Problem
NT-B1-M01-P016 Divisibility of a Quadratic Expression
#16
★★★☆☆ Level 3 of 5
Find all positive integers \(n\) such that \(2n+1\mid n^2+n+3\).
Multiply the expression by \(4\) so that \((2n+1)^2\) appears.
If \(2n+1\mid n^2+n+3\), then \(2n+1\mid 4(n^2+n+3)\). But \(4(n^2+n+3)=4n^2+4n+12=(2n+1)^2+11\). Hence \(2n+1\mid 11\). Since \(n\) is positive, \(2n+1\ge3\), so \(2n+1=11\), giving \(n=5\). Check: \(n^2+n+3=33\) and \(2n+1=11\), so it works.
A good problem on turning a variable divisor into a divisor of a constant.