The following two alternative functional equations are investigated \(\begin{aligned} [f(x+y)+f(x-y)-2f(x)-2f(y)]\times [g(x+y)+g(x-y)-2g(x)-2g(y)]=0, \end{aligned}\) and \(\begin{aligned} [f(x+y)-f(x)-f(y)]\times [g(x+y)+g(x-y)-2g(x)-2g(y)]=0, \end{aligned}\) where \(f,g:\mathbb {R} \rightarrow \mathbb {R}\) ; assuming that in the first equation f and g are in \(C^2(\mathbb {R})\) and in the second one that f is in \(C^1(\mathbb {R})\) and g in \(C^2(\mathbb {R})\) , it is proved that both equations have only trivial solutions, that is for the first equation either f is quadratic on \(\mathbb {R}\) or g is quadratic on \(\mathbb {R}\) ; for the second equation either f is additive on \(\mathbb {R}\) or g is quadratic on \(\mathbb {R}\) .
