy)+(f+g)(x\sigma (y))=2(f+g)(x)+2g(\sigma (y)),\ x,y\in S. \end{aligned}$$
(3.2)

Subtracting (3.2) from (1.8), we deduce easily that g is even (because H is uniquely 2-divisible). In view of (1.8), for all \( x,y,z\in S\), we have

$$\begin{aligned} (f+g)(xyz)+(f+g)(xy\sigma (z))=2(f+g)(xy)+2g(z), \end{aligned}$$
(3.3)
$$\begin{aligned} (f+g)(x\sigma (y)z)+(f+g)(x\sigma (y)\sigma (z))=2(f+g)(x\sigma (y))+2g(z), \end{aligned}$$
(3.4)
$$\begin{aligned} (f+g)(xyz)+(f+g)(x\sigma (y)\sigma (z))=2(f+g)(x)+2g(yz), \end{aligned}$$
(3.5)

and

$$\begin{aligned} (f+g)(x\sigma (y)z)+(f+g)(xy\sigma (z))=2(f+g)(x)+2g(\sigma (y)z). \end{aligned}$$
(3.6)

Summing up equalities (3.3) and (3.4) side by side, and subtracting from the equality thus obtained the sum of equalities (3.5) and (3.6), we infer that

$$\begin{aligned} (f+g)(xy)+(f+g)(x\sigma (y))+2g(z) =2(f+g)(x)+g(yz)+g(\sigma (y)z). \end{aligned}$$

In view of (1.8), we get that

$$\begin{aligned} g(yz)+g(\sigma (y)z)=2g(y)+2g(z),\ y,z\in S. \end{aligned}$$
(3.7)

Replacing z by \(\sigma (z)\) in (3.7), we obtain

$$\begin{aligned} g(yz)+g(y\sigma (z))=2g(y)+2g(z),\ y,z\in S, \end{aligned}$$
(3.8)

because g is \(\sigma \)-even. Going back to (1.8), we see that

$$\begin{aligned} f(xy)+f(x\sigma (y))=2f(x),\ x,y\in S. \end{aligned}$$

The converse is straightforward. \(\square \)

As an application of Theorem 3.1, by using [13, Theorems 2 and 3] , we derive the following result.

Corollary 3.2

Assume that S is an abelian semigroup. Then the pair of functions \( f,g:S\rightarrow H\) satisfies Eq. (1.8) if and only if

$$\begin{aligned} f(x)=a(x)+c\ \text {and}\ g(x)=B(x,x)+b(x), \end{aligned}$$

where \(c\in H\) is a constant, \(a\in \mathcal {A}^{-}(S)\), \(b\in \mathcal {A} ^{+}(S),\) and \(B:S\times S\rightarrow H\) is a symmetric bi-additive function such that \(B(x,\sigma (y))=-B(x,y)\).

4 Alienation of the cauchy and the Drygas equations

In this section we describe solutions \(f,g:S\rightarrow H\) of Eq. (1.9), namely

$$\begin{aligned} f(xy)+g(xy)+g(x\sigma (y))=f(x)+f(y)+2g(x)+g(y)+g(\sigma (y)),\ x,y\in S, \end{aligned}$$

in terms of quadratic maps, elements of \(\mathcal {N}(S,H,\sigma ),\ \sigma \) -odd Jensen maps, and solutions of the following version of Jensen’s equation

$$\begin{aligned} \varphi (yx)+\varphi (\sigma (y)x)=2\varphi (x),\ x,y\in S. \end{aligned}$$
(4.1)

Theorem 4.1

The solutions \(f,g:S\rightarrow H\) of (1.9) are the functions of the form

$$\begin{aligned} \left\{ \begin{array}{c} f(x)=2\varphi (x)+2\theta (x) \\ g(x)=Q(x)+J(x)-\varphi (x)-\theta (x) \end{array} \right. ,\ x\in S, \end{aligned}$$
(4.2)

where \(Q:S\rightarrow H\) is a quadratic map, \(J:S\rightarrow H\) is a \(\sigma \)-odd Jensen map, \(\varphi \) is a \(\sigma \)-odd solution of (4.1), and \( \theta \in \mathcal {N}(S,H,\sigma )\) such that \(\theta \circ \sigma =\theta \) .

Proof

Assume that \(f,g:S\rightarrow H\) satisfy (1.9). With simple computations we show that \(f_{e}\) and \(g_{e}\) are solutions of (1.6). From Theorem 2.1, we deduce that

$$\begin{aligned} \left\{ \begin{array}{l} f_{e}(x)=a(x)+2\theta (x) \\ g_{e}(x)=Q(x)-\theta (x) \end{array} \right. ,\ x\in S, \end{aligned}$$
(4.3)

where \(\theta \in \mathcal {N}(S,H,\sigma )\), \(a\in \mathcal {A}^{-}(S)\) and \( Q:S\rightarrow H\) is a quadratic map. Since \(g_{e}\) and Q are \(\sigma \) -even, we have \(\theta \circ \sigma =\theta \) and hence \(a=0\) because \(a\in \mathcal {A}^{-}(S)\). Therefore we infer that

$$\begin{aligned} \left\{ \begin{array}{l} f_{e}(x)=2\theta (x) \\ g_{e}(x)=Q(x)-\theta (x) \end{array} \right. ,\ x\in S. \end{aligned}$$

