Turkey — Team Selection Test

$f(x) = 2x$, $x \in \mathbb{R}$ is the only solution. Assume that there exists a real number $c \neq 0$ such that $c = f(z) - 2z$ for some $z$. Let $P(x, y)$ be the assertion of the given equation.
$$P(f(y) - 2y, y) \rightarrow (f(y) - 2y) f(f(y) - 2y) = 0, \forall y.$$
This shows that $f(c) = 0$. Since $-2c = f(c) - 2c \neq 0$, we also obtain that $f(-2c) = 0$.
$$\begin{align*} P(x, c) &\rightarrow f(c^2) = f((x+c)^2) - x f(x), \\ P(x, -2c) &\rightarrow f(4c^2) = f((x-2c)^2) - x f(x), \\ P(c, c) &\rightarrow f(c^2) = f(4c^2). \end{align*}$$
Therefore, we obtain that $f((x+c)^2) = f((x-2c)^2)$. (1)
$$\begin{align*} P(x - 2c, c) &\rightarrow f(c^2) = f((x-c)^2) - (x-2c)f(x-2c), \\ P(x + c, -2c) &\rightarrow f(4c^2) = f((x-c)^2) - (x+c)f(x+c). \end{align*}$$
Hence, we conclude that $(x-2c)f(x-2c) = (x+c)f(x+c)$. (2)
Now for all $x$ and $y$
$$P(x+c, y) \rightarrow f(xf(y) + y^2 + cf(y)) = f((x+y+c)^2) - (x+c)f(x+c)$$
$$P(x-2c, y) \rightarrow f(xf(y) + y^2 - 2cf(y)) = f((x+y-2c)^2) - (x-2c)f(x-2c)$$
Using (1) and (2), we get
$$f(xf(y) + y^2 + cf(y)) = f(xf(y) + y^2 - 2cf(y)).$$
Since the function is surjective, for any two real numbers $u \neq v$, there exist two real numbers $x, y$ such that
$$f(y) = \frac{u-v}{3c} \neq 0, \quad x = \frac{u-y^2-cf(y)}{f(y)}$$
This means that $f(u) = f(v)$ for any $u \neq v$, and $f$ is constant which is not possible since it is surjective. Therefore such a $c$ does not exist and $f(x) = 2x$, $x \in \mathbb{R}$. This function satisfies the equation:
$$$$
f(xf(y) + y^2) = 4xy + 2y^2 = 2(x+y)^2 - 2x^2 = f((x+y)^2) - x f(x).