parmenides51 wrote: Find all functions $f: R \to R$ such that $f (f (x) f (y)) = xy$ and there is no $k \in R -\{0,1,-1\}$ such that $f (k) = k$. Let $P(x,y)$ be the assertion $f(f(x)f(y))=xy$ $f(x)$ is bijective and so $\exists u$ such that $f(u)=1$ and $v$ such that $f(v)=0$ $P(x,v)$ $\implies$ $f(0)=vx$ $\forall x$ and so $v=0$ and $f(0)=0$ $P(u,u)$ $\implies$ $f(1)=u^2$ $P(1,1)$ $\implies$ $f(u^4)=1=f(u)$ and so, since injectiive, $u^4=u$ and $u\in\{-1,0,1\}$ $u\ne 0$ since $f(0)=0$ $u\ne-1$, else previously got $f(1)=u^2$ becomes $f(1)=1=f(u)$ and so, since injective, $u=1$ So $u=1$ $P(x,1)$ $\implies$ $f(f(x))=x$ $P(f(x),f(y))$ $\implies$ $f(xy)=f(x)f(y)$ This implies (since injective) $f(-1)=-1$, $f(-x)=-f(x)$ and $f(x)>0$ $\forall x>0$ (and so $f(x)<0$ $\forall x<0$) Then $f(xf(x))=f(x)f(f(x))=xf(x)$ and so $xf(x)\in\{-1,0,1\}$ $xf(x)=0$ implies $x=0$ (since $f(0)=0$ and $f(x)$ injective) $xf(x)=-1$ is impossible since $f(x)$ always has same sign that $x$ when $x\ne 0$ And so $xf(x)=1$ $\forall x\ne 0$ And unique solution $\boxed{f(0)=0\text{ and }f(x)=\frac 1x\quad\forall x\ne 0}$, Which indeed fits.