AMC 12 Mock Exam 5 Problem 16

Let $d=\gcd(n^2+1,\,n^2+2n+2)$. Then $d$ divides the difference $2n+1$. It also divides $4(n^2+1)-(2n+1)(2n-1)=5$, so $d\in\{1,5\}$.

If $5\mid 2n+1$, then $n\equiv2\pmod5$, and $n^2+1\equiv5\equiv0$ and $(n+1)^2+1\equiv10\equiv0\pmod5$. So $d=5$. Conversely, $d=5$ forces $5\mid2n+1$. Hence $d>1$ exactly when $n\equiv2\pmod5$.

The values $n=2,7,12,\dots,2022$ number $\frac{2022-2}{5}+1=\boxed{405}$.