Problem
NT-B2-M07-P010 The Divisor \(2n+1\)
#10
★★★☆☆ Level 3 of 5
Find all positive integers \(n\) such that \(2n+1\mid n^2+n+10\).
Multiply \(n^2+n+10\) by \(4\) so that \((2n+1)^2\) appears.
If \(2n+1\mid n^2+n+10\), then \(2n+1\mid 4(n^2+n+10)\). But \(4n^2+4n+40=(2n+1)^2+39\). Hence \(2n+1\mid 39\). Since \(2n+1>1\) and is odd, \(2n+1=3,13,39\). Thus \(n=1,6,19\), and all three work.
The key move is choosing the multiplier \(4\) to avoid fractional coefficients.