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Find all functoins $f: \mathbb R \rightarrow \mathbb R$ such that $\forall x,y \in \mathbb R$ the equality:: $$f(xy+f(x))=xf(y)+f(x)$$

My work so far:

1) $f(0)=0$;

Let $f(a)=f(b)\not=0$

$x=a, y=b \Rightarrow f(ab+f(a))=af(b)+f(a)$;

$x=b, y=a \Rightarrow f(ba+f(b))=bf(a)+f(b)$;

Then $af(b)+f(a)=bf(a)+f(b) \Rightarrow a=b$

$x=0: f(f(0))=f(0) \Rightarrow f(0)=0$

2) $f(x)=x -$ solution. I have tried to prove that $g(x)=f(x)-x = 0 (\forall x \in \mathbb R)$ $$g(xy+x+g(x))=xg(y)$$

Roman83
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1 Answers1

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So we have $f(0)=0$ and using this fact it is easy to prove that $f(f(x))=f(x)$ for any $x$, which means that at any point $x$, $f(x)=x$ or $f(x)=0$. Now suppose that at some point $a\ne0$ $f(a)\ne0$, this means that $f(a)=a$ and we have $$ f(ay+f(a))=af(y)+f(a)=>f(ay+a)=af(y)+a. $$ This means that either $f(ay+a)=ay+a$ or $f(ay+a)=a$. The only case when $f(ay+a)=a$ is when $ay+a=a=>y=0$. This means that $f(x)=x$. So there two functions that satisfy above functional equality: $f(x)=0$ and $f(x)=x$.

pointer
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