Problem
NT-B1-M02-P009 Quadratic and \(n\)
#9
★★☆☆☆ Level 2 of 5
Prove that \(\gcd(n^2+n+1,n)=1\) for every positive \(n\).
Subtract \(n(n+1)\) from \(n^2+n+1\).
If \(d\mid n\) and \(d\mid n^2+n+1\), then \(d\mid n(n+1)\). Hence \(d\) divides the difference \((n^2+n+1)-n(n+1)=1\). Thus \(d=1\).
Shows how to eliminate higher-degree terms.