4. Let positive integers $a, b$ satisfy $1 \leqslant a < b \leqslant 100$. If there exists a positive integer $k$, such that $a b \mid\left(a^{k}+b^{k}\right)$, then the pair $(a, b)$ is called a "good pair". Find the number of all good pairs.
（Xiong Bin's problem）
4. Let $(a, b)=d, a=s d, b=t d,(s, t)=1(t>s)$.

Thus, $s t d^{2} \mid d^{k}\left(s^{k}+t^{k}\right)$.
Since $\left(s t, s^{k}+t^{k}\right)=1$, it follows that $s t \mid d^{k}$.
Therefore, all prime factors of $st$ can divide $d$.
If $s$ or $t$ has a prime factor $p$ not less than 11, then $p \mid d$. Thus, $p^{2} \mid a$ or $p^{2} \mid b$.
Since $p^{2}>100$, this is a contradiction.
Therefore, the prime factors of $st$ can only be $2,3,5,7$.
If $st$ has at least three of the prime factors $2,3,5,7$, then
$d \geqslant 2 \times 3 \times 5=30$.
Thus, $\max \{a, b\} \geqslant 5 d>100$, which is a contradiction.
If the set of prime factors of $st$ is $\{3,7\}$, then
$\max \{a, b\} \geqslant 7 \times 3 \times 7>100$,

which is a contradiction;
Similarly, the set of prime factors of $st$ cannot be $\{5,7\}$.
If the set of prime factors of $st$ is $\{3,5\}$, $d$ can only be 15, at this time, $s=3, t=5$, so, $(a, b)=(45,75)$. There is 1 good pair.

If the set of prime factors of $st$ is $\{2,7\}$, $d$ can only be 14, at this time, $(s, t)=(2,7),(4,7)$, so, $(a, b)=$ $(28,98),(56,98)$. There are 2 good pairs.
If the set of prime factors of $st$ is $\{2,5\}$, $d$ can only be 10,20.
When $d=10$,
$(s, t)=(2,5),(1,10),(4,5),(5,8)$;
When $d=20$, $(s, t)=(2,5),(4,5)$.
There are 6 good pairs.
If the set of prime factors of $st$ is $\{2,3\}$, $d$ can only be 6, 12, 18, 24, 30.
When $d=6$,
\[
\begin{aligned}
(s, t)= & (1,6),(1,12),(2,3),(2,9), \\
& (3,4),(3,8),(3,16),(4,9), \\
& (8,9),(9,16) ;
\end{aligned}
\]

When $d=12$,
$(s, t)=(1,6),(2,3),(3,4),(3,8)$;
When $d=18$, $(s, t)=(2,3),(3,4)$;
When $d=24$, $(s, t)=(2,3),(3,4)$;
When $d=30$, $(s, t)=(2,3)$.
There are 19 good pairs.
If the set of prime factors of $st$ is $\{7\}$, then $(s, t)=$ $(1,7)$, $d$ can only be 7,14. There are 2 good pairs.

If the set of prime factors of $st$ is $\{5\}$, then $(s, t)=$ $(1,5)$, $d$ can only be $5,10,15,20$. There are 4 good pairs.
If the set of prime factors of $st$ is $\{3\}$, then
When $(s, t)=(1,3)$, $d$ can only be $3,6, \cdots, 33$;
When $(s, t)=(1,9)$, $d$ can only be $3,6,9$;
When $(s, t)=(1,27)$, $d$ can only be 3.
There are 15 good pairs.
If the set of prime factors of $st$ is $\{2\}$, then
When $(s, t)=(1,2)$, $d$ can only be $2,4, \cdots, 50$;
When $(s, t)=(1,4)$, $d$ can only be $2,4, \cdots, 24$;
When $(s, t)=(1,8)$, $d$ can only be $2,4, \cdots, 12$;
When $(s, t)=(1,16)$, $d$ can only be $2,4,6$;
When $(s, t)=(1,32)$, $d$ can only be 2.
There are 47 good pairs.
Therefore, the total number of good pairs is
$1+2+6+19+2+4+15+47=96$ (pairs).