Indija mo 2011

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

$$\begin{array}{rl}{a}_n{u}^n+a_{n-1}{u}^{n-1}v+\cdots +a_{1}u{v}^{n-1}+a_0{v}^n& =0,\\ b_n{u}^n+b_{n-1}{u}^{n-1}v+\cdots +b_{1}u{v}^{n-1}+b_0{v}^n& =0.\end{array}$$

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

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 $$u^n+(a_{n-2}-b_{n-2})u^{n-2}v+\cdots +(a_1-b_1)u{v}^{n-2}+(a_0-b_0)v^{n-1}=0.$$

(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:

$$\begin{array}{rl}{a}_{1}u+a_{0}v& =0,\\ b_{1}u+b_{0}v& =0.\end{array}$$

This forces $a_1{b}_0-a_0{b}_1=0$ contradicting $a_n{b}_0-a_0{b}_n\ne 0$. (Note: The condition $a_n{b}_0-a_0{b}_n\ne 0$ 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.)