21. N2 (USA) Each positive integer $\alpha$ undergoes the following procedure in order to obtain the number $d=d(a)$ :
(1) move the last digit of $a$ to the first position to obtain the number $b$;
(2) square $b$ to obtain the number $c$;
(3) move the first digit of $c$ to the end to obtain the number $d$.
(All the numbers in the problem are considered to be represented in base
10.) For example, for $a=2003$, we have $b=3200, c=10240000$, and $d=02400001=2400001=d(2003)$.
Find all numbers $a$ for which $d(a)=a^{2}$.
21. Let $a$ be a positive integer for which $d(a)=a^{2}$. Suppose that $a$ has $n+1$ digits, $n \geq 0$. Denote by $s$ the last digit of $a$ and by $f$ the first digit of $c$. Then $a=\overline{* \ldots * s}$, where * stands for a digit that is not important to us at the moment. We have $\overline{f \ldots * s}^{2}=a^{2}=d=\overline{* \ldots * f}$ and $b^{2}=\overline{s * \ldots *}{ }^{2}=$ $c=\overline{f * \ldots *}$.
We cannot have $s=0$, since otherwise $c$ would have at most $2 n$ digits. while $a^{2}$ has either $2 n+1$ or $2 n+2$ digits. The following table gives all possibilities for $s$ and $f$ :
\begin{tabular}{c|c|c|c|c|c|c|c|c|c}
$s$ & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
\hline$f=$ last digit of $\overline{x \ldots * 5}^{2}$ & 1 & 4 & 9 & 6 & 5 & 6 & 9 & 4 & 1 \\
$f=$ first digit of $\overline{3 * \ldots *}^{2}$ & $1,2,3$ & $4-8$ & 9,1 & 1,2 & 2,3 & 3,4 & $4,5,6$ & $6,7,8$ & 8,9
\end{tabular}

We obtain from the table that $s \in\{1,2,3\}$ and $f=s^{2}$, and consequently $c=b^{2}$ and $d$ have exactly $2 n+1$ digits each. Put $a=10 x+s$, where $x<10^{n}$. Then $b=10^{n} s+x, c=10^{2 n} s^{2}+2 \cdot 10^{n} s x+x^{2}$, and $d=$ $2-10^{n+1} s x+10 x^{2}+s^{2}$, so from $d=a^{2}$ it follows that $x=2 s \cdot \frac{10^{n}-1}{9}$. Thus $a=\underbrace{6 \ldots 6}_{n} 3, a=\underbrace{4 \ldots 4}_{n} 2$ or $a=\underbrace{2 \ldots 2}_{n}$. For $n \geq 1$ we see that $a$ cannot be $a=6 \ldots 63$ or $a=4 \ldots 42$ (otherwise $a^{2}$ would have $2 n+2$ digits). Therefore $a$ equals $1,2,3$ or $\underbrace{2 \ldots 2}_{n} 1$ for $n \geq 0$. It is easy to verify that these numbers have the required property.