7. Given that $M$ is a four-digit perfect square. If the thousand's digit of $M$ is reduced by 3 and the unit's digit is increased by 3, another four-digit perfect square can be obtained. Then the value of $M$ is $\qquad$
7.4761 .

Let $M=\overline{a b c d}=A^{2}$ and $M$'s thousands digit decreases by 3 and the units digit increases by 3 to equal $B^{2}$. Then
$$
\left\{\begin{array}{l}
A^{2}=1000 a+100 b+10 c+d, \\
B^{2}=1000(a-3)+100 b+10 c+(d+3) .
\end{array}\right.
$$

Therefore, $A^{2}-B^{2}=2997$.
Hence $(A-B)(A+B)=3^{4} \times 37$.
Since $A+B \leqslant 2 \times 99=198$, therefore,
$$
\left\{\begin{array}{l}
A-B=3^{3}, 37, \\
A+B=3 \times 37,3^{4} .
\end{array}\right.
$$

Solving yields $(A, B, M)=(69,42,4761)$ or $(59, 22,3481)$ (not valid).