, 2011

Let $r = u/v$ where $\gcd(u, v) = 1$. Then we get

$$\begin{align*} a_n u^n + a_{n-1} u^{n-1} v + \dots + a_1 u v^{n-1} + a_0 v^n &= 0, \\ b_n u^n + b_{n-1} u^{n-1} v + \dots + b_1 u v^{n-1} + b_0 v^n &= 0. \end{align*}$$

Subtraction gives
$$(a_n - b_n)u^n + (a_{n-2} - b_{n-2})u^{n-2}v^2 + \dots + (a_1 - b_1)u v^{n-1} + (a_0 - b_0)v^n = 0,$$

since $a_{n-1} = b_{n-1}$. This shows that $v$ divides $(a_n - b_n)u^n$ and hence it divides $a_n - b_n$. Since $a_n - b_n$ is a prime, either $v=1$ or $v = a_n - b_n$. Suppose the latter holds. The relation takes the form
$$u^n + (a_{n-2} - b_{n-2})u^{n-2}v + \dots + (a_1 - b_1)u v^{n-2} + (a_0 - b_0)v^{n-1} = 0.$$

(Here we have divided throughout by $v$.) If $n > 1$, this forces $v \mid u$, which is impossible since $\gcd(v, u) = 1$ ($v > 1$ since it is equal to the prime $a_n - b_n$). If $n = 1$, then we get two equations:

$$\begin{align*} a_1 u + a_0 v &= 0, \\ b_1 u + b_0 v &= 0. \end{align*}$$

This forces $a_1 b_0 - a_0 b_1 = 0$ contradicting $a_n b_0 - a_0 b_n \neq 0$. (Note: The condition $a_n b_0 - a_0 b_n \neq 0$ is extraneous. The condition $a_{n-1} = b_{n-1}$ forces that for $n=1$, we have $a_0 = b_0$. Thus we obtain, after subtraction
$(a_1 - b_1)u = 0.$
This implies that $u = 0$ and hence $r = 0$ is an integer.)