$2 \cdot 26$ Find a pair of positive integers $a, b$, satisfying:
(1) $ab(a+b)$ is not divisible by 7;
(2) $(a+b)^{7}-a^{7}-b^{7}$ is divisible by $7^{7}$. Verify your answer.
(25th International Mathematical Olympiad, 1984)
[Solution 1] Let positive integers $a, b$ satisfy the problem's requirements, and set $b=1$. Since
$$
(a+b)^{7}-a^{7}-b^{7}=7 a b(a+b)\left(a^{2}+a b+b^{2}\right)^{2} .
$$

Because $7 \nmid a b(a+b)$, it follows that $7^{3} \mid a^{2}+a b+b^{2}$.
To find such an $a$, we need $7^{3} \mid a^{2}+a+1$.
Let $a=7 k+r, r=0, \pm 1, \pm 2, \pm 3$.
Since $7 \nmid a(a+1)$, then $r \neq 0,-1$.
$a^{2}+a+1=49 k^{2}+7 k(2 r+1)+r^{2}+r+1$.
It is easy to see that when $r=-3$, $7 \mid r^{2}+r+1$, thus $7 \mid a^{2}+a+1$.
At this point, $a^{2}+a+1=7\left(7 k^{2}-5 k+1\right), a=7 k-3$.
Now let $k=7 h+r_{1}, \quad r_{1}=0, \pm 1, \pm 2, \pm 3$, then
$$
\begin{aligned}
& 7 k^{2}-5 k+1 \\
= & 7\left(49 h^{2}+14 h r_{1}\right)-35 h+7 r_{1}^{2}-5 r_{1}+1 .
\end{aligned}
$$

It is easy to see that when $r_{1}=3$, $71-5 r_{1}+1$, thus $7 \mid 7 k^{2}-5 k+1$.
At this point, $7 k^{2}-5 k+1=7\left(7^{2} h^{2}+37 h+7\right), \quad k=7 h+3$.
To ensure $7 \mid 7^{2} h^{2}+37 h+7$, we only need $7 \mid h$, let $h=7 t$, then we must have
$7\left|7^{2} h^{2}+37 h+7, \quad 7^{2}\right| 7 k^{2}-5 k+1,7^{3} \mid a^{2}+a+1$, at this point $a=7^{3} t+18, \quad b=1$.
In particular, when $t=0$, $a=18, \quad b=1$ is a solution.
[Solution 2] From $7^{7} \mid (a+b)^{7}-a^{7}-b^{7}=7 a b(a+b)\left(a^{2}+a b+b^{2}\right)^{2}$, and
$$
\begin{array}{l}
7 \nmid a b(a+b), \\
7^{3} \mid a^{2}+a b+b^{2} .
\end{array}
$$

We know that there should be
Next, we find the positive integer pairs $(a, b)$ that satisfy the equation $a^{2}+a b+b^{2}=7^{3}=343$,
Since $(a+b)^{2}-a b=a^{2}+a b+b^{2} \geqslant 3 a b$, then
$$
\begin{aligned}
3 a b & \leqslant 343, \\
a b & \leqslant 114 .
\end{aligned}
$$

Thus,
$$
343 \leqslant(a+b)^{2} \leqslant 343+114=457 \text {. }
$$

Solving this, we get
$$
19 \leqslant a+b \leqslant 21 \text {. }
$$

Let $a+b=19$, at this point we get $a b=18$, solving this gives $a=18, b=1$, which is easily verified to be a pair of positive integers that satisfy the problem's conditions.