7. (1994 Shanghai Senior High School Competition Question) Two two-digit numbers, their difference is 52, and the last two digits of their squares are the same. Then these two numbers are
7.76, 24. Let the two numbers be $x, y$, then $\left\{\begin{array}{l}x-y=52 \\ x^{2}-y^{2}=100 r\end{array}\right.$, thus $52(x+y)=(x-y)(x+y)=100 r, 13(x+$ $y)=25 r$, hence $13 \mid r$.

Since $x+y=x-y+2 y=52+2 y$ is even, we get $2 \mid r$, therefore $26 \mid r$. Let $r=26 t, t$ be a positive integer, then $x+y=50 t$, $x=25 t+26, y=25 t-26$, by $10 \leqslant x, y \leqslant 99$, we get $\left\{\begin{array}{l}10 \leqslant 25 t+26 \leqslant 99 \\ 10 \leqslant 25 t-26 \leqslant 99\end{array}\right.$, thus $\frac{36}{25} \leqslant t \leqslant \frac{73}{25}$, hence $t=2$.
Therefore, the two numbers are $x=25 \times 2+26=76, y=25 \times 2-26=24$.