Macedonian Mathematical Olympiad

Let the function $f$ is such that for every $x,y\in \mathbb{R}$ it is fulfilled the equation (1). If in (1) we put $y=x$, we obtain $$f(x+f(x))=2x,\text{ for every }x\in \mathbb{R}.(2)$$ Furthermore, for $x=0$ from (2) follows $f(f(0))=0$, and for $x=y=\frac{f(0)}{2}$ from (1) we obtain $f(f(0))=f(0)$. Now, from the last two equations it follows that $f(0)=0$. Now, if in (1) we put $y=0$ and if we use $f(0)=0$, we obtain $$f(\max \{x,0\}+\min \{f(x),0\})=x,\text{ for every }x\in \mathbb{R}.(3)$$ We will consider two cases: Case 1. $x>0$. Then from (3) it follows $$f(x+\min \{f(x),0\})=x.$$ From the last equality follows that or $f(x+f(x))=x$ or $f(x)=x$. If $f(x+f(x))=x$, then for (2) it follows $2x=x$, which is not possible when $x>0$. Hence, $f(x)=x$. Case $2^{\circ }$ $x<0$. Then from (3) it follows $$f(\min \{f(x),0\})=x.$$ Now, from the last equality we obtain or $f(0)=x$ or $f(f(x))=x$. If $f(0)=x$, then $0>x=f(0)=0$, which is a contradiction. Hence, $f(f(x))=x$. Now, if in (2) instead of $x$ we put $f(x)$ and if first we use that $f(f(x))=x$, and after that we use the equality (2), we obtain $$2f(x)=f(f(x)+f(f(x)))=f(f(x)+x)=2x,$$ from where we obtain $f(x)=x$, for $x<0$. Finally, $1^{\circ }$ and $2^{\circ }$ implies that $f(x)=x$ for every $x\in \mathbb{R}$. It is not difficult to check that this function is a solution of the problem.