8.60 Is there a natural number $A$ such that when it is appended to its right, the resulting number is a perfect square?
(39th Moscow Mathematical Olympiad, 1976)
[Solution] Suppose $A$ is an $n$-digit number that satisfies the condition, then $\overline{A A}$ should be a perfect square $k^{2}$, thus we have
$$
\overline{A A}=A\left(10^{n}+1\right)=k^{2} .
$$

Obviously, $10^{n}+1$ should be a multiple of a prime square.
It is easy to see that when $n$ is odd, $10^{n}+1$ is a multiple of 11.
We can prove that $10^{11}+1$ is a multiple of $11^{2}=121$.
In fact,
$$
\begin{aligned}
10^{11}+1= & 11\left(10^{10}-10^{9}+10^{8}-10^{7}+10^{6}-10^{5}+10^{4}-10^{3}\right. \\
& \left.+10^{2}-10+1\right)
\end{aligned}
$$

Since when $t$ is even,
$$
10^{t} \equiv 1(\bmod 11),
$$

and when $t$ is odd, $10^{t} \equiv 10(\bmod 11)$.
Thus we have
$$
\begin{aligned}
& 10^{10}-10^{9}+10^{8}-10^{7}+10^{6}-10^{5}+10^{4}-10^{3}+10^{2}-10+1 \\
\equiv & 1-10+1-10+1-10+1-10+1-10+1 \\
\equiv & 0 \quad(\bmod 11) .
\end{aligned}
$$

Therefore, $10^{11}+1$ is divisible by $11^{2}$.
However, $\quad \frac{10^{11}+1}{121}=826446281$ is not an 11-digit number; its smallest square multiple is
$$
16 \cdot \frac{10^{11}+1}{121}=13223140496 \text {. }
$$

This is an 11-digit number.
For this, take $A=13223140496$.
Then $\overline{A A}$ is a perfect square, which equals $(4 \cdot 826446281)^{2}$.