TITLE: Functional Equation : $f(x) = f(x + y^2 + f(y))$
QUESTION [5 upvotes]: This problem is from my textbook:
Given : $f:\mathbb R\to\mathbb R$
Solve this functional equation :
$f(x) = f(x + y^2 + f(y))$
I think this function is just a simple constant, so I try all my best to prove this, but no result. I'm not sure this way is wrong, or not. Can give me a hint ?
Thanks :)

REPLY [4 votes]: Let $\Pi$ be the group of all periods of $f$. Then for all $y \in \mathbb{R}$ we have $ y^2+f(y) \in \Pi$. Let $x \in \Pi$. Then $\Pi \ni (y+x)^2+f(x+y)=(y+x)^2+f(y)$. $\Pi$ is a group, so $2xy+x^2=(y+x)^2+f(y)-y^2-f(y) \in \Pi$. Thus if there exists nonzero $x \in \Pi$ then $\Pi=\mathbb{R}$, and $f$ is constant. If $\Pi=\{0\}$ then for all $y$ we have $y^2+f(y) \in \Pi$ so $f(y)=-y^2$.