19.1.13 ** Find all $n$ such that the number $144 \cdots 4$ is a perfect square.
Parse $144=12^{2}, 1444=38^{2}$. Let $n>3$, and $1 \underbrace{44 \cdots 4}_{n \uparrow}=k^{2}, k \in \mathbf{N}$. Then $k$ is even. So
$$
36 \underbrace{11 \cdots 1}_{n-2 \uparrow}=1 \underbrace{44 \cdots 4}_{n \uparrow} \div 4=\left(\frac{k}{2}\right)^{2},
$$

is also a square of a natural number. But this is impossible. Because the remainder of a square number divided by 4 is 0 or 1, while $36 \underbrace{11 \cdots 1}$ divided by 4 leaves a remainder of 3.
Therefore, $n$ is 2 or 3.