Problem
NT-B1-M12-P008 Set 2. GCD Without Computation
#8
★★☆☆☆ Level 2 of 5
Find \(\gcd(n^2+n+1,n+1)\).
Replace \(n\) by \(-1\) modulo \(n+1\).
Modulo \(n+1\), \(n\equiv-1\). Then \(n^2+n+1\equiv1-1+1=1\). Hence the common divisor divides \(1\), so the GCD is \(1\).
Very common Euclidean move for expressions.