Solution. The only two-digit numbers that are perfect squares are 16, 25, $36, 49, 64$ and 81. They start with the digits $1, 2, 3, 4, 6$ and 8, and end with the digits $1, 4, 5, 6$ and 9. Therefore, the digits $b, c, d$ can only be 1, 4, and 6. Since $\overline{b c}$ and $\overline{c d}$ are perfect squares, this is only possible for $\overline{b c}=16$ and $\overline{c d}=64$. Finally, $\overline{a b}$ is a perfect square for $a=8$ and $\overline{d e}$ is a perfect square for $e=9$. Therefore, there is only one number that satisfies the conditions of the problem, and that number is 81649.
1. Determine all five-digit numbers $\overline{a b c d e}$ such that $\overline{a b}, \overline{b c}, \overline{c d}$, and $\overline{d e}$ are perfect squares. Provide a detailed explanation of your answer!