Iranian Mathematical Olympiad

Let $P(x)$ be a non-zero polynomial and $d=\mathrm{deg}P$. We shall prove that $f(P(x))=Cd$, for some constant $C$, furthermore $f(0)=0$. It is easy to check that such a function properly works. Letting $(P(x),Q(x))=(P(x),1)$, $(b,P(x))$ in the second equation as well as $(P(x),Q(x))=(a,1)$ in the first equation for some non-zero constants $a,b$ we conclude that $f(a)=0$ and $f(bP(x))=f(P(x))$. Letting $(P(x),Q(x))=(bx,\frac{x}{b}+1)$ to obtain $f(x+b)=f(x+1)$ for all $b\ne 0$. Let $c,d$ be two arbitrary real numbers such that $cd\ne 0$ it follows that $$f\left(x+\frac{d}{c}\right)=f\left(c\left(x+\frac{d}{c}\right)\right)=f(cx+d)=f(x+1).$$ Finally, letting $(P(x),Q(x))=(cx,x-\frac{1}{c})$ for some $c\ne 0,1$ in the first equation yields $$f(x)=f(cx)=f\left(cx+1-\frac{1}{c}\right)=f(x+1).$$ Hence, if $P(x)$ is a linear polynomial then $f(P(x))=f(x+1)$, i.e., a constant function, say $C$. We then finish our proof based on the induction on the degree of polynomial. The base is true for $d=0,1$. Assume that the statement holds true for all polynomials of degree less than $d$. Choose a non-zero rational number $r$ such that $P(r)\ne 0$. Letting $R(x)=\frac{r}{P(r)}P(x+r)$ it follows that $R(0)=r$. Hence, $R(x)=r+xS(x)$, for some polynomial $S(x)$ of degree $d-1$. Putting $(P(x),Q(x))=(R(x),x-r)$ in the first equation yielding $$\begin{array}{rl}f(R(x)-r)=f(xS(x))=f(x)+f(S(x))& =f(R(x-r))\\ & =f\left(\frac{r}{P(r)}P(x)\right)=f(P(x)).\end{array}$$ According to the induction hypothesis $f(S(x))=C(d-1)$ and $f(x)=C$. Hence, $$f(P(x))=Cd.$$