On the equation \(f(n^2+Dm^2)=f(n)^2+Df(m)^2\)

Abstract. Let \(D=2\) or 3, \(E:=\{n^2+Dm^2\vert n,m\in{\Bbb N}\}\), \(\epsilon(n)=1\) if \(n\in E\) and \(\epsilon(n)\in\{-1,1\}\) if \(n\in {\Bbb N}\setminus E\). Let \(f:{\Bbb N}\to{\Bbb C}\) be such a function for which $$ f(n^2+Dm^2)=f(n)^2+Df(m)^2 \quad\text{for every}\quad n,m\in{\Bbb N}. $$ Then either \(f(n)=0\), or \(f(n)=\dfrac{\epsilon(n)}{D+1}\), or \(f(n)=\epsilon(n)n\) for every \(n\in{\Bbb N}\).