Let the four-digit number $w_{1}=\overline{a b c d}$ be a perfect square. By splitting it in the middle, we get two two-digit numbers $x_{1}=\overline{a b}$ and $y_{1}=\overline{c d} ; w_{2}=3 x_{1} y_{1}+1$ is a perfect square. By splitting $w_{2}$ in the middle, we get two two-digit numbers $x_{2}$ and $y_{2} ; w_{3}=$ $2 x_{2} y_{2}$ is a perfect square. By splitting $w_{3}$ in the middle, we get two two-digit numbers $x_{3}$ and $y_{3} ; w_{4}=x_{3} y_{3}+1$ is a perfect square, and 9 times $w_{4}$ is the four-digit number $w_{5},$ which is also a perfect square. Find the four-digit numbers $w_{i}(i=1,2,3,4,5)$.
Solution: From the fact that 9 times $w_{4}$ is the four-digit number $w_{5}$, and $w_{5}$ is also a perfect square, we have
$$
\begin{array}{l}
w_{4}=33^{2}=1089, \\
w_{5}=9 \times 33^{2}=99^{2}=9801 .
\end{array}
$$

From $w_{4}=x_{3} y_{3}+1$ being the perfect square 1089, we have
$$
\begin{array}{l}
1089-1=33^{2}-1 \\
=34 \times 32=17 \times 64=16 \times 68 .
\end{array}
$$

Among the six four-digit numbers 1668, 6816, 6417, 1764, 3432, and 3234, only $1764=42^{2}$ is a perfect square, so $w_{3}=1764$.
From $w_{3}=2 x_{2} y_{2}$ being the perfect square 1764, we have
$$
x_{2} y_{2}=882=21 \times 42=14 \times 63=18 \times 49 \text {. }
$$

Upon inspection, only $1849=43^{2}$ is a perfect square, so $T J w_{2}=1.849$.

From $w_{2}=3 x_{1} y_{1}+1$ being the perfect square 1849, we have $3 x_{1} y_{1}=43^{2}-1=44 \times 42$, thus
$$
x_{1} y_{1}=44 \times 14=22 \times 28=11 \times 56 .
$$

Upon inspection, only 1156 and 1444 are perfect squares, so $w_{1}=1156$ or 1444.
In summary, $w_{1}=1156=34^{2}$ or $1444=38^{2}$,
$$
\begin{array}{l}
w_{2}=1849=43^{2}, w_{3}=1764=42^{2}, \\
w_{4}=1089=33^{2}, w_{5}=9801=99^{2} .
\end{array}
$$
(Tian Yonghai, Suihua Educational Institute, Heilongjiang Province, 152054)