We take the number 49 and insert the number 48 between its digits. Then, we insert 48 again between the 4 and the 8, and repeat this process several times (resulting in $44 \ldots 48 \ldots 89$). Is it true that we always get a perfect square?
The numbers obtained by the procedure described in the task can all be written in the form

$$
\underbrace{44 \ldots 4}_{n} \underbrace{88 \ldots 8}_{n-1} 9
$$

Let

$$
x=\underbrace{11 \ldots 1}_{n}=\frac{1}{9} \cdot \underbrace{99 \ldots 9}_{n}=\frac{1}{9}\left(10^{n}-1\right)
$$

Then $10^{n}=9 x+1$ and the numbers are

$$
\underbrace{44 \ldots 4}_{n} \underbrace{88 \ldots 8}_{n-1} 9=4 x(9 x+1)+8 x+1=36 x^{2}+12 x+1=(6 x+1)^{2}
$$

Thus, the numbers obtained in this way are indeed perfect squares.