So it remains to characterize the \(\sigma \)-odd parts \(f_{o}\) and \(g_{o}\ \) of f and g respectively. From (1.9), we have

$$\begin{aligned} f_{o}(xy)+g_{o}(xy)+g_{o}(x\sigma (y))=f_{o}(x)+f_{o}(y)+2g_{o}(x),\ x,y\in S. \end{aligned}$$
(4.4)

Replacing y by \(\sigma (y)\) in (4.4), we get

$$\begin{aligned} f_{o}(x\sigma (y))+g_{o}(xy)+g_{o}(x\sigma (y))=f_{o}(x)-f_{o}(y)+2g_{o}(x),\ x,y\in S. \end{aligned}$$
(4.5)

If we add (4.4) and (4.5), we get that \(2J:=f_{o}+2g_{o}\) is a \( \sigma \)-odd Jensen’s map. Subtracting (4.4) from (4.5), we see that \(f_{o}\) satisfies (4.1). We put \(f_{o}=2\varphi \), where \( \varphi \) is a solution of (4.1). Hence we infer that

$$\begin{aligned} \left\{ \begin{array}{l} f_{o}(x)=2\varphi (x) \\ g_{o}(x)=J(x)-\varphi (x) \end{array} \right. ,\ x\in S. \end{aligned}$$
(4.6)

Therefore, from (4.3) and (4.6), we get

$$\begin{aligned} \left\{ \begin{array}{l} f(x)=2\varphi (x)+2\theta (x) \\ g(x)=Q(x)+J(x)-\theta (x)-\varphi (x) \end{array} \right. ,\ x\in S, \end{aligned}$$

where \(Q:S\rightarrow H\) is a quadratic map, \(J:S\rightarrow H\) is a \(\sigma \)-odd Jensen’s map, \(\varphi \) is a \(\sigma \)-odd solution of (4.1), and \(\theta \in \mathcal {N}(S,H,\sigma )\) such that \(\theta \circ \sigma =\theta \).

Conversely, simple computations prove that the formula for the pair (f, g) in (4.2) is a solution of (1.9). \(\square \)

Assume \(f=g\) and that the abelian group \((H,+)\) is divisible by 2 and 3. As a consequence of Theorem 4.1, in the following corollary we show that the additive Cauchy equation (1.2) and Drygas’ functional equation (1.4) are alien in the sense of Dhombres on any semigroup (since any additive map is a solution (1.4)).

Corollary 4.2

Assume additionally that \((H,+)\) is uniquely divisible by 2 and 3. The solutions \(f:S\rightarrow H\) of the functional equation

$$\begin{aligned} 2f(xy)+f(x\sigma (y))=3f(x)+2f(y)+f(\sigma (y)),\ x,y\in S, \end{aligned}$$
(4.7)

are the additive functions.

Proof

Applying Theorem 4.1 with \(f=g\), we infer that

$$\begin{aligned} f(x)=Q(x)+J(x)-\varphi (x)-\theta (x)=2\varphi (x)+2\theta (x),\ x\in S, \end{aligned}$$
(4.8)

where \(Q:S\rightarrow H\) is a quadratic map, \(J:S\rightarrow H\) is a \(\sigma \)-odd Jensen map, \(\varphi \) is a \(\sigma \)-odd solution of (4.1), and \( \theta \in \mathcal {N}(S,H,\sigma )\) such that \(\theta \circ \sigma =\theta \) . This implies that

$$\begin{aligned} Q(x)+J(x)=3\varphi (x)+3\theta (x),\ x\in S, \end{aligned}$$
(4.9)

which yields \(Q+J-3\varphi \in \mathcal {N}(S,H,\sigma ).\) Then we have

$$\begin{aligned} \left( Q(xy)+J(xy)-3\varphi (xy)\right) -\left( Q(x\sigma (y))+J(x\sigma (y))-3\varphi (x\sigma (y))\right) =0, \end{aligned}$$

which shows that

$$\begin{aligned} \left( Q(xy)-Q(x\sigma (y))\right) +\left( J(xy)-J(x\sigma (y))\right) =3(\varphi (xy)-\varphi (x\sigma (y))). \end{aligned}$$

and so by definition of \(\varphi \) we get

$$\begin{aligned} \left( Q(xy)-Q(x\sigma (y))\right) +\left( J(xy)-J(x\sigma (y))\right) =6\varphi (y). \end{aligned}$$
(4.10)

Using the fact that the left hand side of (4.10) is independent of x, the \(\sigma -\)eveness of Q and the \(\sigma -\)oddness of J, we obtain

$$\begin{aligned}&\left( Q(xy)-Q(x\sigma (y))\right) +\left( J(xy)-J(x\sigma (y))\right) \\&\quad \left. =\left( Q(\sigma (x)y)-Q(\sigma (x)\sigma (y))\right) +\left( J(\sigma (x)y)-J(\sigma (x)\sigma (y))\right) \right. \\&\quad \left. =\left( Q(x\sigma (y))-Q(xy)\right) +\left( -J(x\sigma (y))+J(xy\right) ).\right. \end{aligned}$$

