For any $n \in Z^{+}$, let $D_{n}$ be the greatest common divisor of all integers of the form $a^{n}+(a+1)^{n}+(a+2)^{n}\left(a \in Z^{+}\right)$, try to find the set $\left\{\mathrm{D}_{\mathrm{n}} \mid \mathrm{n} \in \mathrm{Z}^{+}\right\}$.
For any $\mathrm{n} \in \mathrm{Z}^{+}$, let $\mathrm{a}=\mathrm{D}_{\mathrm{n}}$, then
$$
D_{n}\left|\left[D_{n}{ }^{n}+\left(D_{n}+1\right)^{n}+\left(D_{n}+2\right)^{n}\right] \Rightarrow D_{n}\right|\left(1^{n}+2^{n}\right)
$$

Let $\mathrm{a}=\mathrm{D}_{\mathrm{n}}+1$, then
$$
D_{n}\left|\left[\left(D_{n}+1\right)^{n}+\left(D_{n}+2\right)^{n}+\left(D_{n}+3\right)^{n}\right] \Rightarrow D_{n}\right|\left(1^{n}+2^{n}+3^{n}\right)
$$

Combining (1) and (2), we know $D_{n} \mid 3^{n}$, thus there exists $k \in N$, such that $D_{n}=3^{k}$.
(2) Next, we prove: for any natural number $k$, there is $3^{k} \in\left\{D_{n} \mid n \in Z^{+}\right\}$.
(1) Notice that $3+\left(1^{2}+2^{2}+3^{2}\right)$, so there must be $\mathrm{D}_{2}=1=3^{0}$.
(2) When $k=1$, take $n=1$, then $a^{n}+(a+1)^{n}+(a+2)^{n}=3(a+1)$. Clearly, $D_{1}=3$.
(3) For any $k \in Z^{+}, k \geq 2$, we take $n=3^{k-1}$, and assume $D_{3^{k-1}}=3^{\alpha}$.

First, for any $a \in Z^{+} ， a 、 a+1 、 a+2$ must include one multiple of 3, and the sum of the other two is a multiple of 3. Without loss of generality, assume $3 \mid(\mathrm{a}+2)$, then $3 \mid[\mathrm{a}+(\mathrm{a}+1)]$. According to the LTE lemma, we know
$$
3^{k} \mid\left[a^{3^{k-1}}+(a+1)^{3^{k-1}}\right]
$$

Since $3^{k-1} \geq k$, we have $3^{k} \mid(a+2)^{3^{k-1}}$, thus $3^{k} \mid\left[a^{3^{k-1}}+(a+1)^{3^{k-1}}+(a+2)^{3^{k-1}}\right]$. Therefore, $3^{k} \mid D_{3}^{k-1}$, which means $3^{k} \mid 3^{\alpha}$.
On the other hand, take $a=3^{\alpha}$, then
$$
\begin{array}{l}
3^{\alpha} \mid\left[\left(3^{\alpha}\right)^{3^{k-1}}+\left(3^{\alpha}+1\right)^{3^{k-1}}+\left(3^{\alpha}+2\right)^{3^{k-1}}\right] \\
\Rightarrow 3^{\alpha} \mid\left(1^{3^{k-1}}+2^{3^{k-1}}\right)
\end{array}
$$

According to the LTE lemma, we know $\alpha \leq v_{3}\left(1^{3^{k-1}}+2^{3^{k-1}}\right)=v_{3}(1+2)+v_{3}\left(3^{k-1}\right)=k$.
Combining the above two aspects, we know $\alpha=k$. That is, $\mathrm{D}_{3^{k-1}}=3^{\mathrm{k}}$.
In summary, $\left\{\mathrm{D}_{\mathrm{n}} \mid \mathrm{n} \in \mathrm{Z}^{+}\right\}=\left\{3^{\mathrm{k}} \mid \mathrm{k} \in \mathrm{N}\right\}$.