55rd Ukrainian National Mathematical Olympiad - Fourth Round

Answer: $f(x)\equiv 1$.

First suppose that there exists $a\ne 0$ such that $f(a)=f(0)=b$. By (0) we denote the main equation. Substituting $y=0$ in (0), we get: $$4f(x+b)=f(x)+b+b+1.$$ (1)

Substituting $y=a$ in (0), we get: $$4f(x+b)=f(x)+b+f(xa)+1.$$ (2)

Combining (1) and (2), we obtain $f(ax)=b$, and we obviously have that $f$ is constant. Supposing that $f(x)=b$ and using (0) we get $4b=3b+1$. The result is $f(x)\equiv 1$.

Now, it follows that $f(a)=f(0)\Rightarrow a=0$. If we replace $x$ by $y$ in (0) and combine the result and (0), we obtain:

$f(x+f(y))=f(y+f(x))$.

Substituting $x=-f(y)$ in the last equation, we get $$\begin{array}{rl}f(0)& =f(-f(y)+f(y))=f(y+f(-f(y)))\Rightarrow y+f(-f(y))=0\\ f(-f(y))& =-y.\end{array}(3)$$

Substituting $x=1$ and $y=-f(t)$ in (3), we obtain: $$4f(1+f(-f(t)))=f(1)+f(-f(t))+f(-f(t))+1.$$ Using (3), we get $$4f(1-t)=f(1)-t-t+1.(4)$$ Replace $t=1-z$ in (4), we have $$4f(z)=f(1)-2(1-z)+1=2z+c_1.\text{ That is }f(z)=\frac{1}{2}z+c.$$ It can be shown in the usual way that such function satisfies the condition of the question.