Find $f:f(xf(x)+f(y))=f(x)^2+y$

Question

Find $f:f(xf(x)+f(y))=f(x)^2+y$
Domain and co-domain is real numbers

I did the following:
Let $s=f(0)$
Then $f(f(y))=s^2+y$ so $f$ is surjective

Also, $f(x)=f(y)\implies f(xf(x)+f(y))=f(xf(x)+f(x))\implies x=y$ so $f$ is injective

So, $f$ is bijective. Letting $f(x)=0, y=0$
we get $f(f(0))=0 \implies s=f(0)=0$
In fact $f(f(x))=x$ for ALL $x$

Letting $x=1,y=0$ we get
$f(f(1))=f(1)^2\implies 1=f(1)^2 \implies f(1)=1$ or $f(1)=-1$

If $f(1)=1$ we let $x=1$ to get
$f(y+1)=f(y)+1$
By induction this leads to $f(x)=x$ for all integers

If $f(1)=-1$ we let $x=1$ to get
$f(y-1)=f(y)+1$
By induction we get $f(x)=-x$ for all integers

How to extend the domain over real numbers I don't know, any help would be appreciated

Answer 1

Let $x=f(u)$, by using $f(x)=f(f(u))=u$ one gets now $f(u f(u)+f(y))=u^2+y$. By replacing $u$ with $x$, it follows that for all $x$, $f(x)^2 = x^2$, hence for all $x$, $f(x) = \pm x$. It remains to see if the sign depends on $x$.