On an Equation of Sophie Germain

We deal with the following functional equation f(x)2+4f(y)2=(f(x+y)+f(y))(f(x-y)+f(y))\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} f(x)^2+4f(y)^2 = \big ( f(x+y)+f(y) \big ) \big ( f(x-y)+f(y) \big ) \end{aligned}$$\end{document}which is motivated by the well known Sophie Germain identity. Some connections as well as some differences between this equation and the quadratic functional equation f(x+y)+f(x-y)=2f(x)+2f(y)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} f(x+y)+f(x-y)=2f(x)+2f(y) \end{aligned}$$\end{document}are exhibited. In particular, the solutions of the quadratic functional equation are expressed in the language of biadditive and symmetric functions, while the solutions of the Sophie Germain functional equation are of the form: the square of an additive function multiplied by some constant. Our main theorem is valid for functions taking values in a unique factorization domain. We present also an example which shows that our main result does not hold in each integral domain.


Introduction
We deal with a functional equation motivated by the identity a 4 + 4b 4 = (a + b) 2 which is attributed to Sophie Germain. In fact, she mentioned only the identities p 2 + 4 = (p 2 − 2) 2 + 4p 2 (which, in view of Fermat's two squares theorem, implies that no number of the form p 2 + 4 is prime) and p 4 + q 4 = (p 2 − q 2 ) 2 + 2p 2 q 2 = (p 2 + q 2 ) 2 − 2p 2 q 2 , for details see [4,6]. Although (1) is very easy to check, it is extremely useful in solving number theory problems. It is also a common tool for contests problems like: show that the number 5 444 + 4 555 is not prime or calculate the sum of the series ∞ n=1 k 4k 4 +1 , for details see for example [7]. Inspired by the identity (1), we consider the following functional equation It is immediately seen that the function f (x) = cx 2 is a solution of (2). We ask if there are any solutions of (2) other than f (x) = cx 2 . Since we do not want to assume any regularity conditions of the functions in question, the first guess is that (2) may be equivalent to the equation of a quadratic function Surprisingly, it will turn out that only some solutions of the quadratic functional equation (3) satisfy (2). It is a rare behavior in the world of functional equations. Dealing with functional equations it is more common that either an equation preserves all solutions of the linear equation it is connected with or it forces the continuity of its solutions (like Aczél equation does, see [1]).

Main Results
Assume that (G, +) is an abelian group, R is an integral domain. (2) for all x, y ∈ G then it is even.
Proof. It is easy to see that f (0) = 0. Put x := 0 into (2) in order to obtain The following example shows that the assumption that R is an integral domain is essential.
Then f satisfies (2) and it is not even.
It is enough to put x := 2x 0 and y := x 0 into (2) in order to get the assertion.
and completes the proof.
Proof. By (2), we have Interchanging the roles of x and y and using Lemma 1 (the evenness of f ) we obtain After subtracting Eq. (5) from (6) side by side we get that is, Substitute now x + y and x − y in the place of x and y, respectively, first in (2) and then in (7). By Lemma 2, we have Suppose that for some x, y ∈ R we have f (x) = f (y) and f (x + y) = f (x − y). Then by (2), and by (8), From (10) and (11), it follows that 4f (x)f (x + y) = 0 which in turn gives Our assertion is now derived from (7) and (9).
Remark 1. It is enough to consider the function f : Z 9 → Z 9 given by f (x) = 3, x ∈ Z 9 , to see that the assumption that R is an integral domain is essential in Theorem 1. Proof. Assume that 4f (x) = A(x, x) for all x ∈ G, where A : G 2 → R is a biadditive and symmetric function satisfying (12). Then we have which shows Eq. (2). For the converse, by Theorem 1, function f is quadratic and by [3] (see also [2]) there exists a biadditive and symmetric function A : y) for all x, y ∈ G and to prove (see [3]) its symmetry, additivity with respect to the first variable and the uniqueness). Substituting the form of f into (2) we obtain which shows condition (12).
Define a : G → R by the formula a(x) = A(x,z0) α for x ∈ G. It is obvious that a is additive. We have also for all x ∈ G, which shows that f = γa 2 . Now assume that f = γa 2 for some additive function a : G → R and a constant γ ∈ R. Then we define A : G 2 → R by the formula A(x, y) = 4γa(x)a(y), x, y ∈ G.
It is obvious that A is biadditive, symmetric and A(x, x) = 4f (x) for x ∈ G. We have also so A satisfies (12) and in view of Theorem 2, f satisfies (2). The following example shows that Theorem 3 may not hold in integral domains that are not unique factorization domains.