8. Given that $18^{2}=324, 24^{2}=576$, they are formed by the permutation of two consecutive digits $2,3,4$ and $5,6,7$ respectively; and $66^{2}=4356$ is formed by the permutation of four consecutive digits $3, 4, 5, 6$. Then the next such square number is $\qquad$


8.5476.

For any square number, its last digit can only be $0, 1, 4, 5, 6, 9$, and
$$
\begin{array}{l}
(10 a)^{2}=100 a^{2}, \\
(10 a+5)^{2}=100 a^{2}+100 a+25, \\
(10 a \pm 4)^{2}=100 a^{2} \pm 80 a+16 ; \\
(10 a \pm 1)^{2}=100 a^{2} \pm 20 a+1, \\
(10 a \pm 3)^{2}=100 a^{2} \pm 60 a+9 ; \\
(10 a \pm 2)^{2}=100 a^{2} \pm 40 a+4 .
\end{array}
$$

If following the above rules, it can only be the case of
$$
3456,4536,5436,5364,3564 \text {. }
$$

But none of the above are square numbers.
The next four consecutive digits are $4,5,6,7$.
If arranged into a square number, the last digit can only be 4 or 6, from smallest to largest:
$$
4756,5476,5764,7564 .
$$

After calculation, only $5476=74^{2}$ is a square number.