Please tell me if I'm wrong Let $P(x,y)$ be the assertion above $P(0,0)=>f(0)=0$ Assume there exists $m$ a nonzero real number such that $f(m)=0$ $P(m/2,m/2)=>f(0)=0=m^2/4 =>m=0$, which means $f$ is injective to zero $P(0,x)=>f(f(x)-f(-x))=0 =>f(x)-f(-x)=0 =>f(x)=f(-x)$ for every real $x$ $P(-x,y) =>-xy=f(f(y-x)-f(-x-y))=f(f(x-y)-f(x+y))=f(f(x+y)-f(x-y))=xy$, which is impossible for all values of $x$ and $y$, therefore no such function exists .