60th Belarusian Mathematical Olympiad

Let the function $f:\mathbb{R}\to \mathbb{R}$ satisfy the equality $$f(f(x))=xf(x)+x-1(\ast )$$ for all $x\in \mathbb{R}$.

a) Set $c=f(0)$. We have $f(c)=f(f(0))=0\cdot f(0)+0-1=-1$. So $$-1=f(c).(1)$$ Therefore, taking into account $(\ast )$ for $x=c$ and (1), we obtain the required value $$f(-1)=f(f(c))=c\cdot f(c)+c-1=-c+c-1=-1.$$

b) Let $a=f(1)$. Taking into account $(\ast )$ for $x=1$, we obtain $$f(a)=f(f(1))=1\cdot f(1)+1-1=a.$$ Since $$a=f(a),(2)$$ it follows that $$f(a)=f(f(a))=a\cdot f(a)+a-1=a^2+a-1.(3)$$ From (2) and (3) it follows that $a^2+a-1=a$, i.e. either $a=-1$ or $a=1$.

Both the values are achieved. For example, $$f(x)=-1\text{and}f(x)=\{\begin{array}{ll}-1,& \text{if }x\ne 1,\\ 1,& \text{if }x=1.\end{array}(3)$$ It is easy to verify that both the functions satisfy the initial equation $(\ast )$ for all real $x$.

Some more examples: $$f(x)=\{\begin{array}{ll}\frac{1}{x},& \text{if }x\ne 0,\\ -1,& \text{if }x=0\end{array}\text{or}f(x)=\{\begin{array}{ll}-1,& \text{if }x\ne 0,\\ \alpha ,& \text{if }x=0,\end{array}$$ where $\alpha$ is any real number different from $0$. Another function: $f(x)=\frac{1}{x}$, if $x\in A$, and $f(x)=-1$ for all other $x$, where $A$ is an arbitrary subset of $\mathbb{R}\setminus \{0\}$, such that $A^{-1}\subset A$.