IMO Problem Shortlist

It is no hard to see that the two functions given by $f(x)=x$ and $f(x)=-x$ for all real $x$ respectively solve the functional equation. In the sequel, we prove that there are no further solutions. Let $f$ be a function satisfying the given equation. It is clear that $f$ cannot be a constant. Let us first show that $f(0)=0$. Suppose that $f(0)\ne 0$. For any real $t$, substituting $(x,y)=\left(0,\frac{t}{f(0)}\right)$ into the given functional equation, we obtain $$\begin{array}{c}f(0)=f(t),\end{array}\text{\hspace{1em}(1)}$$ contradicting the fact that $f$ is not a constant function. Therefore, $f(0)=0$. Next for any $t$, substituting $(x,y)=(t,0)$ and $(x,y)=(t,-t)$ into the given equation, we get $$f(tf(t))=f(0)+t^2=t^2$$ and $$f(tf(0))=f(-tf(t))+t^2$$ respectively. Therefore, we conclude that $$\begin{array}{c}f(tf(t))=t^2,f(-tf(t))=-t^2,\text{ for every real }t.\end{array}\text{\hspace{1em}(2)}$$ Consequently, for every real $v$, there exists a real $u$, such that $f(u)=v$. We also see that if $f(t)=0$, then $0=f(tf(t))=t^2$ so that $t=0$, and thus 0 is the only real number satisfying $f(t)=0$. We next show that for any real number $s$, $$\begin{array}{c}f(-s)=-f(s)\end{array}\text{\hspace{1em}(3)}$$ This is clear if $f(s)=0$. Suppose now $f(s)<0$, then we can find a number $t$ for which $f(s)=-t^2$. As $t\ne 0$ implies $f(t)\ne 0$, we can also find number $a$ such that $af(t)=s$. Substituting $(x,y)=(t,a)$ into the given equation, we get $$f(tf(t+a))=f(af(t))+t^2=f(s)+t^2=0$$ and therefore, $tf(t+a)=0$, which implies $t+a=0$, and hence $s=-tf(t)$. Consequently, $f(-s)=f(tf(t))=t^2=-\left(-t^2\right)=-f(s)$ holds in this case. Finally, suppose $f(s)>0$ holds. Then there exists a real number $t\ne 0$ for which $f(s)=t^2$. Choose a number $a$ such that $tf(a)=s$. Substituting $(x,y)=(t,a-t)$ into the given equation, we get $f(s)=f(tf(a))=f((a-t)f(t))+t^2=f((a-t)f(t))+f(s)$. So we have $f((a-t)f(t))=0$, from which we conclude that $(a-t)f(t)=0$. Since $f(t)\ne 0$, we get $a=t$ so that $s=tf(t)$ and thus we see $f(-s)=f(-tf(t))=-t^2=-f(s)$ holds in this case also. This observation finishes the proof of (3). By substituting $(x,y)=(s,t),(x,y)=(t,-s-t)$ and $(x,y)=(-s-t,s)$ into the given equation, we obtain $$\begin{array}{r}f(sf(s+t)))=f(tf(s))+s^2\\ f(tf(-s))=f((-s-t)f(t))+t^2\end{array}$$ and $$f((-s-t)f(-t))=f(sf(-s-t))+(s+t{)}^2$$ respectively. Using the fact that $f(-x)=-f(x)$ holds for all $x$ to rewrite the second and the third equation, and rearranging the terms, we obtain $$\begin{array}{rl}f(tf(s))-f(sf(s+t))& =-s^2\\ f(tf(s))-f((s+t)f(t))& =-t^2\\ f((s+t)f(t))+f(sf(s+t))& =(s+t{)}^2\end{array}$$ Adding up these three equations now yields $2f(tf(s))=2ts$, and therefore, we conclude that $f(tf(s))=ts$ holds for every pair of real numbers $s,t$. By fixing $s$ so that $f(s)=1$, we obtain $f(x)=sx$. In view of the given equation, we see that $s=\pm 1$. It is easy to check that both functions $f(x)=x$ and $f(x)=-x$ satisfy the given functional equation, so these are the desired solutions.

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Alternative solution.

As in Solution 1 we obtain (1), (2) and (3). Now we prove that $f$ is injective. For this purpose, let us assume that $f(r)=f(s)$ for some $r\ne s$. Then, by (2) $$r^2=f(rf(r))=f(rf(s))=f((s-r)f(r))+r^2$$ where the last statement follows from the given functional equation with $x=r$ and $y=s-r$. Hence, $h=(s-r)f(r)$ satisfies $f(h)=0$ which implies $h^2=f(hf(h))=f(0)=0$, i.e., $h=0$. Then, by $s\ne r$ we have $f(r)=0$ which implies $r=0$, and finally $f(s)=f(r)=f(0)=0$. Analogously, it follows that $s=0$ which gives the contradiction $r=s$. To prove $|f(1)|=1$ we apply (2) with $t=1$ and also with $t=f(1)$ and obtain $f(f(1))=1$ and $(f(1){)}^2=f(f(1)\cdot f(f(1)))=f(f(1))=1$. Now we choose $\eta \in \{-1,1\}$ with $f(1)=\eta$. Using that $f$ is odd and the given equation with $x=1,y=z$ (second equality) and with $x=-1,y=z+2$ (fourth equality) we obtain $$\begin{array}{rl}& f(z)+2\eta =\eta (f(z\eta )+2)=\eta (f(f(z+1))+1)=\eta (-f(-f(z+1))+1)\\ & =-\eta f((z+2)f(-1))=-\eta f((z+2)(-\eta ))=\eta f((z+2)\eta )=f(z+2)\end{array}\text{\hspace{1em}(4)}$$ Hence, $$f(z+2\eta )=\eta f(\eta z+2)=\eta (f(\eta z)+2\eta )=f(z)+2.$$ Using this argument twice we obtain $$f(z+4\eta )=f(z+2\eta )+2=f(z)+4$$ Substituting $z=2f(x)$ we have $$f(2f(x))+4=f(2f(x)+4\eta )=f(2f(x+2)),$$ where the last equality follows from (4). Applying the given functional equation we proceed to $$f(2f(x+2))=f(xf(2))+4=f(2\eta x)+4$$ where the last equality follows again from (4) with $z=0$, i.e., $f(2)=2\eta$. Finally, $f(2f(x))=f(2\eta x)$ and by injectivity of $f$ we get $2f(x)=2\eta x$ and hence the two solutions.