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.

  • $\begingroup$ 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 $0=f(f(0)+f(y))$. $\endgroup$ – lulu Jun 23 at 10:56
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    $\begingroup$ What is a source of this problem? $\endgroup$ – Aqua Jun 23 at 12:29
  • $\begingroup$ Sorry I don't know. My sister asked me... $\endgroup$ – user628755 Jun 23 at 13:38
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    $\begingroup$ Ask you sister :) $\endgroup$ – Aqua Jun 23 at 18:07
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    $\begingroup$ 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 $\endgroup$ – 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: $af(1+af(y)) = f(f(y))=0$

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

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