Help with the functional equation $f(x - f(y)) = f(f(y)) + 2 x f(y) + f(x) - 1$ I am asked to find all $f : \mathbb{R} \to \mathbb{R}$ such that $f(x - f(y)) = f(f(y)) + 2 x f(y) + f(x) - 1$, which I solved and got that $f(x) = 1 - x^2$ -- a correct solution and everything.
What puzzles me is how do I find all solutions. I wrote that all functions $g : \mathbb{R} \to \mathbb{R}$ which map in the same way as $f$ are solutions, but is there something I'm missing here?
Have an awesome day, all of you.
 A: Since you wrote $f:\newcommand{\R}{\mathbb R}\R\rightarrow\R$ I think it is fair to assume that the domain of the function is all of $\R$. Now, let $A=f(\R)$ be the codomain of the function. For any $k\in A$ we then have
$$
f(x-k)=f(k)+2x\ k+f(x)-1
$$
Not a bounded function
First note that $f$ cannot be a bounded function for if $A\subseteq[-M,M]$ and $k\in A$ we get
$$
M\geq |f(x-k)|=|f(k)+2x\ k+f(x)-1|\geq |2x\ k-1|-2M
$$
so then we see that $|2x\ k-1|\leq 3M$ which is impossible since $2x\ k$ is not bounded unless $k=0$. Of course then we must also check that $A=\{0\}$ is impossible, but that is directly seen from
$$
f(x)=f(x-0)=f(0)+2x\cdot 0+f(x)-1
$$
implying that $f(0)=1$ thus contradicting $A=\{0\}$. It follows that $f$ must be unbounded. This tells us that $A$ must contain infinitely many function values. In particular $f$ cannot be a constant function.
Analyzing if codomain contains zero
If $0\in A$ it follows as stated above that $f(0)=1$. Now for any $k$ in the infinite set $A$ we then have
$$
1=f(k-k)=2 f(k)+2k^2-1\\
\iff\\
f(k)=1-k^2
$$
so in this case $f$ acts exactly like the function you proposed on the set $A$.
If we also know (I do not know if we do) that $f$ is a polynomial expression this then implies that $1-x^2-f(x)$ is a polynomial having all elements of $A$ as roots so it follows that this is the zero polynomial and thus $f(x)=1-x^2$ in that case.
If codomain does not contain zero
So what happens if $A$ does not contain zero. Then we cannot be certain that $f(0)=1$. Then define $f(0)=2q+1$ and let us see what happens when analyzing this. We then have for $k\in A$ that
$$
2q+1=f(k-k)=2 f(k)+2k^2-1\\
\iff\\
f(k)=(q+1)-k^2
$$
so if again we assume $f$ to be a polynomial it follows that $f(x)=(q+1)-x^2$. But then we must have $q<-1$ [and thus $f(0)=2q+1<-1$] for otherwise $f$ does have zeros. But if we plug this $f$ into the functional equation and expand both sides we see that this is impossible.
Conclusion:
If $f$ is a polynomial expression then $f(x)=1-x^2$ is the unique solution to the functional equation. I do not know whether it be possible for $f$ NOT to be a polynomial but still satisfying the equation.
