[Example 1.6.1] Find a pair of positive integers $a, b$, satisfying:
(1) $a b(a+b)$ is not divisible by 7;
(2) $(a+b)^{7}-a^{7}-b^{7}$ is divisible by $7^{7}$.
We first simplify $(a+b)^{7}-a^{7}-b^{7}$.
$$
\begin{aligned}
& (a+b)^{7}-a^{7}-b^{7} \\
= & 7 a^{6} b+21 a^{5} b^{2}+35 a^{4} b^{3}+35 a^{3} b^{4}+21 a^{2} b^{5}+7 a b^{6} \\
= & 7 a b\left(a^{5}+3 a^{4} b+5 a^{3} b^{2}+5 a^{2} b^{3}+3 a b^{4}+b^{5}\right) \\
= & 7 a b(a+b)\left(a^{4}+b^{4}+3 a^{2} b^{2}+2 a^{3} b+2 a b^{3}\right) \\
= & 7 a b(a+b)\left(a^{2}+a b+b^{2}\right)^{2} .
\end{aligned}
$$

From (1), we know $7+a b(a+b)$, so we only need to find $a, b \in \mathbf{N}^{*}$, such that
$$
7^{6} \mid\left(a^{2}+a b+b^{2}\right)^{2},
$$

which means
$$
7^{3} \mid a^{2}+a b+b^{2} \text {. }
$$

At this point, the expression $a^{2}+a b+b^{2}$ involves two variables, so we might as well assign a special value to one of them. For example, let $b=1$, then the problem is transformed into finding a positive integer $a$ such that
$$
7^{3} \mid a^{2}+a+1 \text {. }
$$

Now we can focus on (1), and concentrate on finding the positive integer $a$.
Let $a=7 k+r, r \in\{-3,-2,-1,0,1,2,3\}$, then
$$
a^{2}+a+1=49 k^{2}+7 k(2 r+1)+r^{2}+r+1 \text {, }
$$

Thus, the problem is transformed into finding $r$ such that $r^{2}+r+1$ is divisible by 7. It is easy to verify that when $r=-3$, $r^{2}+r+1=(-3)^{2}-3+1=7$ is divisible by 7, at this time
$$
a^{2}+a+1=7\left(7 k^{2}-5 k+1\right) .
$$

The problem is then transformed into finding $k$ such that
$$
7^{2} \mid 7 k^{2}-5 k+1 .
$$

Let $k=7 h+r_{1}, r_{1} \in\{-3,-2,-1,0,1,2,3\}$, then
$$
7 k^{2}-5 k+1=7\left(49 h^{2}+14 h r_{1}\right)-7 \times 5 h+7 r_{1}^{2}-5 r_{1}+1 .
$$

Upon verification, when $r_{1}=3$, $-5 r_{1}+1=-14$ is divisible by 7,
at this time $k=7 h+3$,
$$
7 k^{2}-5 k+1=7\left(7^{2} h^{2}+37 h+7\right) .
$$

The problem is then transformed into finding $h$ such that
$$
7 \mid 7^{2} h^{2}+37 h+7 \text {. }
$$

Since
$$
7 \mid 7^{2} h^{2}+7,
$$

it is sufficient that $7 \mid h$, so let $h=7 l$.
At this time
$$
\begin{array}{l}
k=49 l+3, \\
a=7^{3} l+18 .
\end{array}
$$

Then assign a special value to $l$, let $l=0$, then $a=18$.
Therefore, $a=18, b=1$ is a pair of positive integer solutions that meet the requirements of the problem.
In fact, for the requirements of the problem, solving $a=18, b=1$ is enough, in fact, when $b=1$, $a=7^{3} l+18$ (where $l$ is an integer) are all solutions that meet the conditions, if we do not set $b=1$, we can also find $b$ and $a=7^{3} l+18 b$ as well as $7+b$ and $a=7^{3} l+324 b$ (where $b, l$ are integers) are solutions that meet the conditions.

In the process of solving this problem, setting $b=1$ played a key role, the substitution of special values led to a special solution method for a general problem.