, 2020

Define $P(a, b)$ to be the substitution $x = a, y = b$ into the functional equation.

(1) $P(0, y)$: $f(0) = f(0)(f(0) + f(y))$. When $f(0) \neq 0$, $f$ must be a constant function, and in this case $f(x) = \frac{1}{2}, \forall x \in \mathbb{R}$ is a solution.

(2) If $f(a) = 0$, then $P(a, y)$: $f(ay) = 0$, which gives that $a = 0$ or $f$ is identically $0$, and $f$ identically $0$ is indeed a solution.

(3) $P(x, -f(x))$: $0 = f(x)(f(x) + f(-f(x)))$, so regardless of whether $x$ is $0$ or not, we have $-f(x) = f(-f(x))$.

(4) $P(x, 0)$: $f(xf(x)) = f(x)^2$. When $x = -1$, combining with (3) gives $f(-1)^2 = f(-f(-1)) = -f(-1)$, so $f(-1) = -1$.

(5) $P(-1, 0)$: $f(1) = 1$.

(6) $P(1, y)$: $f(y + 1) = f(y) + 1$.

(7) $P(x, y + 1) - P(x, y)$: $f(xy + x + xf(x)) = f(x) + f(xy + xf(x))$, that is, the Cauchy equation
$$f(x) + f(y) = f(x + y), \forall x, y$$

(8) $f(xy + xf(x)) = f(xy) + f(xf(x)) = f(xy) + f(x)^2 = f(x)(f(x) + f(y))$, hence we obtain $f(xy) = f(x)f(y)$, and combining this with the Cauchy equation gives $f(x) = x, \forall x \in \mathbb{R}$.

(9) In summary, the functions $f$ satisfying the condition are exactly three
$$f(x) = x, \forall x \in \mathbb{R}$$
$$f(x) = 0, \forall x \in \mathbb{R}$$
$$f(x) = \frac{1}{2}, \forall x \in \mathbb{R}$$