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

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.
• but isn't $f(0)=0$? If $x=0$ we get $1 \cdot f(0)=0$ – Vasya Apr 9 '19 at 18:34

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.