Problem
NT-B1-M11-P023 Divisor \(2n+1\)
#23
★★★★☆ Level 4 of 5
Find all positive integers \(n\) such that \(2n+1\mid n^2+n+3\).
Multiply the expression by \(4\).
If \(2n+1\mid n^2+n+3\), then \(2n+1\mid4(n^2+n+3)\). But \(4(n^2+n+3)=(2n+1)^2+11\). Hence \(2n+1\mid11\). For \(n\ge1\), \(2n+1\ge3\), so \(2n+1=11\), \(n=5\). Check: \(11\mid33\).
Two-move problem: guess the factor \(4\), then bound the divisor.