TST for BMO

Let $f(0) = a, a \in \mathbb{R}$. We set $x = 0$ and obtain $f(a) = 0$. Next, $y = a$ yields
$$f(f(x)) = x f(a + 1) \quad (1).$$
Assume first that $f(a+1) \neq 0$. Then $f$ is an injection. Indeed, assuming $f(x_1) = f(x_2)$ for some $x_1 \neq x_2$, after putting these values in (1) we get contradiction. Set $x=1$ in the initial condition. Since $f$ is injective, it follows
$$f(1) + f(y) = y + 1 \quad (2)$$
Setting $y = 1$ yields $2f(1) = 2$, hence $f(1) = 1$. $f(y) = y + 1$. Now, (2) gives us $f(y) = y, \forall y \in \mathbb{R}$, which indeed is solution.

Let now consider the case $f(a + 1) = 0$. Then
$f(f(x)) = 0, \forall x \in \mathbb{R}.$
This means that $f(y) = 0, \forall y \in \text{Im}(f)$. Assume that there exists a point $y_0$, for which $f(y_0 + 1) \neq 0$. Setting $y = y_0$ in the initial condition yields
$$f(f(x) + x f(y_0)) = x f(y_0 + 1) \quad (3).$$
For any $x_0 \in \mathbb{R}$ putting $x := x_0/(f(y_0 + 1))$ in (3), we get,
$$f(f(x) + x f(y_0)) = x_0.$$
Since $x_0$ is arbitrary, it means that $f$ is surjective, which implies $\text{Im}(f) = \mathbb{R}$.
Therefore, $f(y) = 0, \forall y \in \mathbb{R}$, which contradicts the assumption $f(y_0 + 1) \neq 0$.
So, in this case we get $f(x) = 0, \forall x \in \mathbb{R}$, which also is solution. $\square$