Japan Mathematical Olympiad

Let $f(0) = a$ and in the given functional equation
$$f(x+y)f(f(x)-y) = xf(x) - yf(y) \quad (1)$$
substitute $x = y = 0$, then we get
$$f(0)f(a) = 0,$$
from which we obtain $f(0) = 0$ or $f(a) = 0$. Since $f(0) = 0$ implies that $a = 0$ by definition, we see that $f(a) = 0$ always holds.
Let $t$ be an arbitrary real number. By substituting $x = 0, y = t$, into (1), we obtain
$$f(t)f(a-t) = -tf(t) \quad (2)$$
from which we can conclude that either $f(t) = 0$ or $f(a-t) = -t$ holds. Therefore, if $u \neq 0$ and $f(u) \neq 0$, then since $f(a-u) = -u \neq 0$, by substituting $t = a-u$ we obtain $f(u) = u-a$. Consequently, we get the implication
$$u \neq 0 \Rightarrow [f(u) = 0 \text{ or } f(u) = u-a] \quad (3)$$
Let us first consider the case where $f(u) = 0$ holds for every $u \neq 0$. We will show that such an $f$ satisfies the functional equation (1). If $a = 0$, then the function $f$ is the identically zero function, and it is obvious that the equation (1) is satisfied in this case. So, let us suppose that $a \neq 0$. Since $vf(v) = 0$ holds for all $v$ by the right-hand side of the functional equation (1) is always 0. On the other hand, the left-hand side of (1) is $\neq 0$ only when both $x+y=0$ and $f(x)-y=0$ are satisfied. We will show this cannot happen. For, if $x=y=0$, then $f(x)-y=f(0)=a \neq 0$, while if $0 \neq x = -y$, then $f(x)-y=-y \neq 0$. Therefore, the left-hand side of (1) is always 0 also, and thus our function $f$ satisfies (1).
Next we consider the case where there exists $u \neq 0$ for which $f(u) \neq 0$. Let us denote one such $u$ by $b$. If we substitute $x = a, y = -b$ in the equation (1), we get $f(a-b)f(b) = bf(-b)$, while by (2) we have $f(b)f(a-b) = -bf(b)$, and since $b \neq 0$, we conclude that
$$f(-b) = -f(b) \neq 0 \quad (4)$$
We also have, in view of (3) that
$$f(b) = b - a, \quad f(-b) = -b - a. \quad (5)$$
Combining (4) and (5), we get that $a = 0$, and therefore, $f(b) = b$. We now show that $f(x) = x$ must hold for all $x$ in this case. So, suppose that there

exists a $c$ for which $f(c) \neq c$. Since $f(0) = a = 0$. $c \neq 0$, and hence we have by (3) $f(c) = 0$. By substituting $x = b$ and $y = c$ in (1), we obtain
$$f(b+c)f(b-c) = b^2. \quad (6)$$
Since, $b \neq 0$, $f(b+c) \neq 0$, $f(b-c) \neq 0$ must hold. As $f(0) = 0$, we also have $b+c \neq 0$ and $b-c \neq 0$. If we apply (3) to $u = b+c$ and $u = b-c$, we conclude that $f(b+c) = b+c$ and $f(b-c) = b-c$ and therefore, $f(b+c)f(b-c) = b^2-c^2$. However, since $c \neq 0$, this contradicts (6). Thus, we conclude that if there exists a $b \neq 0$ for which $f(b) \neq 0$, then $f(x) = x$ for all $x$. It is easy to check that the function $f(x) = x$ satisfies the functional equation (1).
As we covered all the possibilities we conclude that the only functions $f$ that satisfy the functional equation (1) are:
$$f(x) = x$$
and
$$f(x) = a \text{ (if } x = 0\text{); } 0 \text{ (if } x \neq 0\text{)},$$
where $a$ can be an arbitrary real number.