One. (20 points) Try to find a four-digit number, the square of the sum of the two-digit numbers formed by its first two digits and the last two digits is exactly equal to this four-digit number.



保留源文本的换行和格式，所以翻译结果如下：

One. (20 points) Try to find a four-digit number, the square of the sum of the two-digit numbers formed by its first two digits and the last two digits is exactly equal to this four-digit number.
One, Solution 1: Let the two two-digit numbers be $x$ and $y$, $10 \leqslant x, y \leqslant 99$.

We have $(x+y)^{2}=100 x+y$, which means
$$
\begin{array}{l}
x^{2}+2(y-50) x+\left(y^{2}-y\right)=0 \\
\text { When } \Delta=4(y-50)^{2}-4\left(y^{2}-y\right) \\
=4(2500-99 y) \geqslant 0,
\end{array}
$$

i.e., $2500-99 y \geqslant 0$, then $y \leqslant 25$, the equation has real solutions
$$
x=50-y \pm \sqrt{2500-99 y} \text {. }
$$

Since $2500-99 y$ must be a perfect square, and the last digit of a perfect square can only be $0,1,4,5,6,9$, hence $y$ can only be 25.
At this time, $x=30$ or $x=20$.

Therefore, the required four-digit number is 2025 or 3025.