$10 \cdot 14$ Let $R_{n}=\frac{1}{2}\left(a^{n}+b^{n}\right), a=3+2 \sqrt{2}, b=3-2 \sqrt{2}, n=0$, $1,2, \cdots$ Then $R_{12345}$ is an integer, its unit digit is
(A) 1 .
(B) 3 .
(C) 5 .
(D) 7 .
(E) 9
(41st American High School Mathematics Examination, 1990)
[Solution 1] Given $a b=1, a+b=6$,
and $a^{n}+b^{n}=(a+b)\left(a^{n-1}+b^{n-1}\right)-a b\left(a^{n-2}+b^{n-2}\right)$
$$
=6\left(a^{n-1}+b^{n-1}\right)-\left(a^{n-2}+b^{n-2}\right).
$$

Let $x_{n} \doteq a^{n}+b^{n}(\bmod 5)$, then from the above equation we know
$$
x_{n} \equiv x_{n-1}-x_{n-2} \quad(\bmod 5).
$$

The sequence (after modulo operation) is a purely periodic sequence with a period of 6, which can be seen by noting that $x_{n+3} \equiv x_{n+2}-x_{n+1} \equiv\left(x_{n+1}-x_{n}\right)-x_{n+1} \equiv-x_{n} \quad(\bmod 5)$,
then
$$
x_{n+6} \equiv-x_{n+3} \equiv-\left(-x_{n}\right) \equiv x_{n} \quad(\bmod 5),
$$

and it is easy to calculate that the first 6 terms of $\left\{x_{n}(\bmod 5)\right\}$ are
$$
1,4,3,4,1,2.
$$

Since $12345=6 \cdot 2057+3$, we know $x_{12345}(\bmod 5)=3$, so the last digit could be 3 or 8.

Since $R_{n}=\frac{1}{2}\left(a^{n}+b^{n}\right)$ is an integer, the last digit of $a^{12345}+b^{12345}$ can only be 8, which means the last digit of $R_{12345}=\frac{1}{2}\left(a^{12345}+b^{12345}\right)$ is 9 or 4.
Therefore, the answer is $(E)$.
[Solution 2] Given $a=3+2 \sqrt{2}, b=3-2 \sqrt{2}$, then
$$
a+b=6, \quad a b=1,
$$

thus, they are the two roots of the equation $x^{2}-6 x+1=0$.
By Newton's identities:
If $u, v$ are the roots of $A x^{2}+B x+C=0(A \neq 0)$, and we denote $S_{n}=u^{n}+v^{n}(n=1,2, \cdots)$, then we always have
$$
A S_{n+2}+B S_{n+1}+C S_{n}=0 \quad(n=1,2, \cdots)
$$

Its proof is not difficult, it only requires substitution and simplification.
Since $R_{n}=\frac{1}{2} S_{n}$, then $R_{n}$ also satisfies Newton's identities, i.e.,
$R_{n+2}-6 R_{n+1}+R_{n}=0$ or $R_{n+2}=6 R_{n+1}-R_{n}(n=1,2, \cdots)$.
The rest of the solution is the same as Solution 1.
Therefore, the answer is $(E)$.