Iranian Mathematical Olympiad

Take any polynomial $Q(x)$ with rational coefficients then the function $f$ meets the condition of the problem if and only if $g(P) = f(P) - P \cdot Q$ satisfies. Where $P(x)$ is a polynomial with rational coefficients. We shall prove this claim in one direction. In the opposite direction, just change $Q$ by $-Q$. Indeed, $g(P+R) = f(P+R) - (P+R) \cdot Q = f(P) - P \cdot Q + f(R) - R \cdot Q = g(P) + g(R)$. The second property would indeed follow from the Euclidean algorithm. Since $\text{gcd}(P, f(P)) = 1$ then $\text{gcd}(P, f(P) - P \cdot Q) = 1$.

We can then assume that $f(1) = 0$. After this, it suffices to show that $f(P) = C \cdot P'$ satisfies the statement of problem for some non-zero rational $C$.

Notice that $f(nP) = nf(P)$ for all integers $n$ and further for all rational numbers $r$ we have $f(rP) = rf(P)$. Then, if we put $P(x) = a_d x^d + \dots + a_0$ it follows that $f(P) = a_d f(x^d) + \dots + a_0 f(1) = a_d f(x^d) + \dots + a_1 f(x)$. Thus, it would only suffices to prove that $f(x^n) = n x^{n-1} f(x)$. For $n = 1$ it is clear. Now consider $(x-r)^n$ for some rational number $r$ and some integer $n \ge 2$. Notice that

$$\begin{align*} f((x-r)^n) &= \sum_{k=0}^{n} \binom{n}{k} f(x^k) (-r)^{n-k} = f(x^n) + \sum_{k=0}^{n-1} \binom{n}{k} k x^{k-1} f(x) (-r)^{n-k} \\ &= f(x^n) + n f(x) \sum_{k=1}^{n-1} \binom{n-1}{k-1} x^{k-1} (-r)^{n-k} \\ &= f(x^n) + n f(x) + \sum_{k=0}^{n-2} \binom{n-1}{k} x^k (-r)^{n-k-1}. \end{align*}$$
$$Hence,$$
f((x - r)^n) = f(x^n) + n f(x) ((x - r)^{n-1} - x^{n-1})
$$$$
That is, $f((x-r)^n) - n f(x)(x-r)^{n-1} = f(x^n) - n f(x)x^{n-1}$. Applying the second condition, it follows that $x-r$ divides the left. Hence, it should divide the right. Since the right side doesn't depend on $r$, it follows that it must have infinitely many linear factors implies that it must be zero. Therefore, $f(x^n) = n f(x)x^{n-1}$. Now, it follows that $f(P) = f(x) P$. We shall then need to prove that $f(x) = C$ for some non-zero rational number $C$. Assume on the contrary, let $f(x) = Q(x)$ for some non-constant polynomial $Q(x)$ with rational coefficients, then consider an irreducible factor $T(x)$ of it with rational coefficients then $T(x)$ divides $f(x)$. Hence, $T(x)$ divides $f(T) = Q(x) \cdot T'(x)$. But, $T(x)$ since $T(x)$ is irreducible, it would also be square-free. Yielding $\text{gcd}(T, f(T)) = 1$. Hence, $f(x) = C$. All in all, the solutions are of the form $C \cdot P'(x) + P(x)Q(x)$, for some non-zero rational number $C$ and any polynomial $Q(x)$ with rational coefficients.