Question

Consider two polynomials

$P\left(x\right)=a_n{x}^n+a_{n-1}{x}^{n-1}+\cdots +a_{1}x+a_0$, and

$Q\left(x\right)=b_n{x}^n+b_{n-1}{x}^{n-1}+\cdots +b_{1}x+b_0$

with integer coefficients such that $a_n-b_n$ is a prime, $a_{n-1}=b_{n-1}$ and $a_n{b}_0-a_0{b}_n\ne 0$.

Suppose there exists a rational number $r$ such that $P\left(r\right)=Q\left(r\right)=0$. Prove that $r$ is an integer.

Answer 1

Let $r=u/v$ where $gcd(u,v)=1$.
Then we get $a_n{u}^n+a_{n-1}un-1v+\cdots +a_{1}u{v}^{n-1}+a_0{v}^n=0$,
$b_n{u}^n+b_{n-1}un-1v+\cdots +b_{1}u{v}^{n-1}+b_0{v}^n=0$.
Substraction 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 $v=1orv=a_n-b_n$.
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 through-out by $v$.) If $n>1$, this forces $v|u$, which is impossible since $gcd(v,u)=1$ $(v>1sinceitisequaltotheprime{a}_n-b_n)$.
If $n=1$, then we get two equations: $a_{1}u+a_{0}v=0$, $b_{1}u+b_{0}v=0$.
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 substraction $(a_1-b_1)u=0$.
This implies that $u=0$ and hence $r=0$ is an integer.)