Find all real functions so that $f(xf(y)+f(x))=f(yf(x))+x$

Question

Find all functions $f:\mathbb{R} \rightarrow \mathbb{R}$ so that $f(xf(y)+f(x))=f(yf(x))+x$

$f(x)=\pm x$ should be the only solution. It's easy to get that $f(f(0))=f(0)$.

Answer 1

First you can show that $f(0)=0$. Assume $f(0)=a$. Then you also have $f(a)=0$. If you use $x=y=a$, you arrive at $a=2a$ and therefore $a=0$. Now you can investigate the case $y=0$: $$f(f(x))=x$$ meaning that $f$ is an involution (and goes through the point (0,0)).

You can probably proceed from here.

Answer 2

Thank to @orion we know now that f is an involution, which means it's a bijection. Bcs f is an onto function we know there exists an b so that $f(b)=1$, and bcs $f(f(x))=x$ we get also $f(1)=b$.

$x\rightarrow 1, y \rightarrow 1 \Rightarrow f(2b)=2 \Rightarrow f(2)=2b$

$x\rightarrow b, y\rightarrow 1 \Rightarrow f(b^2+1)=2b$ Now we have $f(b^2+1)=2b=f(2)$ and there we get that $b=\pm 1$.

Let $f(1)=1$: $x\rightarrow 1, y\rightarrow t \Rightarrow f(f(t)+1)=f(t)+1$ and if we change $f(t)+1=k$ we get $f(k)=k$.

If $f(1)=-1$ we get $f(k)=-k$.