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.