$f(x+y)=[f(x)]^2+y$
Prove whether or not this functional equation works for some function and, if it does, specify the domain(s) and co-domain(s).
P.S.: I've invented this problem from scratch just a couple of minutes ago. I hope to see some interesting results and proofs.
EDIT: I was thinking this while trying to solve this equation. Assume the function is constant. This means that i can write $f$ as $f(a)=c$. Therefore i can rewrite the equation as $c=c^2+y$ and so $y=c-c^2$. But that is a contradiction, since I've written a variable ($y$) in terms of a constant ($c$). As a consequence of this, let's try $y=0$: $f(x) = [f(x)]^2$. The only solutions to this are $f(x)=1$ or $f(x)=0$, but both are impossible, since we just proved that the function cannot be constant. If my reasoning is correct, this should imply that the function does not allow $y=0$ in its domain.
The functional equation [for reference] is: $$f(x+y)=f(x)^2+y\ . \tag{1} $$ I'll be assuming both the domain and codomain are subsets of the reals, to be determined by $(1).$ If in this we put $y=0$ and keep $x,$ we get to $$f(x)=f(x)^2. \tag{2}$$ This immediately implies the range of $f$ is a subset of $\{0,1\}.$
Next going back to $(1)$ we now put $x=0$ and keep $y,$ and get $$f(y)=f(0)^2+y\ . \tag{3}$$ At this point we know as noted that $f(0)=0$ [''case A''] or else $f(0)=1$ [''case B''].
In case A $(3)$ becomes $f(y)=y.$ But since then $y$ is in the range of $f$ it is $0$ or $1$ and we now know in case A the domain must be a subset of $\{0,1\}.$
In case B $(3)$ becomes $f(y)=1+y.$ Here from $1+y \in \{0,1\}$ we see in case B the domain is a subset of $\{-1,0\}.$
Already things are quite restricted. But we can go one more step by returning to $(1)$ and this time putting $y=x$ keeping both variable to get to $$ f(2x)=f(x)^2+x. \tag{3}$$ Going back to cases A,B we now see that in case A, $x$ cannot be $1$ else $2$ is in domain of $f.$ And in case B, $x$ cannot be $-1$ else $-2$ ends up in the domain of $f.$ The upshot is that in either case A or B the domain of $f$ is the single number $0.$ Finally the range of $f$ is in either case only $0$ by putting $x=0$ into $(3).$ I think the codomain could be just $0$ or could be $\{0,1\}.$