The integers $ n^{2}-n+1$ and $ n^{2}-1$ lie in $ (n(n-1),n^{2})$ and are relatively prime to $ n$. Any common divisor between these two divides their difference, $ n-2$, and then must also divide $ n^{2}-1-(n+2)(n-2)=3$. So if $ n\not\equiv 2\pmod{3}$, $ n^{2}-n+1$ and $ n^{2}-1$ are relatively prime. In the case that $ n\equiv 2\pmod{3}$, $ n(n-1)$ is not divisible by 3, so $ n^{2}-n+3$ is relatively prime to $ n$, and it too lies in the interval $ (n(n-1),n^{2})$. But then $ n^{2}-n+1$ and $ n^{2}-n+3$ are relatively prime, since any common divisor between them must divide their difference, 2, but $ n^{2}-n+1$ is always odd.