APMO

It is clear that if $f$ is a constant function which satisfies the given equation, then the constant must be $0$. Conversely, $f(x)=0$ clearly satisfies the given equation, so, the identically $0$ function is a solution. In the sequel, we consider the case where $f$ is not a constant function. Let $t\in \mathbf{R}$ and substitute $(x,y,z)=(t,0,0)$ and $(x,y,z)=(0,t,0)$ into the given functional equation. Then, we obtain, respectively, $$\begin{array}{rl}& f(f(t)+2f(0))=f(f(t)-f(0))+f(f(0))+2f(0),\\ & f(f(t)+2f(0))=f(f(0)-f(t))+f(f(0))+2f(0),\end{array}$$ from which we conclude that $f(f(t)-f(0))=f(f(0)-f(t))$ holds for all $t\in \mathbf{R}$. Now, suppose for some pair $u_1,u_2$, $f\left(u_1\right)=f\left(u_2\right)$ is satisfied. Then by substituting $(x,y,z)=\left(s,0,u_1\right)$ and $(x,y,z)=\left(s,0,u_2\right)$ into the functional equation and comparing the resulting identities, we can easily conclude that $$\begin{array}{c}f\left(s{u}_1\right)=f\left(s{u}_2\right)\end{array}\text{\hspace{1em}(*)}$$ holds for all $s\in \mathbf{R}$. Since $f$ is not a constant function there exists an $s_0$ such that $f\left(s_0\right)-f(0)\ne 0$. If we put $u_1=f\left(s_0\right)-f(0),u_2=-u_1$, then $f\left(u_1\right)=f\left(u_2\right)$, so we have by $(\ast )$ $$f\left(s{u}_1\right)=f\left(s{u}_2\right)=f\left(-s{u}_1\right)$$ for all $s\in \mathbf{R}$. Since $u_1\ne 0$, we conclude that $$f(x)=f(-x)$$ holds for all $x\in \mathbf{R}$. Next, if $f(u)=f(0)$ for some $u\ne 0$, then by $(\ast )$, we have $f(su)=f(s0)=f(0)$ for all $s$, which implies that $f$ is a constant function, contradicting our assumption. Therefore, we must have $f(s)\ne f(0)$ whenever $s\ne 0$. We will now show that if $f(x)=f(y)$ holds, then either $x=y$ or $x=-y$ must hold. Suppose on the contrary that $f\left(x_0\right)=f\left(y_0\right)$ holds for some pair of non-zero numbers $x_0,y_0$ for which $x_0\ne y_0,x_0\ne -y_0$. Since $f\left(-y_0\right)=f\left(y_0\right)$, we may assume, by replacing $y_0$ by $-y_0$ if necessary, that $x_0$ and $y_0$ have the same sign. In view of $(\ast )$, we see that $f\left(s{x}_0\right)=f\left(s{y}_0\right)$ holds for all $s$, and therefore, there exists some $r>0,r\ne 1$ such that $$f(x)=f(rx)$$ holds for all $x$. Replacing $x$ by $rx$ and $y$ by $ry$ in the given functional equation, we obtain $$\begin{array}{c}f(f(rx)+f(ry)+f(z))=f(f(rx)-f(ry))+f\left(2r^{2}xy+f(z)\right)+2f(r(x-y)z)\end{array}\text{\hspace{1em}(i)}$$ and replacing $x$ by $r^{2}x$ in the functional equation, we get $$\begin{array}{c}f\left(f\left(r^{2}x\right)+f(y)+f(z)\right)=f\left(f\left(r^{2}x\right)-f(y)\right)+f\left(2r^{2}xy+f(z)\right)+2f\left(\left(r^{2}x-y\right)z\right)\end{array}\text{\hspace{1em}(ii)}$$ Since $f(rx)=f(x)$ holds for all $x\in \mathbf{R}$, we see that except for the last term on the right-hand side, all the corresponding terms appearing in the identities (i) and (ii) above are equal, and hence we conclude that $$\begin{array}{c}f(r(x-y)z)=f\left(\left(r^{2}x-y\right)z\right))\end{array}\text{\hspace{1em}(iii)}$$ must hold for arbitrary choice of $x,y,z\in \mathbf{R}$. For arbitrarily fixed pair $u,v\in \mathbf{R}$, substitute $(x,y,z)=\left(\frac{v-u}{r^2-1},\frac{v-r^{2}u}{r^2-1},1\right)$ into the identity (iii). Then we obtain $f(v)=f(ru)=f(u)$, since $x-y=u,r^{2}x-y=v,z=1$. But this implies that the function $f$ is a constant, contradicting our assumption. Thus we conclude that if $f(x)=f(y)$ then either $x=y$ or $x=-y$ must hold. By substituting $z=0$ in the functional equation, we get $$f(f(x)+f(y)+f(0))=f(f(x)-f(y)+f(0))=f((f(x)-f(y))+f(2xy+f(0))+2f(0).$$ Changing $y$ to $-y$ in the identity above and using the fact that $f(y)=f(-y)$, we see that all the terms except the second term on the right-hand side in the identity above remain the same. Thus we conclude that $f(2xy+f(0))=f(-2xy+f(0))$, from which we get either $2xy+f(0)=-2xy+f(0)$ or $2xy+f(0)=2xy-f(0)$ for all $x,y\in \mathbf{R}$. The first of these alternatives says that $4xy=0$, which is impossible if $xy\ne 0$. Therefore the second alternative must be valid and we get that $f(0)=0$. Finally, let us show that if $f$ satisfies the given functional equation and is not a constant function, then $f(x)=x^2$. Let $x=y$ in the functional equation, then since $f(0)=0$, we get $$f(2f(x)+f(z))=f\left(2x^2+f(z)\right)$$ from which we conclude that either $2f(x)+f(z)=2x^2+f(z)$ or $2f(x)+f(z)=-2x^2-f(z)$ must hold. Suppose there exists $x_0$ for which $f\left(x_0\right)\ne x_0^2$, then from the second alternative, we see that $f(z)=-f\left(x_0\right)-x_0^2$ must hold for all $z$, which means that $f$ must be a constant function, contrary to our assumption. Therefore, the first alternative above must hold, and we have $f(x)=x^2$ for all $x$, establishing our claim. It is easy to check that $f(x)=x^2$ does satisfy the given functional equation, so we conclude that $f(x)=0$ and $f(x)=x^2$ are the only functions that satisfy the requirement.