Problem
NT-B2-M02-P009 One Class Modulo \(11\)
#9
★★★☆☆ Level 3 of 5
Find all \(n\) such that \(11\mid n^2+3n+5\).
Multiply the expression by \(4\) and complete the square.
We have \(4(n^2+3n+5)=(2n+3)^2+11\). Divisibility by \(11\) is equivalent to \((2n+3)^2\equiv0\pmod{11}\), so \(2n+3\equiv0\pmod{11}\). Since \(2^{-1}\equiv6\pmod{11}\), we get \(n\equiv-18\equiv4\pmod{11}\).
A good place to show completing the square modulo a prime.