parmenides51 wrote: Find all triples $(a,b, c)$ of real numbers for which there exists a non-zero function $f: R \to R$, such that $$af(xy + f(z)) + bf(yz + f(x)) + cf(zx + f(y)) = 0$$for all real $x, y, z$. Let $P(x,y,z)$ be the assertion $af(xy+f(z))+bf(yz+f(x))+cf(zx+f(y))=0$ If $a+b+c=0$, any non allzero constant fits. If $a+b+c\ne 0$ : No non allzero constant solution fits. $P(x,0,1)$ $\implies$ $af(f(1))+bf(f(x))+cf(x+f(0))=0$ $P(x,1,0)$ $\implies$ $af(x+f(0))+bf(f(x))+cf(f(1))=0$ Subtracting, we get $(a-c)(f(x+f(0))-f(f(1)))=0$ And so, since $f(x)$ is non constant, we get $a=c$ And we get in same manner $a=b=c$ and since $a+b+c\ne 0$ we get Assertion $Q(x,y,z)$ : $f(xy+f(z))+f(yz+f(x))+f(zx+f(y))=0$ $\forall x,y,z$ $Q(0,0,0)$ $\implies$ $f(f(0))=0$ $Q(x,0,0)$ $\implies$ $f(f(x))=0$ $\forall x$ $Q(x-f(0),1,0)$ $\implies$ $f(x)=0$ $\forall x$, which does not fits Hence the answer $\boxed{a+b+c=0}$