, 2023

Let $P(a, b)$ denote the result of substituting $x = a, y = b$.
$$P(0,0) : f(f(0))f(0) = f(0) \Rightarrow f(f(0)) = 1 \text{ or } f(0) = 0$$

Case 1. $f(f(0)) = 1$
$$P(f(0), 0) : f(f(0))^2 = f(0)^3 + f(0) \Rightarrow f(0)^3 + f(0) = 1 \cdots (1)$$
$$P(0, f(0)) : f(1)f(0) = f(0) \Rightarrow f(1) = 1$$
$$P(-1, 0) : f(f(0))f(-1) = 2f(0) \Rightarrow f(-1) = 2f(0)$$
$$P(-1, 1) : f(0)f(-1) = 1 + f(-1) \Rightarrow 2f(0)^2 = 1 + 2f(0) \cdots (2)$$
From equation (2) we obtain
$$2f(0)^2 - 2f(0) - 1 = 0 \Rightarrow f(0) = \frac{1 \pm \sqrt{3}}{2}.$$
But substituting this back into equation (1) shows it does not hold, so $f(0)$ has no solution, hence $f(f(0)) \neq 1$.

Case 2. $f(0) = 0$
$$\begin{aligned} P\left(\frac{-f(x)}{x}, x\right) : & f(0)f\left(\frac{-f(x)}{x}\right) = \frac{f(x)^3}{x^2} + f(-f(x)) \\ & \Rightarrow 0 = \frac{f(x)^3}{x^2} + f(-f(x)) \\ & \Rightarrow f(x)^3 = -x^2 f(-f(x)) \cdots (3) \end{aligned}$$

Proof. If there exists $c \neq 0, f(c) = 0$:
$$P(c, 1) : 0 = c^2 f(1) \Rightarrow f(1) = 0$$
$$P(1, y) : 0 = 2f(y) \Rightarrow f(y) = 0, \forall y \in \mathbb{R}$$
Hence $\forall x \neq 0, f(x) \neq 0$ ($f$ has no other zeros), otherwise $f(x) = 0, \forall x \in \mathbb{R}$ (which is one of the solutions).

Claim: $f(-1) = 2$ or $f(-1) = -1$.
Proof.
$$P(-1, x) : f(-x + f(x)) f(-1) = f(x) + f(-x)$$
$$P(-1, -x) : f(x + f(-x)) f(-1) = f(x) + f(-x)$$
Combining the above two equations gives
$$\begin{aligned} f(-x + f(x)) &= f(x + f(-x)) \\ \Rightarrow -x + f(x) &= x + f(-x) \\ \Rightarrow f(x) - f(-x) &= 2x \cdots (4) \end{aligned}$$
Then
$$P(1, 1) : f(1 + f(1)) = 2$$
$$P(-1, -1) : f(1 + f(-1)) f(-1) = f(-1) + f(1) = 2 - 2f(1)$$
$$P(-1, 1 + f(1)) : f(-1 - f(1) + 2) f(-1) = f(1 + f(1)) + f(-1 - f(1))$$
From equation (4) we obtain $f(1) - f(-1) = 2$ and $f(1 + f(1)) - f(-1 + f(1)) = 2(1 + f(1))$, combining with the above equation gives
$$\begin{aligned} f(1 - (2 + f(-1))) f(-1) &= 2 + (f(1 + f(1)) - 2(1 + f(1))) \\ \Rightarrow f(-1 + f(-1)) f(-1) &= 2 - 2f(1). \end{aligned}$$
And
$$\begin{align*} & f(1+f(-1)) - f(-(1+f(-1))) = 2(1+f(-1)) \\ \Rightarrow \quad & \frac{2+2f(-1)}{f(-1)} - \frac{2-2f(1)}{f(-1)} = 2(1+f(-1)) \\ \Rightarrow \quad & 2(f(-1)+f(1)) = 2(1+f(-1))f(-1) \\ \Rightarrow \quad & f(1)+f(-1) = f(-1)^2 + f(-1) \\ \Rightarrow \quad & 2+2f(-1) = f(-1)^2 + f(-1) \\ \Rightarrow \quad & f(-1)^2 - f(-1) - 2 = 0 \\ \Rightarrow \quad & f(-1) = 2 \text{ or } f(-1) = -1. \end{align*}$$

