Problem
ALG-B1-M04-P011 Two Permuted Trinomials
#11
★★★★★ Level 5 of 5
Integers \(a,b,c\) are such that the values of \(b x^2+c x+a\) and \(c x^2+a x+b\) at \(x=1000\) are equal. Can the first trinomial take the value \(2024\) at \(x=1\)?
Inspired by regional olympiad method · 2010 · Grade 11 · Problem 7
Use the equality at \(1000\) to express \(a\) through \(b,c\), then inspect the value at \(1\).
The condition gives \(1000^2b+1000c+a=1000^2c+1000a+b\). Moving terms and dividing by \(999\), we get \(1001b-1000c-a=0\), so \(a=1001b-1000c\).
The value of the first trinomial at \(1\) is \(a+b+c=1002b-999c=3(334b-333c)\). It is always divisible by \(3\). The number \(2024\) is not divisible by \(3\), so this is impossible.
Architecture preserved: equality at a distant point gives a linear relation between coefficients, then divisibility appears.