Seventeenth ROMANIAN MASTER OF MATHEMATICS

Observe that if $x_1,y_1,x_2,y_2\in S$ with $x_1{y}_1=x_2{y}_2$ then $$f(x_1)+f(y_1)=f(x_2)+f(y_2).(1)$$

This tells us that for $u,v,w\in S$, $f(uv)+f(w)=f(u)+f(vw)$, so $f(uv)-f(u)-f(v)=f(vw)-f(w)-f(v)$. Notice the RHS is independent of $u$ so the same must be true of the LHS. By replicating the argument with $u$ and $v$ switched, we also see the LHS is independent of $v$ so in fact $$f(uv)-f(u)-f(v)=k\text{for some constant }k\in \mathbb{Z}(2)$$

Using (1) again we have, for $y,z\in S$ $$\begin{array}{c}f(cz)+f(y)=f(z)+f(cy),\\ \text{so }f(cy)-f(y)=l,\text{for some constant }l\in \mathbb{Z}.\end{array}(3)$$

Setting $x=cz$ in the original functional equation for $z\in S$ shows $$\begin{array}{c}f(c(yz+1))\stackrel{(3)}{=}f(yz+1)+l=f(cz)+f(y)\stackrel{(3)}{=}f(z)+f(y)+l,\\ \text{so }f(yz+1)=f(y)+f(z).\end{array}$$

Let $x\in S$ and set $y=x,z=x+2$ in the above to get $$\begin{array}{c}f((x+1{)}^2)\stackrel{(2)}{=}2f(x+1)+k=f(x)+f(x+2)\\ \Rightarrow f(x)+f(x+2)-2f(x+1)=-k=\text{constant}\end{array}$$ which forces $f$ to be a quadratic. By setting $x=y$ in the original functional equation and considering the degree of both sides, we see $f$ must be in fact be constant. The only constant function that satisfies the condition is $f\equiv 0$.