Problem
NT-B1-M01-P009 A Divisor with a Parameter
#9
★★☆☆☆ Level 2 of 5
Find all integers \(n\) such that \(n+3\mid n^2+n+1\).
Modulo \(n+3\), replace \(n\) by \(-3\).
We have \(n\equiv -3\pmod{n+3}\). Hence \(n^2+n+1\equiv 9-3+1=7\pmod{n+3}\). Therefore \(n+3\mid7\). Thus \(n+3=\pm1,\pm7\), so \(n=-2,-4,4,-10\). Substitution confirms all four values.
A standard technique of reducing modulo a variable divisor.