Team selection tests for BMO 2018

Denote ($\ast$) as the given condition. Replacing $x=y=-\frac{1}{2}$ in ($\ast$), we get $$f\left(2{\left(f\left(-\frac{1}{2}\right)\right)}^2-\frac{1}{2}\right)=0$$ If there exists some $a\ne 0$ such that $f(a)=0$. Substituting $x=a$ in ($\ast$), we get $f(y)=0$ for any $y$, which is obviously a solution.

Let us consider now the case $f(x)\ne 0$ for any nonzero real number $x$. Hence, $f(0)=0$ and $f\left(-\frac{1}{2}\right)=\pm \frac{1}{2}$. On the other hand, it is easy to see that if $f$ is solution, then $-f$ is also a solution. Therefore, we only need to consider the case $f(0)=0$ and $f\left(-\frac{1}{2}\right)=-\frac{1}{2}$. From this, we have $$\begin{array}{rl}f\left(-\frac{1}{2}-f(y)\right)& =-f(y)-\frac{1}{2},\forall y\in \mathbb{R}\\ f\left(\frac{1}{2}\right)& =\frac{1}{2}\end{array}$$ and $$f(x-f(x))=f(x)-x,\forall x\in \mathbb{R}$$ Therefore, $$\begin{array}{rl}f\left(x-\frac{1}{2}-f(x)\right)& =f\left(-\frac{1}{2}-f(x-f(x))\right)\\ & =-\frac{1}{2}-f(x-f(x))=x-\frac{1}{2}-f(x)\end{array}$$ which implies $f\left(-f\left(\frac{1}{2}\right)\right)=-f\left(\frac{1}{2}\right)$. Now, substituting $x=\frac{1}{2}$ and $y=-f\left(\frac{1}{2}\right)$ in ($\ast$), we get $$f\left(\frac{1}{2}-2{\left(f\left(\frac{1}{2}\right)\right)}^2\right)=0.$$ Thus $f\left(\frac{1}{2}\right)=\pm \frac{1}{2}$. If $f\left(\frac{1}{2}\right)=-\frac{1}{2}$, then $\frac{1}{2}=-f\left(\frac{1}{2}\right)=f\left(-f\left(\frac{1}{2}\right)\right)=f\left(\frac{1}{2}\right)=-\frac{1}{2}$, contradiction. Therefore, $f\left(\frac{1}{2}\right)=\frac{1}{2}$.

From here, it follows that $$f\left(f(y)+\frac{1}{2}\right)=f(y)+\frac{1}{2},\forall y\in \mathbb{R}$$ In this equation, substituting $y=x-\frac{1}{2}-f(x)$, we get $$f(x-f(x))=x-f(x),\forall x\in \mathbb{R}$$ Comparing this result with $f(x-f(x))=f(x)-x$, we get $f(x)=x$ for any $x$, which is clearly a solution.

In conclusion, there are three solutions: $f(x)=0$, $f(x)=x$ and $f(x)=-x$ for all real number $x$.