Problem
NT-B1-M11-P016 Divisor \(n+2\)
#16
★★★☆☆ Level 3 of 5
Find all positive integers \(n\) such that \(n+2\mid n^2+5\).
Modulo \(n+2\), \(n\) equals \(-2\).
We have \(n^2+5\equiv4+5=9\pmod{n+2}\). Hence \(n+2\mid9\). Since \(n\ge1\), \(n+2\ge3\), so \(n+2=3\) or \(9\). Thus \(n=1\) or \(n=7\). Both work.
Very useful template for parameter divisibility.