, 2023

Solution:
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 $\frac{\nu_{p}(a)}{\nu_{p}(b)} \in \{3, \frac{1}{3}\}$ 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, $b \geq 2^{2} \cdot 3^{3} > 100$). Hence, the optimal pair is $(a, b) = (2^{3} \cdot 3^{1}, 2^{1} \cdot 3^{3}) = (24, 54)$, 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 $78$.