Example 5 There is a four-digit number
$$
N=\overline{(a+1) a(a+2)(a+3)} \text {, }
$$

which is a perfect square. Find $a$.
Solution: Note that the last digit of a perfect square can only be $0, 1, 4, 5, 6, 9$, thus
$$
a+3 \equiv 0,1,4,5,6,9(\bmod 10) .
$$

Also, $a \geqslant 0, a+3 \leqslant 9$, so $a=1,2,3,6$.
Therefore, the last two digits of $N$ are
$$
\overline{(a+2)(a+3)}=34,45,56,89 \text {. }
$$

Since the last two digits of a perfect square can only be even 0, even 1, even 4, even 9, 25, odd 6, only $\overline{(a+2)(a+3)}=56$, 89 meet the conditions, i.e., $a=$ 3,6.
When $a=6$, $N=7689$ is not a perfect square, so it is discarded.
When $a=3$, $N=4356=66^{2}$, which meets the conditions.
Therefore, $a=3$ is the solution.