10. If a four-digit number satisfies the following conditions:
（1）The thousands digit is the same as the hundreds digit, and the tens digit is the same as the units digit;
（2）The difference between this four-digit number and 99 is a perfect square. Find this four-digit number.
10. Let the four-digit number be
$A=1000 x+100 x+10 y+y(1 \leqslant x \leqslant 9,0 \leqslant y \leqslant 9, x, y$ are integers), then $A-99=11(100 x+y)-99=11(100 x+y-9)$
is a perfect square,
and since 11 is a coefficient, then $100 x+y-9=11 \cdot k^{2}$ ( $k$ is a positive integer),
i.e., $11 \cdot k^{2}=99 x+(x+y-9)$.
Thus, $11 \mid(x+y-9)$, and since $-8 \leqslant x+y-9 \leqslant 9$,
then $x+y-9=0$,
i.e., $k^{2} \cdot 11=99 x$,
which means $k^{2}=x \cdot 3^{2}$.
Therefore, $x$ is a perfect square,
i.e., $x=1,4,9, y=9-x=8,5,0$.
The required four-digit numbers are $1188,4455,9900$.