Three, (15 points) Find all four-digit numbers $\overline{a b c d}$ that satisfy the following condition:
$$
\overline{a b c d}=(\overline{a b}+\overline{c d})^{2} \text {, }
$$

where the digit $c$ can be 0.
Three, let $\overline{a b}=x, \overline{c d}=y$. Then
$$
x, y \in \mathbf{Z}, \text{ and } 10 \leqslant x \leqslant 99, 0 \leqslant y \leqslant 99.
$$

From the problem, we know
$$
(x+y)^{2}=100 x+y,
$$

which simplifies to $x^{2}-2(50-y) x+y^{2}-y=0$.
Since $x, y \in \mathbf{Z}$, we have
$$
\Delta=4(2500-99 y)
$$

is also a perfect square.
Thus, 2500-99y is a perfect square.
Let $2500-99 y=k^{2}(k \in \mathbf{Z}, 0 \leqslant k \leqslant 50)$. Then $(50+k)(50-k)=99 y$ holds.
Since $50 \leqslant 50+k \leqslant 100, 0 \leqslant 50-k \leqslant 50$, we have
$$
50+k=55,66,77,88,99,
$$

or $50-k=0,11,22,33,44$.
Thus, $99 y=55 \times 45, 66 \times 34, 77 \times 23$,
$$
88 \times 12, 99 \times 1,
$$

or $99 y=100 \times 0, 89 \times 11, 78 \times 22$,
$$
67 \times 33, 56 \times 44.
$$

Therefore, the values of $y$ that are non-negative integers are
$$
y=25, y=1, y=0.
$$

Substituting back into equation (1), we get
$x=30$ or $20, x=98$ or $0, x=0$ or 100.
Clearly, $x=0$ or 100 do not meet the conditions.
Thus, there are 3 four-digit numbers that meet the conditions:
$$
3025, 2025, 9801.
$$