$f:\mathbb{R} \to \mathbb{R}$ such that $f(x+y^2+f(y)) = f(x-y^2-f(y))$ Find all functions $f:\mathbb{R} \to \mathbb{R}$ such that $f(x+y^2+f(y)) = f(x-y^2-f(y))$.
There is actually a proof that $f\equiv c$ and $f(x) = -x^2$ are the only ones - it goes through the not too natural substitution $g(x) = f(x) + x^2$ and using that in the resulting equation every real number can be written as a difference of the form $g(y) - g(z)$.
I am looking for a different and a bit more natural approach, e.g. that $f$ must be periodic (if it is different from the abovementioned solutions). Any help appreciated!
 A: I will assume $f$ is continuous.
Now if $f(y)+y^2$ is nonconstant then the image $I:=\{f(y)+y^2:y\in\mathbb R\}$ contains some open interval $(a,b)$. Now for any $z\in (a,b)\subseteq I$ and any $x\in\mathbb R$ we have that $f(x-z)=f(x+z)$, by our assumption. Thus, $f$ has period $2z$. Now, every real number in $(2a,2b)$ is a period of $f$, which shows that $f$ must be constant.
A: We want to find all functions $f:\mathbb{R} \rightarrow \mathbb{R}$ such that for every $x,y \in \mathbb{R}$ following holds
$$ f(x) = f(x+2y^2+2f(y)) \label{eq:main} $$
If $\exists x \in \mathbb{R}$ such that $f(x) \neq -x^2$, then $f$ is periodic. If $\mathfrak{p} \neq 0$ is period of $f$, then for all $x, y \in \mathbb{R}$
\begin{align}
f(x) &= f(x + 2(y+\mathfrak{p})^2 + 2f(y + \mathfrak{p})) \\
&= f(x + 4y\mathfrak{p} + 2\mathfrak{p}^2 + 2y^2 + 2f(y)) \\
&= f(x + 4y\mathfrak{p} + 2\mathfrak{p}^2)
\end{align}
This implies that $f$ is constant.
Any constant function and $f(x) = -x^2$ satisfy the equation, so they are the only solutions.
