$f:\mathbb{R_{\geq 0}} \to \mathbb{R_{\geq 0}}$ such that for all $x$ we have $xf(1+xf(y))=f(f(x)+f(y))$

Find all nonnegative real number $$a$$, such that $$f(a)=0$$ for any function $$f$$ satisfying: $$xf(1+xf(y))=f(f(x)+f(y))$$ with all $$x,y$$ are nonnegative real number.

I don't know why this problem only ask the number $$a$$ for $$f(a)=0$$ instead of the function $$f(x)$$. Is there any special thing here? Is it possible to show the bijection? I will show what I get from trying to solve this: $$f(f(x)+f(0))=f(1+xf(0))=0$$. How to find $$t=f(0)$$, show me your solution or any idea please, thank all.

• I don't understand your calculation. If we let $y=0$ we get $xf(1+xf(0))=f(f(x)+f(0))$ which is not what you wrote. Better, I think, to let $x=0$, which yields $0=f(f(0)+f(y))$. – lulu Jun 23 at 10:56
• What is a source of this problem? – Aqua Jun 23 at 12:29
• Sorry I don't know. My sister asked me... – user628755 Jun 23 at 13:38
• Ask you sister :) – Aqua Jun 23 at 18:07
• One might think that $f(x)\equiv 0$ is the only solution. As a counterexample, consider the function $$f(x)=\begin{cases} 1 \text{ if 0<x<1} \\ 0 \text{ else }\end{cases}$$ which works – user574848 Jun 24 at 1:00

I have started writing couse I thought I will solve it. But no, sorry. Perhaps you will get some idea from here. If you don't like it I will delete it.

We have: $$xf(1+xf(y))=f(f(x)+f(y))$$

Suppose there exists $$a$$ such that $$f(a)=0$$.

• If $$y=a$$ we get: $$bx = f(f(x))$$ where $$b=f(1)$$. So if $$\boxed{b\ne 0}$$ then $$f$$ is injective (so $$f$$ has only one zero), so for $$x=1$$ we get $$f(1+f(y)) = f(b+f(y))\implies b=1$$. Now for $$x=0$$ we get $$f(f(0)+f(y))=0$$, so $$f(0)+f(y)=a$$ for all $$y$$, thus $$f$$ is constant. A contradiction. So $$\boxed {b=0}$$ and $$f(f(x))=0$$ for all $$x$$. So in special case (if we put $$x=a$$) we get $$f(0)=0$$.

• If $$x=a$$ we get: $$af(1+af(y)) = f(f(y))=0$$

• Not sure if it helps but the right side is symmetric in x, y so we have $xf(1+xf(y))=yf(1+yf(x))$ – kingW3 Jun 23 at 11:56
• @Aqua Do you know the solution for function $f$, It does not seem to be enough..., can you tell me the solution for $a$? – user628755 Jun 24 at 2:03
• You asked your sister? – Aqua Jun 24 at 7:27
• @Aqua. It is a problem in $belarus$ training for imo 2018 – user628755 Jun 26 at 2:56
• Help me please everyone... – user628755 Jun 26 at 3:20