Problem
NT-B2-M07-P019 A Cubic Against a Quadratic
#19
★★★★☆ Level 4 of 5
Find all positive integers \(n\) such that \(n^2+3n+3\mid n^3-1\).
Reduce \(n^3-1\) modulo \(n^2+3n+3\), then compare sizes.
Let \(D=n^2+3n+3\). From \(n^2\equiv -3n-3\pmod D\), we get \(n^3\equiv -3n^2-3n\equiv 9n+9-3n=6n+9\pmod D\). Thus \(n^3-1\equiv 6n+8\pmod D\), so \(D\mid 6n+8\). For \(n\ge 5\), \(D=n^2+3n+3>6n+8\), impossible. Checking \(n=1,2,3,4\), only \(n=1\) works, since \(7\mid 0\). The answer is \(n=1\).
Method: degree reduction modulo a variable divisor followed by a final bound.