NT-B2-M01-P009 Coprime Polynomials

Put \(m=n+2\). The expressions become more symmetric.

Let \(m=n+2\). Then \(n^2+3n+3=m^2-m+1\), and \(n^2+5n+7=m^2+m+1\). A common divisor divides the difference \(2m\). But \(m^2-m+1\) is coprime to \(m\), because modulo \(m\) it has remainder \(1\). Hence the common divisor can only divide \(2\). Since \(m^2-m+1\) is always odd, the common divisor is not \(2\). Therefore the gcd is \(1\).

The change of variable is not mandatory, but it reveals the structure.