Problem
NT-B1-M02-P011 When the GCD Is Greater Than One
#11
★★☆☆☆ Level 2 of 5
Find all integers \(n\) for which \(\gcd(n+2,n^2+3n+5)>1\).
Reduce the second expression modulo \(n+2\).
Modulo \(n+2\), we have \(n\equiv -2\). Then \(n^2+3n+5\equiv4-6+5=3\). Therefore \(\gcd(n+2,n^2+3n+5)=\gcd(n+2,3)\). It is greater than \(1\) exactly when \(3\mid n+2\), i.e. \(n\equiv1\pmod3\).
The problem requires not only bounding the GCD, but finding the exact condition.