amparvardi wrote: Find all functions $f: [0, +\infty) \to [0, +\infty)$ satisfying the equation \[(y+1)f(x+y) = f\left(xf(y)\right)\] For all non-negative real numbers $x$ and $y.$ Setting $x=0$ in this equality, we get $(y+1)f(y)=f(0)$ and so $f(x)=\frac a{x+1}$ Plugging this in original equation, we get $a=0$ or $a=1$ Hence the answer : $f(x)=0$ $\forall x\ge 0$ $f(x)=\frac 1{x+1}$ $\forall x\ge 0$

I think this Slovenia paper was too easy.....isn't it? I want to pose a problem Find all functions $f:\mathbb{R}\rightarrow \mathbb{R}$ such that $f(x+y)=f(x)f(y)f(xy)$ for all $x,y$ in $\mathbb{R}$ Maths is the doctor of science.... Sayantan...

sayantanchakraborty wrote: ... I want to pose a problem ... Post as a new thread, not as a second new problem in an old thread.

Why are you posting all the Indian MO problems which are already discussed? http://www.artofproblemsolving.com/Forum/viewtopic.php?p=344412&sid=76a84a7b75c6d8d3bfe2d332309154f3#p344412

Let $P(x,y):=(y+1)f(x+y) = f\left(xf(y)\right)$ $P(x,0)$ yields $f(x)=f(xf(0))$ so now let $f(0)=c$ Furthermore notice that from $P(0,x)$ we obtain $(x+1)f(x)=c\Longrightarrow f(x)=\frac{c}{x+1}$ Moreover when we plug in our result we obtain $\frac{(y+1)c}{x+y+1}=\frac{(y+1)c}{cx+y+1}$ which implies $c=0\text{ or }1$ Therefore $f(x)=\frac{1}{x+1}\text{ and }f(x)=0$ So, in conclusion $\boxed{f(x)=\frac{1}{x+1}\text{ and }f(x)=0, \forall x\in\mathbb{R}_{\ge0}}$ $\blacksquare$.