China Team Selection Test

Determine all $f: \mathbb{R} \rightarrow \mathbb{R}$ such that
$$f(xf(y) + y^3) = yf(x) + f(y)^3.$$

Let $P(x, y)$ denote the original proposition.

First, we consider the constant solution. Clearly, the only constant solution is:
$$\boxed{f(x) = 0 \ \ \forall x \in \mathbb{R}}.$$

Now, assume $f$ is non-constant.

From $P(x, 0)$:
$$f(xf(0)) = f(0)^3 \quad \forall x \implies f(0) = 0 \text{ since } f \text{ is non-constant}.$$

Claim 1: $f(y^3) = f(y)^3$ for all $y$.

Proof: From $P(0, y)$:
$$f(y^3) = f(y)^3. \quad \blacksquare$$

Claim 2: $f$ is injective at 0.

Proof: Suppose $f(a) = 0$ for some $a \neq 0$. Then from $P(x, a)$:
$$af(x) = f(a^3) \implies f \text{ is constant, contradiction!} \quad \blacksquare$$

Claim 3: $f(1) = 1$.

Proof: From Claim 1, putting $y = 1$ gives $f(1) = 0$, $-1$, or $1$. $0$ can be ruled out by Claim 2. If $f(1) = -1$, then from $P(x, 1)$:
$$f(1 - x) = f(x) - 1 \quad \forall x \implies f(x) = f(1 - (1 - x)) = f(1 - x) - 1 = f(x) - 2,$$
which is absurd. Therefore, $f(1) = 1$. $\quad \blacksquare$

Claim 4: $f(x + 1) = f(x) + 1$ for all $x$.

Proof: From $P(x, 1)$. $\quad \blacksquare$

Now, for any $y, z$ with $y \neq 0$, we can choose an $x$ such that $z = xf(y) + y^3$ by Claim 2. Then from $P(x + 1, y)$:
$$\begin{align*} f(xf(y) + y^3 + f(y)) &= yf(x + 1) + f(y)^3 \\ \implies f(z + f(y)) &= yf(x) + f(y)^3 + y \ \ \dots \ \ \text{from Claim 4} \\ &= f(xf(y) + y^3) + y \ \ \dots \ \ \text{from } P(x, y) \\ \implies f(z + f(y)) &= f(z) + y. \end{align*}$$

Clearly, the above holds when $y = 0$ as well, so call it $Q(z, y)$.

Claim 5: $f(f(y)) = y$ for all $y$.

Proof: From $Q(0, y)$. $\quad \blacksquare$

Now, $Q(z, f(y))$ along with Claim 5 implies:
$$f(y + z) = f(y) + f(z) \quad \forall y, z \implies f \text{ is additive}.$$

Using the additivity in $P(x, y)$, we get:
$$f(xf(y)) + f(y^3) = yf(x) + f(y)^3 \implies f(xf(y)) = yf(x),$$
by Claim 1. Replacing $y$ by $f(y)$ in the above and using Claim 5, we get:
$$f(xy) = f(x)f(y) \quad \forall x, y \implies f \text{ is multiplicative as well}.$$

It is well known that the only non-constant function which is both additive and multiplicative is:
$$\boxed{f(x) = x \ \ \forall x \in \mathbb{R}}.$$

$\quad \blacksquare$