Example 5 Find a pair of positive integers $a, b$ that satisfy: (1) $ab(a+b)$ is not divisible by 7; (2) $7^{7} \mid \left[(a+b)^{7}-a^{7}-b^{7}\right]$.
(25th IMO)
Explanation: Substitution
$$
(x, y, z)=(a+b,-a,-b) \text {. }
$$

Then $x+y+z=0$, and
$$
\begin{array}{l}
\left(\sigma_{1}, \sigma_{2}, \sigma_{3}\right)=(0, x y+y z+z x, x y z) . \\
\text { Let } S_{n}=x^{n}+y^{n}+z^{n}(n \in \mathbf{N}) \text {. Then } \\
S_{0}=3, S_{1}=0, S_{2}=-2 \sigma_{2}, \\
S_{n+3}=-\sigma_{2} S_{n+1}+\sigma_{3} S_{n}(n \in \mathbf{N}) .
\end{array}
$$

Calculations yield
$$
\begin{array}{l}
S_{3}=-\sigma_{2} S_{1}+\sigma_{3} S_{0}=3 \sigma_{3}, \\
S_{4}=-\sigma_{2} S_{2}+\sigma_{3} S_{1}=2 \sigma_{2}^{2}, \\
S_{5}=-\sigma_{2} S_{3}+\sigma_{3} S_{2}=-3 \sigma_{2} \sigma_{3}-2 \sigma_{2} \sigma_{3} \\
=-5 \sigma_{2} \sigma_{3}, \\
S_{7}=-\sigma_{2} S_{5}+\sigma_{3} S_{4}=5 \sigma_{2}^{2} \sigma_{3}+2 \sigma_{2}^{2} \sigma_{3} \\
=7 \sigma_{2}^{2} \sigma_{3} . \\
\text { Also, }\left(\sigma_{1}, \sigma_{2}, \sigma_{3}\right) \\
=\left(0,-a^{2}-a b-b^{2}, a b(a+b)\right),
\end{array}
$$

Thus, $(a+b)^{7}-a^{7}-b^{7}$
$$
=7 a b(a+b)\left(a^{2}+a b+b^{2}\right)^{2} \text {. }
$$

From (1) and (2), we have $7^{3} I\left(a^{2}+a b+b^{2}\right)$.
Taking $a^{2}+a b+b^{2}=7^{3}=343$, we get
$$
(a+b)^{2}-a b=343 \text {. }
$$

Since $343=(a+b)^{2}-a b$
$$
=a^{2}+a b+b^{2} \geqslant 3 a b,
$$

we have $a b \leqslant 114$, and $(a+b)^{2}=a b+343$.
Thus, $343 \leqslant(a+b)^{2} \leqslant 343+114=457$, which means $19 \leqslant a+b \leqslant 21$.
Solving, we get $(a, b)=(18,1),(1,18)$.