Problem 1736

First, assume that $f(0) = 0$. Inserting $x = 0$ into the functional equation we get $f(y) = 0$ for all $y \in \mathbb{R}$. This function satisfies the equation.

Now, let $f(0) \neq 0$. Inserting $y = 0$ into the equation we get
$$(x - 2)f(0) + f(2f(x)) = f(x)$$
for all $x \in \mathbb{R}$. Obviously, this means that $f$ is injective.

Now, put $x = 2$ into the equation to get
$$f(y + 2f(2)) = f(2 + yf(2)) \quad \text{for all } y \in \mathbb{R}.$$
Since $f$ is injective, we have $y + 2f(2) = 2 + yf(2)$ for all $y \in \mathbb{R}$. If we insert $y = 0$, then we get $f(2) = 1$.

Since $f(2) = 1$ and $f$ is injective, we have $f(3) \neq 1$. Insert $x = 3$ and $y = \frac{3}{1-f(3)}$ into the equation to get
$$f\left(\frac{3}{1-f(3)} + 2f(3)\right) = 0.$$
We have shown that $f$ has a zero. Let $a$ be a zero of $f$, $f(a) = 0$. Inserting $y = a$ into the initial equation yields
$$f(a + 2f(x)) = f(x + af(x)) \quad \text{for all } x \in \mathbb{R}.$$
Since $f$ is injective, we have $a + 2f(x) = x + af(x)$ for all $x \in \mathbb{R}$. Since $a \neq 2$ (remember that $f(2) = 1 \neq 0$), we get $f(x) = \frac{x-a}{2-a}$.

Finally, inserting $f(x) = \frac{x-a}{2-a}$ into the equation we get $a = 1$. The only two solutions to this functional equation are the functions $f(x) = 0$ and $f(x) = x - 1$.