XVIII-th Macedonian mathematical Olympiad

If we choose $x=0$, then $$f(yf(0))=f(f(0)).$$ If $f(0)\ne 0$, then for $y=\frac{1}{f(0)}$, we get $$f(t)=f(f(0))=c,$$ i.e. $f$ is a constant function. But then, substituting in (1), the equation gets the form $$c=c+cx,x\in \mathbb{R}.$$ As $x$ is arbitrary, we get that $c=0$. Due to the resulting contradiction, $f(0)=0$.

II. Suppose that $f(x)=0$ for some $x\ne 0$. Then, substituting in the initial equation $$xf(y)=0,$$ for each $y\in \mathbb{R}$. Therefore, $f(y)=0$ for each $y\in \mathbb{R}$. Due to the resulting contradiction, $f(x)=0$, $x\in \mathbb{R}$.

It is not hard to check that $f(x)=0$, $x\in \mathbb{R}$ is a solution of the initial equation.

III. If $y=0$, then $f(x)=f(f(x))$. Now, substituting in the first equation, $$f(x+yf(x))=f(x)+xf(y).$$

IV. Let $f(1)=a\ne 0$. Then $$\begin{array}{rl}x=1,y=1& \Rightarrow f(1+a)=2a\\ x=1,y=-1& \Rightarrow f(1-a)=a+f(-1)\\ x=1+a,y=-1& \Rightarrow f(1-a)=f(1+a-2a)=2a+(1+a)f(-1)\end{array}$$ Now from (1) and (2) we get $$\begin{array}{rl}a+f(-1)& =2a+(1+a)f(-1)\\ -a& =af(-1)\\ f(-1)& =-1\\ f(1-a)& =a-1\end{array}$$ On the other hand $$\begin{array}{rl}x& =1-a,y=1\\ x& =1\\ x& =-1\end{array}\begin{array}{rl}0& =f(1-a+a-1)=a-1+(1-a)a=-(a-1{)}^2\\ & =a-1\\ & =f(1+y)=1+f(y)\\ & =f(-1-y)=-1-f(y)\end{array}$$ Now from (3) and (4) $-f(y)=1+f(-1-y)=f(-y)$. Therefore, for $y=-1$ we have $$f(x-f(x))=f(x)-x.$$ Finally, $$\begin{array}{rl}f(x-f(x))& =f(f(x-f(x)))=f(f(x)-x)=-f(x-f(x))\\ f(x-f(x))& =0\\ x-f(x)& =0\\ f(x)& =x\end{array}$$ This function satisfies the equation, so the required solutions are: $f(x)=0$ and $f(x)=x$.