Example 22([39.4]) Find all positive integer pairs $\{a, b\}$, such that $a^{2} b+a+b$ is divisible by $a b^{2}+b+7$.


(i) First, analyze the size relationship between $a$ and $b$. Since it must be that $a^{2} b + a + b \geqslant a b^{2} + b + 7$, we have
$$a - 7 \geqslant a b (b - a).$$

This implies that (why)
$$a \geqslant b \geqslant 1.$$

We need to subtract a multiple of $a b^{2} + b + 7$ from $a^{2} b + a + b$ so that the absolute value of the remainder is intuitively smaller than $a b^{2} + b + 7$. This way, the remainder must be 0.
(ii) Use the division algorithm. Let
$$a = q b + r, \quad 0 \leqslant r < b.$$

From (1), we have
$$a b^{2} + b + 7 \mid a b r + b + r - 7 q.$$

Since $a b^{2} + b + 7 > a b r + b + r - 7 q$, we have
$$a b^{2} + b + 7 > |b - 7 q|.$$

From this and equation (2), we deduce that $b - 7 q = 0$, so the solution is
$$\{a, b\} = \{7 t^{2}, 7 t\}, \quad t \text{ is any positive integer.}$$
(iv) If $r > 0$, then $b > 1$. We will prove that there is no solution in this case.
(a) When $b = 2$, we must have $r = 1, a = 2 q + 1$. In this case, the right side of equation (1) becomes
$$8 q + 13 = a b^{2} + b + 7 \mid a b r + b + r - 7 q = 5 - 3 q.$$

Clearly, $5 - 3 q \neq 0$ and $8 q + 13 > |5 - 3 q|$. Therefore, the above equation cannot hold, so there is no solution.
(b) When $b \geqslant 3$, we have (why)
$$\begin{aligned}
a b^{2} + b + 7 & \geqslant a b (r + 1) + b + 7 = a b r + b + b^{2} q + b r + 7 \\
& > |a b r + b + r - 7 q|
\end{aligned}$$

Therefore, the right side of equation (1) holds if and only if
$$a b r + r + b - 7 q = 0.$$

However, $a b r + r + b > b^{2} q \geqslant 9 q > 7 q$, so the above equation cannot hold, and thus there is no solution.