Japan Mathematical Olympiad

Let $f(0)=a$ and in the given functional equation $$f(x+y)f(f(x)-y)=xf(x)-yf(y)(1)$$ 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 $$f(t)f(a-t)=-tf(t)(2)$$ from which we can conclude that either $f(t)=0$ or $f(a-t)=-t$ holds. Therefore, if $u\ne 0$ and $f(u)\ne 0$, then since $f(a-u)=-u\ne 0$, by substituting $t=a-u$ we obtain $f(u)=u-a$. Consequently, we get the implication $$u\ne 0\Rightarrow [f(u)=0\text{ or }f(u)=u-a](3)$$ Let us first consider the case where $f(u)=0$ holds for every $u\ne 0$. 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 $a\ne 0$. 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 $\ne 0$ 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 $f(x)-y=f(0)=a\ne 0$, while if $0\ne x=-y$, then $f(x)-y=-y\ne 0$. 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 $u\ne 0$ for which $f(u)\ne 0$. 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 $b\ne 0$, we conclude that $$f(-b)=-f(b)\ne 0(4)$$ We also have, in view of (3) that $$f(b)=b-a,f(-b)=-b-a.(5)$$ 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 $f(c)\ne c$. Since $f(0)=a=0$. $c\ne 0$, and hence we have by (3) $f(c)=0$. By substituting $x=b$ and $y=c$ in (1), we obtain $$f(b+c)f(b-c)=b^2.(6)$$ Since, $b\ne 0$, $f(b+c)\ne 0$, $f(b-c)\ne 0$ must hold. As $f(0)=0$, we also have $b+c\ne 0$ and $b-c\ne 0$. 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 $c\ne 0$, this contradicts (6). Thus, we conclude that if there exists a $b\ne 0$ for which $f(b)\ne 0$, 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 $$f(x)=a\text{ (if }x=0\text{); }0\text{ (if }x\ne 0\text{)},$$ where $a$ can be an arbitrary real number.