parmenides51 wrote: Find all functions ${f : (0, +\infty) \rightarrow R}$ satisfying $f(x) - f(x+ y) = f \left( \frac{x}{y}\right) f(x + y)$ for all $x, y > 0$. (Austria) Let $P(x,y)$ be the assertion $f(x)-f(x+y)=f(\frac xy)f(x+y)$ If $f(u)=-1$ for some $u>0$, then $P(u,1)$ implies contradiction So $f(x)\ne -1$ $\forall x>0$ So $P(x,y)$ may be written as New assertion $Q(x,y)$ : $f(x+y)=\frac{f(x)}{f(\frac xy)+1}$ Subtracting $Q(x,y)$ from $Q(y,x)$, we get $f(x)(f(\frac yx)+1)=f(y)(f(\frac xy)+1)$ (1) $Q(x+y,z)$ $\implies$ $f(x+y+z)=\frac{f(x+y)}{f(\frac xz+\frac yz)+1}$ $=\frac{\frac{f(x)}{f(\frac xy)+1}}{\frac{f(\frac xz)}{f(\frac xy)+1}+1}$ And so $f(x+y+z)=\frac{f(x)}{f(\frac xz)+f(\frac xy)+1}$ Swapping there $x,y$ and subtracting, we get $f(x)f(\frac yz)+f(x)(f(\frac yx)+1)=f(y)f(\frac xz)+f(y)(f(\frac xy)+1)$ Using (1) above, we conclude $f(x)f(\frac yz)=f(y)f(\frac xz)$ Replacing there $(x,y,z)$ by $(x,xy,x)$, we get $f(x)f(y)=f(xy)f(1)$ If $f(1)=0$, this implies $\boxed{\text{S1 : }f(x)=0\quad\forall x>0}$ which indeed is a solution If $f(1)=c\ne 0$, we get $f(xy)=\frac 1cf(x)f(y)$ and $f(x)\ne 0$ $\forall x>0$ $Q(x,y)$ becomes then $f(x+y)=\frac{f(x)f(y)}{cf(x)+f(y)}$ Swapping $x,y$, we get $c=1$ and so $f(xy)=f(x)f(y)$ and $f(x+y)=\frac{f(x)f(y)}{f(x)+f(y)}$ This means that $\frac 1{f(x)}$ is both additive and multiplicative and so is identity (since allzero is here impossible) And so $\boxed{\text{S2 : }f(x)=\frac 1x\quad\forall x>0}$ which indeed is a solution.