(i) $f(-1) = 2:$
$\therefore f(1) - f(-1) = 2 \Rightarrow f(1) = 4.$
$$P(1, -1) : f(-1 + 2) f(1) = 2f(-1)$$
Therefore $f(1)^2 = 2f(-1) \Rightarrow 16 = 4$ (contradiction), so $f(-1) \neq 2$.

(ii) $f(-1) = -1:$
$f(1) - f(-1) = 2 \Rightarrow f(1) = 1.$
$$P(x, 1) : f(x + 1) = \frac{x^2}{f(x)} + 1$$
$$\begin{align*} & P(-x-1, -1) : f(x)f(-x-1) = -x^2 - 2x - 1 + \left( \frac{x^2}{f(x)} + 1 \right) \\ \Rightarrow \quad & f(-x-1) = \frac{x^2}{f(x)^2} - \frac{x^2}{f(x)} - \frac{2x}{f(x)}. \end{align*}$$
From equation (4) we get $f(x+1) - f(-x-1) = 2x+2$, so
$$\begin{aligned} & \frac{x^2}{f(x)} + 1 - \frac{x^2}{f(x)^2} + \frac{x^2}{f(x)} + \frac{2x}{f(x)} = 2x+2 \quad (\text{multiplying both sides by } f(x)^2) \\ \Rightarrow & 2x^2 f(x) - 2xf(x)^2 + 2xf(x) - 2f(x)^2 + f(x)^2 - x^2 = 0 \\ \Rightarrow & (x - f(x))(2xf(x) + 2f(x) - x - f(x)) = 0 \\ \Rightarrow & x - f(x) = 0 \quad \text{or} \quad 2xf(x) + f(x) - x = 0 \\ \Rightarrow & f(x) = x \quad \text{or} \quad f(x) = \frac{x}{2x+1} \left( \text{when } x = -\frac{1}{2} \text{, we must have } f(x) = x \right). \end{aligned}$$
If there exists $p \neq 0, f(p) = \frac{p}{2p+1}$
$$P(1, p) : f(p + f(p)) = 2f(p) \Rightarrow f\left(\frac{2p^2 + 2p}{2p+1}\right) = \frac{2p}{2p+1}.$$
Since $f\left(\frac{2p^2+2p}{2p+1}\right) = \frac{2p^2+2p}{2p+1}$ or $\frac{\frac{2p^2+2p}{2p+1}}{2 \times \frac{2p^2+2p}{2p+1} + 1} = \frac{2p^2+2p}{2p+1} \neq \frac{2p}{2p+1}$, therefore
$$\begin{aligned} \frac{2p}{2p+1} &= \frac{\frac{2p^2+2p}{2p+1}}{2 \times \frac{2p^2+2p}{2p+1} + 1} = \frac{2p^2+2p}{4p^2+6p+1} \\ \Rightarrow \frac{1}{2p+1} &= \frac{p+1}{4p^2+6p+1} \\ \Rightarrow (2p+1)(p+1) &= 4p^2+6p+1 \\ \Rightarrow 2p^2+3p &= 0 \\ \Rightarrow p &= -\frac{3}{2} \\ \Rightarrow f\left(-\frac{3}{2}\right) &= \frac{3}{4}. \end{aligned}$$
$P(-\frac{3}{2}, 1) : f(-\frac{3}{2} + 1) = \frac{\left(-\frac{3}{2}\right)^2}{f\left(-\frac{3}{2}\right)} + 1 = 4$, which contradicts $f(-\frac{1}{2}) = -\frac{1}{2}$, so no such $p$ exists
⇒ $\forall x, f(x) = x$