Team selection tests for BMO 2018

Let $a=f(0)$ and denote ($\ast$) as the given condition. Assume that $a\ne 0$, substituting $x=2a$ and $y=0$ in ($\ast$), we get $$2a^2=\frac{(f(2a){)}^2}{2a}+a$$ or $$4a-2={\left(\frac{f(2a)}{2a}\right)}^2.$$ This follows that $4a-2$ is a perfect square which is a contradiction since there is no perfect square congruent to $2$ modulo $4$. Thus $a=0$, or $f(0)=0$. Now, substituting $y=0$ in ($\ast$), we get $$x^{2}f(-x)=(f(x){)}^2,\forall x\in \mathbb{Z},x\ne 0$$ It follows that $$x^{2}f(x)=(f(-x){)}^2,\forall x\in \mathbb{Z},x\ne 0$$ Therefore $$(f(x){)}^4=x^4(f(-x){)}^2=x^{6}f(x),\forall x\in \mathbb{Z},x\ne 0$$ From this and $f(0)=0$, we get $f(x)=0$ or $f(x)=x^2$ for any integer $x$. Clearly, the function $f(x)=x^2$ satisfies the original equation. Let us consider now the case there exists some integer $c\ne 0$ such that $f(c)=0$. Replacing $y=c$ in ($\ast$), we get $$c^{2}f(2x)=0,\forall x\in \mathbb{Z},x\ne 0$$ Thus, for any even integer $x$, we have $f(x)=0$. Now, if there exists some odd number $d$ such that $f(d)=d^2$, taking $x$ even and $y=d$ in ($\ast$), we get $$d^{2}f\left(2x-d^2\right)=f\left(d^3\right),\forall x\in \mathbb{Z},x\ne 0,2\mid x$$ If $f\left(d^3\right)\ne 0$, then $f\left(2x-d^2\right)\ne 0$ and hence $f\left(d^3\right)=d^6,f\left(2x-d^2\right)=(2x-d^2{)}^2$. Plug back into the above equation, we get a contradiction. Therefore, $f\left(d^3\right)=0$ and hence $f\left(2x-d^2\right)=0$ for any nonzero even integer $x$. It follows that $f(x)=0$ for any integer $x$ such that $x\equiv -1(\mathrm{mod}4)$ and $x\ne -d^2$. In $x^{2}f(-x)=(f(x){)}^2$, taking $x\equiv -1(\mathrm{mod}4),x\ne -d^2$, we get $f(-x)=0$. It follows that $f(x)=0$ for any integer $x$ such that $x\equiv 1(\mathrm{mod}4)$ and $x\ne d^2$. Thus $f(x)=0$ for any integer $x\ne \pm d^2$. However, we have $f(d)=d^2\ne 0$, so $d=\pm d^2$ or $d=\pm 1$. It follows that $d=d^3$ and so $f(d)=f\left(d^3\right)=0$, which is a contradiction. Therefore, in this case, we have $f(x)=0$ for any integer $x$. In conclusion, the equation has two solution: $f(x)=0$ and $f(x)=x^2$.