Let $P_1(x) = x^2 + a_1x + b_1$ and $P_2(x) = x^2 + a_2x + b_2$ be two quadratic polynomials with integer coeffcients. Suppose $a_1 \ne a_2$ and there exist integers $m \ne n$ such that $P_1(m) = P_2(n), P_2(m) = P_1(n)$. Prove that $a_1 - a_2$ is even.
Problem
Source: CRMO 2015 region 1 p2
Tags: algebra, polynomial, Integer Polynomial, Even
khan.academy
30.09.2018 06:58
$$m(a_1-a_2)=n(a_2-a_1)+2(b_2-b_1) \implies (m+n)(a_1-a_2)=2(b_2-b_1)$$Since, $m \not = n$, we can span them over integers such that $m+n$ is not even.So, $2|(a_1-a_2)$, and we are done.
math_and_me
30.09.2018 09:10
khan.academy wrote: $$m(a_1-a_2)=n(a_2-a_1)+2(b_2-b_1) \implies (m+n)(a_1-a_2)=2(b_2-b_1)$$Since, $m \not = n$, we can span them over integers such that $m+n$ is not even.So, $2|(a_1-a_2)$, and we are done. $m$ and $n$ are arbitrary constants. You cannot vary them over integers as you did in your solution.
TheDarkPrince
30.09.2018 09:18
We have \begin{align*} m^2+ma_1 + b_1 &= n^2 + na_2 + b_2\\ m^2+ma_2+b_2 &= n^2 + na_1 + b_1.\\ \end{align*} We have \[m^2-n^2 + (m-n)a_1 = n^2 - m^2 + (n-m)a_2 \implies m+n+a_1 = - (m+n+a_2) \implies a_1 + a_2 = - 2(m+n).\]Thus $a_1+a_2$ is even giving $a_1-a_2$ is also even.
NTistrulove
25.01.2022 18:08
Why isn't anyone doing the most obvious steps? Anyway here's my solution,
Since $P_1(m)=P_2(n), P_1(n)=P_2(m)$, we have
\[P_1(m)-P_1(n)=P_2(n)-P_2(m)\]\[a_1(m-n)=a_2(n-m)\]\[(a_1+a_2)(m-n)=0\]Since $m\neq n$, we have $a_1=-a_2$ and $a_1-a_2=2a_1$.
