Romania — NMO Selection Tests for the Balkan and International Mathematical Olympiads

Since $\deg f > \deg g$, $|g(z)/f(z)| < 1$ for all complex numbers $z$ of large enough absolute value. Consequently, as $p$ runs through the infinite set of primes under consideration, the roots of the $pf + g$ all lie in some disc $|z| < R$, where $R$ does not depend on $p$; for if $|z|$ is large enough, then
$$|pf(z) + g(z)| \ge |f(z)|(p - |g(z)/f(z)|) > 0.$$
For each prime $p$ such that $pf + g$ has a rational root, by Gauss' lemma, $pf + g$ is the product of two integral polynomials, one of degree 1 and the other of degree $\deg f - 1$. Now the condition that $p$ be prime comes in to imply that the leading coefficient of one of these factors is a divisor of the leading coefficient of $f$.
For each such prime $p$, it is therefore possible to choose an integral factor $h_p$ of $pf+g$, $\deg h_p = 1$ or $\deg h_p = \deg f - 1$, whose leading coefficient divides the leading coefficient of $f$. Hence the leading coefficients of the $h_p$ form a bounded set. Since the roots of the $h_p$ all lie in the same disc, $|z| < R$, Vieta's relations imply that the coefficients of the $h_p$ all form a bounded set; and since they are all integral, this set is finite, so $h_p = h$ for infinitely many of these primes $p$.
Finally, if $p$ and $q$ are two such, then $h$ is a divisor of $(p-q)f$. Since $\deg h_p = 1$ or $\deg h_p = \deg f - 1$, the conclusion follows.