Japanese Mathematical Olympiad - The Final Round

The given equation clearly holds if $f(x) = 0$ for all $x$. Assume that there exists $a$ such that $f(a) \neq 0$.
Substituting $y = -f(x)$ into the given equation and letting $c = f(0)$, we obtain
$$f(-f(x)) = c + f(x)^2. \quad (1)$$
Substituting $y = -f(y)$ into the equation and using (1),
$$\begin{aligned} f(f(x) - f(y)) &= f(x)^2 - 2f(x)f(y) + f(-f(y)) \\ &= f(x)^2 - 2f(x)f(y) + f(y)^2 + c \\ &= (f(x) - f(y))^2 + c. \end{aligned} \quad (2)$$

$$f(a)^2 + 2y f(a) = f(y + f(a)) - f(y).$$
Since the left-hand side runs through $\mathbb{R}$ when $y$ runs through $\mathbb{R}$, the right-hand side runs through $\mathbb{R}$. Hence, $f(x) - f(y)$ runs all real numbers when $x$ and $y$ run through $\mathbb{R}$.
Therefore, according to (2), $f(x) = x^2 + c$ for all $x$. This meets the given equation for any constant $c$.
Therefore, $f(x) = x^2 + c$ ($c$ any constant) or $f(x) = 0$ (for all $x$).