, 2023

Solution:
Note that if $a$ and $c$ are divisible by more than one distinct prime, then we can just take the prime powers of a specific prime. Thus, assume $a$ and $c$ are powers of a prime $p$. Assume $a = 4^{x}$ and $c = 4^{y}$. Then $x b = y d$.

Because $b$ and $d$ are squares, the ratio of $x$ to $y$ is a square, so assume $x = 1$ and $y = 4$. We can't take $b = 4$ and $c = 1$, but we instead can take $b = 36$ and $c = 9$. It can be checked that other values of $x$ and $y$ are too big. This gives $4^{36} = 256^{9}$, which gives a sum of $305$.

If $a$ and $c$ are powers of $9$, then $\max(a, c) \geq 9^{4}$, which is already too big. Thus, $305$ is optimal.