China Southeastern Mathematical Olympiad

Let $(a,b)=d$, $a=sd$, $b=td$, $(s,t)=1$, $t>1$, then $st{d}^2=d^k(s^k+t^k)$. So $k\ge 2$ and $st=d^{k-2}(s^k+t^k)$. Since $(st,s^k+t^k)=1$, we have $st=d^{k-2}$. Therefore, any prime factor of $st$ can be divided by $d$. If there is a prime factor $p$ of $s$ or $t$ no less than $11$, then $p$ divides $d$. So $p^2$ divides $a$ or $p^2$ divides $b$, but $p^2>100$, which is a contradiction. So the prime factor of $st$ may be $2$, $3$, $5$ or $7$. If there are at least three prime factors of $st$ among $2$, $3$, $5$, $7$, then there is a prime factor of $s$ or $t$ no less than $5$. And $d>2\times 3\times 5=30$, so that $a$ or $b\ge 5d>100$, which is a contradiction. The prime factor set of $st$ cannot be $\{3,7\}$, otherwise, $a$ or $b\ge 7\times 3\times 7>100$, which is a contradiction. Similarly, the prime factor set of $st$ cannot be $\{5,7\}$. Therefore, the prime factor set of $st$ can only be $\{2\}$, $\{3\}$, $\{5\}$, $\{7\}$, $\{2,3\}$, $\{2,5\}$, $\{2,7\}$ or $\{3,5\}$.

i) If the prime factor set of $st$ is $\{3,5\}$, then $d$ can only be $15$. Then, $s=3$, $t=5$. So there is one good pair $(a,b)=(45,75)$.

ii) If the prime factor set of $st$ is $\{2,7\}$, then $d$ can only be $14$. Then $s=2$, $t=7$ or $s=4$, $t=7$. So there are two good pairs $(a,b)=(28,98)$ and $(56,98)$.

iii) If the prime factor of $st$ is $\{2,5\}$, then $d$ can only be $10$ or $20$. For $d=10$, then $s=2$, $t=5$; $s=1$, $t=10$; $s=4$, $t=5$; $s=5$, $t=8$. For $d=20$, then $s=2$, $t=5$; $s=4$, $t=5$. There are six good pairs.

iv) If the prime factor of $st$ is $\{2,3\}$, then $d$ can only be $6$, $12$, $18$, $24$ or $30$. For $d=6$, $s=1$, $t=6$; $s=1$, $t=12$; $s=2$, $t=3$; $s=2$, $t=9$; $s=3$, $t=4$; $s=3$, $t=8$; $s=3$, $t=16$; $s=4$, $t=9$; $s=8$, $t=9$; $s=9$, $t=16$. For $d=12$, $s=1$, $t=6$; $s=2$, $t=3$; $s=3$, $t=4$; $s=3$, $t=8$. For $d=18$, $s=2$, $t=3$; $s=3$, $t=4$. For $d=24$, $s=2$, $t=3$; $s=3$, $t=4$. $d=30$, $s=2$, $t=3$. There are $19$ good pairs.

v) If the prime factor set of $st$ is $\{7\}$, then $s=1$, $t=7$, $d$ can only be $7$ or $14$. So, there are two good pairs.

vi) If the prime factor set of $st$ is $\{5\}$, then $s=1$, $t=5$, $d$ can only be $5$, $10$, $15$ or $20$. So, there are four good pairs.

vii) If the prime factor set of $st$ is $\{3\}$ then we have the following: when $s=1$, $t=3$, $d$ can only be $3$, $6$, $\dots$, or $33$; when $s=1$, $t=9$, $d$ can only be $3$, $6$ or $9$; when $s=1$, $t=27$, $d$ can only be $3$. There are $15$ good pairs.

viii) If the prime factor set of $st$ is $\{2\}$, then we have the following: when $s=1$, $t=2$, $d$ can only be $2$, $4$, $\dots$, or $50$; when $s=1$, $t=4$, then $d$ can only be $2$, $4$, $\dots$, or $24$; when $s=1$, $t=8$, then $d$ can only be $2$, $4$, $\dots$, or $12$; when $s=1$, $t=16$, then $d$ can only be $2$, $4$ or $6$; when $s=1$, $t=32$, then $d$ can only be $2$. There are $47$ good pairs.

Therefore, there are all together $1+2+6+19+2+4+15+47=96$ good pairs. $\boxed{96}$