Problem
NT-B1-M04-P014 Divisibility by \(2n+1\)
#14
★★★☆☆ Level 3 of 5
Find all positive \(n\) such that \(2n+1\mid n^2+n+7\).
Multiply the expression by \(4\).
If \(2n+1\mid n^2+n+7\), then \(2n+1\mid4(n^2+n+7)\). But \(4(n^2+n+7)=(2n+1)^2+27\). Hence \(2n+1\mid27\). For positive \(n\), \(2n+1=3,9,27\), so \(n=1,4,13\). All three work.
A problem from the module structure: the congruence must be discovered.