For any prime $p$ and a positive integer $n$, let $\nu_{p}(n)$ be the largest nonnegative integer $k$ for which $p^{k}$ divides $n$. Taking $\nu_{p}$ on both sides of the given equation, we get $$\nu_{p}(a)+\nu_{p}(b)=2 \cdot\left|\nu_{p}(a)-\nu_{p}(b)\right|$$ which means $\frac{\nu_{p}(a)}{\nu_{p}(b)} \in\left\{3, \frac{1}{3}\right\}$ for all primes $p$. Using this with $a, b \leq 100$, we get that - We must have $\left(\nu_{2}(a), \nu_{2}(b)\right) \in\{(0,0),(1,3),(3,1),(2,6),(6,2)\}$ because $a$ and $b$ cannot be divisible by $2^{7}$. - We must have $\left(\nu_{3}(a), \nu_{3}(b)\right) \in\{(0,0),(1,3),(3,1)\}$ because $a$ and $b$ cannot be divisible by $3^{6}>100$. - $a$ and $b$ cannot be divisible by any prime $p \geq 5$, because if not, then one of $a$ and $b$ must be divisible by $p^{3} \geq 5^{3}>100$. If $\left(\nu_{2}(a), \nu_{2}(b)\right)=(2,6)$ (and similarly with $(6,2)$ ), then we must have $(a, b)=(4,64)$, so the sum is 68 . If $\left(\nu_{3}(a), \nu_{3}(b)\right)=(1,3)$ (and similarly with $(3,1)$ ), then we must have $\nu_{2}(b) \leq 1$ (otherwise, $b \geq$ $\left.2^{2} \cdot 3^{3}>100\right)$. Hence, the optimal pair is $(a, b)=\left(2^{3} \cdot 3^{1}, 2^{1} \cdot 3^{3}\right)=(24,54)$, so the sum is $24+54=78$. If neither of the above happens, then $a+b \leq 2^{1}+2^{3} \leq 10$, which is clearly not optimal. Hence, the optimal pair is $(24,54)$, and the answer is 78 .