Iranian Mathematical Olympiad

We shall firstly prove that the set of integer roots of the $(P + nQ)_{n=1,...}$ would be unbounded. For sake of this, notice that if $m \ne n$ the polynomials $P + mQ$, $P + nQ$ have no common integer root; otherwise $P$ and $Q$ would have. Now, let $r_n$ be an integer root of $P + nQ$. It follows that $P(r_n)/Q(r_n) = -n$. Divide $P$ by $Q$, then, there would be polynomials $T$ and $R$ such that $P(x) = T(x)Q(x) + R(x)$, $\deg R(x) < \deg Q(x)$ and $T(x)$, $R(x)$ would be of integer coefficients. Plugging $x = r_n$ yielding $T(r_n) + R(r_n)/Q(r_n) = -n$. Thus, $R(r_n)/Q(r_n)$ would be an integer. Since

we can assign a unique integer root to each of $P(x) + nQ(x)$, we can find an arbitrary large root $r_n$. But, the degree condition on $R$ implies that $|R(r_n)/Q(r_n)| < 1$ for all large enough $r_n$. Hence, it must be zero and therefore, $R(x)/Q(x) = 0$ for infinitely many $x$. Yielding $R(x) = 0$ and hence, $P(x)$ would be divisible by $Q$.

Let $P(x) = Q(x)S(x)$ for some polynomial $S(x)$ with integer coefficients. It follows that $P+nQ = Q(S+n)$ and $r_n$ would be an integer root of $S(x)+n$. It follows that the line $y = -n$ must cut the graph of $y = S(x)$ at integer points. This would be a well-known result that this can only happen if and only if $S(x) = Ax + B$ for some integers $A, B$.