fun fact: i did the (a) one in a hard way lol

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are the solutions

(a) $n=1\Rightarrow 1(1+2022)+2=2025=45^2$ (b) $a=1\Rightarrow n(n+1)+2$, because we have $n=1$ is a perfect square, this is not satisfy $a=2\Rightarrow n(n+2)+2$, by bounding between squares we have $n^2<n(n+2)+2<(n+1)^2$, so for $a=2$ it cannot be a perfect square $a=3 \Rightarrow n(n+3)+2$, by bounding between squares we have $(n+1)^2<n(n+2)+2<(n+2)^2$, so for $a=3$ it cannot be a perfect square $a=4k \Rightarrow n(n+4k)+2\equiv 0,1+2\equiv 2,3$ (mod 4), so for $a=4k$ it cannot be a perfect square For $a\geq 5, a \not\equiv 0$ (mod 4), let $a=4k+m, 1\leq m\leq 3$ then we have $n(n+a)+2=n^2+4kn+mn+2\equiv 0,1,2,3$ (mod 4), so it could be a perfect square $a=\{2,3,4k\}$