*8. In decimal, find the smallest natural number: its square number starts with 19 and ends with 89.
The last digit of the required number can only be 3 or 7. We write out all two-digit numbers ending in 3 and 7, and we get $17, 33, 67, 83$. To ensure that the first two digits of the natural number $x$ are 19, the inequality $19 \leqslant x^{2} \cdot 10^{-N}<20$ must hold, where $N$ is a natural number. When $N = 2k$ and $N=2k+1$, we get the inequalities $19 \leqslant x^{2} \cdot 10^{-2k}<20$ and $190 \leqslant x^{2} \cdot 10^{-2k}<200$. Taking the square root, in the first case we get the inequality $4.3588989 \cdots \leqslant x \cdot 10^{-k}<4.4721359 \cdots$; and in the second case, $13.784048 \cdots \leqslant x \cdot 10^{-k}<14.142135 \cdots$.

It is not difficult to see that, due to the variation of $k$, there will be infinitely many possibilities. We examine $k=1$ and $k=2$. If $4.358 \leqslant x \cdot 10^{-1}<4.472$, then $x=44$; if $4.358 \leqslant x \cdot 10^{-2}<4.472$, then $436 \leqslant x \leqslant 447$. If $13.784 \cdots \leqslant x \cdot 10^{-1}<14.142 \cdots$, then $x=14$; if $13.784 \cdots \leqslant x \cdot 10^{-2}<14.142 \cdots$, then $138 \leqslant x \leqslant 141$. When $k=3$, we similarly examine the two possible cases, and so on. Among the $x$ values we obtain, the smallest number whose last two digits are $17, 33, 67, 83$ is $1383\left(1383^{2}=1912689\right)$.