On an Equation of Sophie Germain

Abstract

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}\t\t\t\t\\usepackage{amsmath}\t\t\t\t\\usepackage{wasysym}\t\t\t\t\\usepackage{amsfonts}\t\t\t\t\\usepackage{amssymb}\t\t\t\t\\usepackage{amsbsy}\t\t\t\t\\usepackage{mathrsfs}\t\t\t\t\\usepackage{upgreek}\t\t\t\t\\setlength{\\oddsidemargin}{-69pt}\t\t\t\t\\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}\t\t\t\t\\usepackage{amsmath}\t\t\t\t\\usepackage{wasysym}\t\t\t\t\\usepackage{amsfonts}\t\t\t\t\\usepackage{amssymb}\t\t\t\t\\usepackage{amsbsy}\t\t\t\t\\usepackage{mathrsfs}\t\t\t\t\\usepackage{upgreek}\t\t\t\t\\setlength{\\oddsidemargin}{-69pt}\t\t\t\t\\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.