Problem 1702

1. Define the Polynomials and Rational Roots:
Let $f(x) = a_n x^n + \cdots + a_0$ and $g(x) = b_m x^m + \cdots + b_0$. We are given that for infinitely many primes $p$, the polynomial $pf + g$ has a rational root. Let $r_p = \frac{u_p}{v_p}$ be a rational root of $pf + g$ for some coprime integers $u_p$ and $v_p$.

2. Boundedness of Rational Roots:
Since $\deg g < \deg f$, there exists a constant $C$ such that for $|x| \geq C$, $\left| \frac{g(x)}{f(x)} \right| < 0.01$. Given $pf(r_p) + g(r_p) = 0$, we have $\left| \frac{g(r_p)}{f(r_p)} \right| = p$. Since $p > 0.01$, it follows that $|r_p| < C$.

3. Rational Root Theorem Application:
By the Rational Root Theorem, $v_p \mid pa_n$ and $u_p \mid pa_0 + b_0$. Therefore, for each valid $p$, one of the following must hold:
- $p \mid v_p$ (call such primes peasants), or
- $v_p \mid a_n$ (call such primes rebels).

4. Infinitely Many Peasant Primes:
Suppose finitely many valid $p$ are peasants. Let $S$ be the set of all valid rebel primes $p$. Since $v_p \mid a_n$, there are finitely many possible values for $v_p$. For each $v_p$, there are finitely many possible values for $u_p$ because $|r_p| = \left| \frac{u_p}{v_p} \right|$ is bounded. Thus, there are finitely many possible values for $r_p$ over all $p \in S$. This contradicts the infinitude of valid $p$ since $p = -\frac{g(r_p)}{f(r_p)}$.

5. Case Analysis:
Consider a sufficiently large peasant prime $p$ such that $p$ is larger than all $a_i$ and $b_i$. Since $r_p = \frac{u_p}{v_p}$, we have:
$$(pf + g)(r_p) = \sum_{i=0}^n pa_i \left( \frac{u_p}{v_p} \right)^i + \sum_{j=0}^m b_j \left( \frac{u_p}{v_p} \right)^j = 0$$
Multiplying through by $v_p^n$ gives:
$$\sum_{i=0}^n pa_i u_p^i v_p^{n-i} + \sum_{j=0}^m b_j u_p^j v_p^{n-j} = 0 \quad (\heartsuit)$$
Taking $(\heartsuit) \mod p^2$, since $p \mid v_p$, the first term reduces to $pa_n u_p^n \mod p^2$, and the second term reduces to $b_m u_p^m v_p^{n-m} \mod p^2$.

6. **Case 1: $n \geq m + 2$:**
$$b_m u_p^m v_p^{n-m} \equiv 0 \pmod{p^2}$$
Thus, $pa_n u_p^n \equiv 0 \pmod{p^2}$, implying $a_n u_p^n \equiv 0 \pmod{p}$. Since $p > a_n$ and $\gcd(u_p, v_p) = 1$, this is impossible.

7. **Case 2: $n = m + 1$:**
$$pa_n u_p^n + b_{n-1} u_p^{n-1} v_p \equiv 0 \pmod{p^2}$$
Since $p \nmid u_p$, we can divide and reduce:
$$a_n u_p + b_{n-1} \left( \frac{v_p}{p} \right) \equiv 0 \pmod{p}$$
Let $d_p = \frac{v_p}{p}$. Then:
$$\frac{u_p}{d_p} \equiv -\frac{b_{n-1}}{a_n} \pmod{p}$$
Hence, $\frac{u_p}{d_p} = C + \frac{pZ_p}{a_n}$ for some integer $Z_p$. Since $r_p = \frac{u_p}{v_p} = \frac{1}{p} \frac{u_p}{d_p}$, we have:
$$r_p = \frac{C}{p} + \frac{Z_p}{a_n}$$
Since $|r_p|$ is bounded, there exists some integer $Z$ such that $Z_p = Z$ for infinitely many $p$. Thus:
$$r_p = \frac{C}{p} + \frac{Z}{a_n}$$
As $p \to \infty$, $r_p \to \frac{Z}{a_n}$, a fixed rational number.

8. Contradiction:
Since $p = -\frac{g(r_p)}{f(r_p)}$ and $r_p \to \frac{Z}{a_n}$, but $f$ has no rational roots, we have $f\left( \frac{Z}{a_n} \right) \neq 0$. Thus:
$$-\frac{g(r_p)}{f(r_p)} \to -\frac{g\left( \frac{Z}{a_n} \right)}{f\left( \frac{Z}{a_n} \right)}$$
This is a fixed number, contradicting the infinitude of valid $p$.

Therefore, $f$ must have a rational root.

$\blacksquare$