Example + (a) For which $n \in \mathbf{N}$, does $x^{2}+x+1 \mid x^{2 n}+x^{n}+1$?
(b) For which $n \in \mathbf{N}$, does $37 \mid 1 \underbrace{0 \cdots 01} \underbrace{0 \cdots 01}$?
Solution 1 (a) Directly use the following relationship for transformation:
$$
x^{3}-1=(x-1)\left(x^{2}+x+1\right), x^{3}-1 \mid x^{3 m}-1 .
$$
( i ) $n=3 k \Leftrightarrow x^{6 k}+x^{3 k}+1=\left(x^{6 k}-1\right)+\left(x^{3 k}-1\right)+3=\left(x^{2}+x+1\right) Q(x)+3$.
( ii ) $n=3 k+1 \Leftrightarrow x^{6 k+2}+x^{3 k+1}+1=x^{2}\left(x^{6 k}-1\right)+x\left(x^{3 k}-1\right)+x^{2}+x+1$
$$
=\left(x^{2}+x+1\right) R(x) \text {. }
$$
(iii)
$$
\begin{aligned}
n & =3 k+2 \Leftrightarrow x^{6 k+4}+x^{3 k+2}+1=x^{4}\left(x^{6 k}-1\right)+x^{2}\left(x^{3 k}-1\right)+x^{4}+x^{2}+1 \\
& =x^{4}\left(x^{6 k}-1\right)+x^{2}\left(x^{3 k}-1\right)+x\left(x^{3}-1\right)+x^{2}+x+1=\left(x^{2}+x+1\right) S(x)
\end{aligned}
$$

The answer is
$$
x^{2}+x+1 \mid x^{2 n}+x^{n}+1 \Leftrightarrow 3 \times n \text {. }
$$
(b) When $x=10$, $x^{2}+x+1=111, x^{2(n+1)}+x^{n+1}+1=1 \underbrace{0 \cdots 01}_{n} \underbrace{0 \cdots 01}_{n}, 111=3 \cdot 37$. Since the sum of the digits of the given dividend is 3, it is a multiple of 3. Thus, $37 \mid 1 \underbrace{0 \cdots 01}_{n} \underbrace{0 \cdots 01}_{n}$, if and only if $n \equiv$
$0(\bmod 3)$ or $n \equiv 1(\bmod 3)$.
Solution 2 Let the roots of $x^{2}+x+1=0$ be $w$, $w^{2}$. Using the relationships $w^{3}=1$ and $w^{2}+w+1=0$, we get
$$
\begin{array}{c}
n=3 k \Rightarrow w^{6 k}+w^{3 k}+1=1+1+1=3 \\
n=3 k+1 \Rightarrow w^{6 k-2}+w^{3 k-1}+1=w^{2}+w+1=0 \\
n=3 k+2 \Rightarrow w^{6 k-4}+w^{3 k-2}+1=w^{4}+w^{2}+1=w+w^{2}+1=0 .
\end{array}
$$