TITLE: Functions satisfying $f\left( f(x)^2+f(y) \right)=xf(x)+y$
QUESTION [11 upvotes]: Find all functions $f:\mathbb{R} \rightarrow \mathbb{R}$ such that $f\left( f(x)^2+f(y) \right)=xf(x)+y$ for all real numbers $x$ and $y$.

Clearly $f(x)=x$ is a solution, check by substitution.
I'm at a loss as to how to show this is the only one or find others.
Any help would be very much appreciated.

REPLY [5 votes]: lemmas in order they were noticed, all easy:
$f$ is injective
$f$ is surjective
$f(f(y)=y$   (let $f(x)=0$)
$f(-x)=-f(x)$ for nonzero $f(x)$  ($f$ is odd)
$f(0)=0$ by bijectivity and oddness
$f(f(x)^2)=xf(x)$  taking $y=0$
$f(u^2)=uf(u)$ by taking $x=f(u)$
$f(f(x)^2 + xf(x)) = xf(x) + x^2$ , (let $y=x^2$)
$f(x)^2=x^2$  (apply $f$ to both sides of preceding +injectivity)
$f(x) = \pm x$
We see now that $f(x)=-x$ is also a solution.  Let $f(x) = s(x)x$ with $|s|=1$.  Note $s(-x)=s(x)$
Taking $x$ with $x^2 > |y|$, $s(x)s(y)=1$  ($x$ sets the sign of each side, then compare the $y$ terms)  so that $s(x)=s(y)$.  Also $s(x^2)=s(x)$ for nonzero $x$ from $f(x^2)=xf(x)$.
Iterating this, the sign is constant on $|x|>1$ and on $|x|<1$.   You can check by calculation whether there is a solution with the sign chosen differently on these intervals.