Sphinx's Riddle (Part 2)

Considering the third row, $G$ can only $1$ , $4$ , or $9$ . However, there's no square for the ninety-odds, so $\overline{GH}$ can only be $16$ or $49$ . Then the sum of two squares can be $164 = 10^2 + 8^2$ or $493 = 18^2 + 13^2$ .

If $493$ is used, there will be 3 odd digits left: $1$ , $5$ , and $7$ . For the conditions in the first row, $A$ must be prime while the number tailing it will also make it prime. Therefore, $B$ & $C$ can't be even as it will be composite, and it can't be $5$ either because it will make the number divisible by $5$ . Thus, the values of $B$ & $C$ are either $1$ or $7$ , and the prime $A$ can only be $5$ or $2$ . However, combining the first two digits will result in: $21$ , $27$ , $51$ , and $57$ . None of them are prime, so for the third row, $493$ isn't applicable.

Now if $164$ is used, the remaining odd numbers will be: $3$ , $5$ , $7$ , & $9$ . From the reasons discussed above, two of these odd numbers will fall upon $B$ and $C$ . Then, the new possibilities for $\overline{ABC}$ will be: $239$ , $293$ , $379$ , $397$ , $539$ , $593$ , or $739$ . Of all these primes, only $239 = 23 + 6^3$ . Hence, $\overline{ABC} = 239$ .

Then the 3 last digits left are $5$ , $7$ , & $8$ . To start with a composite digit, $D$ must be $8$ , and $875 = 30^2 - 5^2$ while $857$ is prime. Therefore, $\overline{DEF} = 875$ .

As a result, $\overline{ABCDEFGHI} = \boxed{239875164}$ .