Three. (25 points) Does there exist a four-digit number $\overline{a b c d}$, such that the last four digits of its square are also $\overline{a b c d}$? Does there exist a five-digit number $\overline{a b c d e}$, such that the last five digits of its square are also $\overline{a b c d e}$? If they exist, find all of them; if not, explain why.
Let $\overline{a b c d}=x$. Since $x^{2}$ and $x$ have the same last four digits, the last four digits of $x^{2}-x$ are 0000, i.e., $10000 \mid x^{2}-x$, or equivalently, $2^{4} \cdot 5^{4} \mid x(x-1)$.
(i) If $2^{4} \mid x$ and $5^{4} \mid x-1$, then $x-1$ is a four-digit odd number and a multiple of 625. Therefore, $x-1$ can only be 1875, 3125, 4375, 5625, 6875, 8125, 9375. Thus, $x$ is 1876, 3126, 4376, 5626, 6876, 8126, 9376. Among these 7 numbers, only 9376 is divisible by $2^{4}$. Verification shows that 9376 is the solution.
(ii) If $2^{4} \mid x-1$ and $5^{4} \mid x$, then $x$ is a four-digit odd number and a multiple of 625. Therefore, $x$ can only be 1875, 3125, 4375, 5625, 6875, 8125, 9375. Thus, $x-1$ is 1874, 3124, 4374, 5624, 6874, 8124, 9374. None of these 7 numbers are divisible by $2^{4}$. Hence, there is no solution in this case.
$\therefore \overline{a b c d}=9376$ is the solution. $\left(9376^{2}=87909376\right)$
Using the same method, we can find that $\overline{a b c d e}=90625$. $\left(90625^{2}=8212890625\right)$