So we infer that

$$\begin{aligned} Q(xy)-Q(x\sigma (y))=Q(x\sigma (y))-Q(xy), \end{aligned}$$

which yields that

$$\begin{aligned} Q(xy)=Q(x\sigma (y))\text { for all }x,y\in S. \end{aligned}$$

Thus

$$\begin{aligned} 2Q(xy)=Q(xy)+Q(x\sigma (y))=2Q(x)+2Q(y), \end{aligned}$$

for all \(x,y\in S\) and so \(A:=Q\) is an additive function with \(A\circ \sigma =A\). Finally, from (4.8) and (4.9), we get

$$\begin{aligned} f=2(\varphi +\theta )=\frac{2}{3}\left( Q+J\right) =\frac{2}{3}\left( A+J\right) . \end{aligned}$$
(4.11)

Replacing f by its expression \(\frac{2}{3}\left( A+J\right) \) in (4.7 ), we find that

$$\begin{aligned}&\frac{2}{3}\left( 3A(x)+3A(y)+2J(x)+J(xy)\right) \\&\quad \left. =\frac{2}{3}\left( 3A(x)+3A(y)+3J(x)+J(y)\right) ,\ x,y\in S,\right. \end{aligned}$$

which implies, by the assumptions on H, that J is an additive function. Finally, from (4.11), we infer that f is additive.

The converse statement can be trivially shown. \(\square \)

As another consequence of Theorem 4.1, we have the following result.

Corollary 4.3

Let \(f,g:S\rightarrow H\) be a solution of (1.9). If \(f_{o}\) is central, then

$$\begin{aligned} \left\{ \begin{array}{l} f(x)=a(x)+2\theta (x) \\ g(x)=Q(x)+J(x)-\theta (x) \end{array} \right. ,\ x\in S, \end{aligned}$$
(4.12)

where \(\theta \in \mathcal {N}(S,H,\sigma )\) such that \(\theta \circ \sigma =\theta \), \(a\in \mathcal {A}^{-}(S),\) J is a \(\sigma \)-odd Jensen’s map, and \(Q:S\rightarrow H\) is a quadratic map.

Conversely, any pair of functions (f, g) described by (4.12) is a solution of (1.9).

Proof

Let \(f,g:S\rightarrow H\) be a solution of (1.9). In the proof of Theorem 4.1, we showed that

$$\begin{aligned} \left\{ \begin{array}{l} f_{e}(x)=2\theta (x) \\ g_{e}(x)=Q(x)-\theta (x) \end{array} \right. ,\ x\in S, \end{aligned}$$

and

$$\begin{aligned} \left\{ \begin{array}{l} f_{o}(x)=2\varphi (x) \\ g_{o}(x)=J_{1}(x)-\varphi (x) \end{array} \right. ,\ x\in S, \end{aligned}$$

where \(Q:S\rightarrow H\) is a quadratic map, \(J_{1}:S\rightarrow H\) is a \( \sigma \)-odd Jensen’s map, \(\theta \in \mathcal {N}(S,H,\sigma )\) and \( \varphi \) is a solution of (4.1) such that \(\theta \circ \sigma =\theta \) and \(\varphi \circ \sigma =-\varphi \). Hence, from [5, Theorem 3.2] we infer that there exists \(a_{1}\in \mathcal {A}^{-}(S)\) such that \( f_{o}=2a_{1} \), which yields that \(\varphi =a_{1}\) (because H is a uniquely 2-divisible group). So we get the claimed result with the \(\sigma \)-odd Jensen’s map \(J(x)=g_{o}(x)=J_{1}(x)-a_{1}^{-}\) and \(a:=2a_{1}\).

The converse statement can be trivially shown. \(\square \)

Finally, we solve Eq. (1.10), namely

$$\begin{aligned} f(xy)+g(xy)+g(xy^{-1})=f(x)+f(y)+2g(x)+g(y)+g(y^{-1}),\ x,y\in S, \end{aligned}$$

where S is a 2-divisible abelian group. This enables us to see that modulo a constant, Jensen’s and Drygas’ equations are strongly alien on 2 -divisible abelian groups.

Corollary 4.4

Let S be a 2-divisible abelian group. The pair \(f,g:S\rightarrow H\) satisfies Eq. (1.10) if and only if there exist a constant \(c\in H\), \(A_{1},A_{2}\in \mathcal {A}^-(S)\) and a quadratic map \(Q:S\rightarrow H \) such that

$$\begin{aligned} \left\{ \begin{array}{l} f(x)=A_{1}(x)+2c \\ g(x)=Q(x)+A_{2}(x)-c \end{array} \right. ,\ x\in S. \end{aligned}$$
(4.13)

Proof

The proof follows from Corollary 4.3, [13, Theorem 2] and Remark 2.3. \(\square \)