Find the number of pairs $$(a, b)$$ of natural numbers such that $$b$$ is a 3-digit number, $$a+1$$ divides $$b-1$$ and $$b$$ divides $$a^2+a+2$$.

Put $$d = a+1$$, so $$a^2+a+2 = d^2-d+2$$ and the conditions say $$b \equiv 1 \pmod d$$ with $$b \mid d^2-d+2$$. Writing $$d^2-d+2 = bk$$ and reducing modulo $$d$$ gives $$k \equiv 2 \pmod d$$, and any $$k \ge d+2$$ makes $$b$$ too small, so $$k = 2$$ and $$b = \frac{d^2-d+2}{2}$$, which satisfies $$b \equiv 1 \pmod d$$ only for odd $$d$$. Requiring $$100 \le b \le 999$$ gives the odd values $$d = 15, 17, \ldots, 45$$, so there are 16 pairs.