Find all real functions of real variable such that $$(x + 1)f(xf(y)) = xf(y(x + 1))$$

Let $a=f(0)$.

  • For $y=0$ we get $(x+1)f(ax) = ax$, so if $a\ne 0$ we get $$f(x) = {ax\over x+a}$$ which is actual solution (check it in a starting equation).

So suppose $a=0$.

  • For $x=0$ we get $0=0$ and we get nothing new. If we put $x=1$ we get $\boxed{2f(f(y)) = f(2y)}$ and if we put $y=1$ we get $\boxed{(x+1)f(bx)=xf(x+1)}$ where $b=f(1)$. And here I'm stuck.
  • 1
    $\begingroup$ but isn't $f(0)=0$? If $x=0$ we get $1 \cdot f(0)=0$ $\endgroup$ – Vasya Apr 9 '19 at 18:34

Eh, actually, it is not that difficult. Sorry friends to bother you.

Clearly $f(x)\equiv 0$ and $f(x) \equiv x$ are solution of FE. Now suppose $f$ is neither of them.

Suppose there is $y_0\ne 0$ such that $f(y_0)= 0$. Then if $y=y_0$ we get $xf(y_0(x+1))=0$ for all $x$, so if $x\ne 0$ we get $ f(y_0(x+1))=0$ which means $f(x)\equiv 0$ since map $x\mapsto y_0(x+1)$ is linear and thus surjective.

So $0$ is the only zero for $f$. Since $f$ is not identity we have some $c$, such that $f(c)\ne c$. Now let $x={c\over f(c)-c}$ (then $f(c)\neq 0$ and $c\ne 0$) and $y=c$, then we get $${f(c)\over f(c)-c}\color{red}{f\Big({cf(c)\over f(c)-c}\Big)} = {c\over f(c)-c} \color{red}{f\Big({cf(c)\over f(c)-c}\Big)} $$

Since red factor can not be $0$ we can cancel it and we get $f(c)=c$ which is a contradiction.

So the only solutions are the one I mentioned at the begining.

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