$f(x - f(y))= f(f(y)) - 2xf(y) + f(x)$ 
Let $x,y \in \mathbb R$. Determine all $f : \mathbb R \to \mathbb R$ that satisfies $$f(x - f(y)) = f(f(y)) - 2xf(y) + f(x)$$

My solution: $f(x) = 0$ is obvious. So, I’ll consider other functions.
$x = f(y)$ $\to$ $f(0) = f(x) - 2x^2 + f(x)$
Which implies $f(x) = 2x^2 + \frac{f(0)}{2}$.
Do we have to find $f(0)$ or I can just put it back to the original statement, then derive $f(0)$ and conclude?
 A: This doesn’t work; you set $x=f(y)$, which means your equation only holds when $x$ is in the range of $f$, and not necessarily all $x$.
A: Here is a solution: Suppose that $f$ solves the functional equation, and write $g(x) = f(x) - x^2$.

Claim. $ g(x-f(y)) = g(x) + \frac{f(0)}{2} $ holds for all $x, y \in \mathbb{R}$.

Proof. Using the functional equation, we get
\begin{align*}
g(x-f(y))
&= \bigl( f(f(y)) - 2xf(y) + f(x) \bigr) - \bigl( x - f(y) \bigr)^2 \\
&= g(f(y)) + g(x)
\end{align*}
Plugging $x = f(y)$, we get $2g(f(y)) = g(0) = f(0)$, hence we get $g(f(y)) = \frac{f(0)}{2}$ for any $y$. Then plugging this to the above identity completes the proof. $\square$
Since $f \equiv 0$ is a trivial solution, it suffices to consider the case where $f$ is not identically zero, and we do so. Let $a$ satisfy $f(a) \neq 0$. Then plugging $y = t - f(a)$,
\begin{align*}
g(x) + \tfrac{f(0)}{2}
&= g\bigl(x - f(t - f(a))\bigr) \\
&= g\bigl(x - (t - f(a))^2 - g(t - f(a))\bigr) \\
&= g\bigl(x - t^2 + 2f(a)t - f(a)^2  - g(t) - \tfrac{f(0)}{2}\bigr) \\
&= g\bigl(x + 2f(a)t - f(a)^2 - \tfrac{f(0)}{2} - f(t)\bigr) \\
&= g\bigl(x + 2f(a)t - f(a)^2 - \tfrac{f(0)}{2}\bigr) + \tfrac{f(0)}{2}.
\end{align*}
Then plugging $x = f(a)^2 + \frac{f(0)}{2}$ shows that $g(2f(a)t)$ does not depend on the value of $t$, hence $g$ is constant. Then the above claim shows that $f(0) = 0$, and therefore $f(x) = x^2$.
A: And you actually proved it is not, since asumming it is surjective, the answer is $2x^2+\frac{f(0)}{2},$ which is not surjective.
