Complicated real to real functional equation Find all $f:\mathbb{R} \rightarrow \mathbb{R}$ satisfying
$$f(f(y)+x^2+1)+2x=y+(f(x)+1)^2$$
for all $x,y \in \mathbb{R}.$
So far I have proved that $f$ is bijective. How should I continue?
 A: Let 
$$ I=\{\,x\in\mathbb R\mid \forall y\in\mathbb R\colon f(y+x)=f(y)+x\}$$
(so that specifically $f(x)=f(0)+x$ for all $x\in I$) and
$$ S=\{\,x\in\mathbb R\mid f(x)=x\,\}.$$
Clearly, $I$ is an additive subgroup of $\mathbb R$.
Let $a=x^2+1$ and $b=(f(x)+1)^2-2x$.
Then for all $y$ we have $ f(f(y)+a)=y+b$,
hence
$$f(y+a+b)= f(f(f(y)+a)+a)=f(y)+b+a$$
so that 
$$\tag1(f(x)+1)^2+(x-1)^2\in I\qquad\text{for all }x\in\mathbb R.$$
Let $x,y\in I$. Then also $-x\in I$ and the functional equation reads
$$ f(y+f(0)+(\pm x)^2+1)\pm 2x=y+ (\pm x+f(0)+1)^2$$
i.e.,
$$f(y+f(0)+x^2+1) = y+f(0)+x^2+1+ f(0)(1\pm 2x+f(0))$$
and by subtracting both variants $f(0)x=0$. From $(1)$ we see that $I$ is not trivial, hence we can pick $x\ne0$ and obtain $f(0)=0$, hence
$$\tag2 I\subseteq S.$$

Remark. If we add the condition that $f$ be continuous, we get immediately from $(1)$ that $I$ contains an unbounded connected set, hence $I=\mathbb R$.
  We try to continue with not necessarily continuous $f$.

Immediately form the functional equation we get 
$$\tag3 x,y\in S\implies y+x^2+1\in S.$$
and also from $(1)$
$$\tag4 x\in S\implies 2x^2+2\in  I.$$
Specifically, $2\in I$ because $0\in S$, so $2\mathbb Z\subseteq I$.
Then from $(2)$ and $(3)$ we have $\mathbb Z\subseteq S$. Thus $1\in S, 2\in I$ allows us to conclude using $x=1$ that
$$\tag5 f(f(y))=f(f(y)+1^2+1)-2=y$$
so that $f$ is an involution.
Plugging  $y=-1\in S$ into the functional equation, we get
$$\tag6f(x^2)=(f(x)+1)^2-2x-1.$$
For $z\in\mathbb R$ let $x=-\frac{z+1}2$.
Using the functional equation for both sides of
$$ f(y+x^2+1)=f((y+z)+(x+1)^2+1)$$
we see that $f(y)-f(y+z)$ depends only on $x$, not on $y$. With $f(0)=0$  we 
obtain
$$ \tag7f(y+z)=f(y)+f(z),$$
i.e., $f$ is additive.
Finally, assume for some $c\in\mathbb R$ that $f(c)\ne c$. Then from $(5)$ and $(7)$ we obtain $f(f(c)-c)=c-f(c)$, i.e., there exist $v=f(c)-c\ne 0$ with $f(v)=-v$. Per additivity we may replace $v$ with an integer multiple of itself and so can assume wlog. thet $v>9$.
Then we can write
$v=x^2$ with $x>3$ and obtain from $(6)$
$$ -x^2=f(x^2)=-2x-1+(f(x)+1)^2$$
and so 
$$ 2=(f(x)+1)^2+(x-1)^2\ge (x-1)^2>4.$$
To resolve this contradiction, we have to drop the assumption that $f(a)\ne a$. We conclude

Proposition. $f(x)=x$ for all $x\in\mathbb R$.

