Example 22 (37th Canadian Mathematical Olympiad) If an ordered triplet of positive integers $\{a, b, c\}$ satisfies $a \leqslant b \leqslant c, (a, b, c)=1, a^{n}+b^{n}+c^{n}$ is divisible by $a+b+c$, then $\{a, b, c\}$ is called $n$-energetic. For example, $\{1, 2, 2\}$ is 5-energetic.
(1) Find all ordered triplets of positive integers that satisfy: for any $n \geq 1$, the ordered triplet of positive integers is $n$-energetic;
(2) Find all ordered triplets of positive integers that are both 2004-energetic and 2005-energetic, but not 2007-energetic.
Number Theory Problem

(1) Since $(a+b+c) \mid (a^{2}+b^{2}+c^{2})$ and $(a+b+c) \mid (a^{3}+b^{3}+c^{3})$, then
Mathematics
Intermediate.
Problem
$(a+b+c) \mid \left[(a+b+c)^{2}-a^{2}-b^{2}-c^{2}\right]$, which means
$(a+b+c) \mid (2ab + 2bc + 2ca)$.
Also, $(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca) = a^{3}+b^{3}+c^{3}-3abc$, so $(a+b+c) \mid 3abc$.
Let a prime $p$ satisfy $p^{\alpha} \mid (a+b+c) (\alpha \geqslant 1)$.
If there exists $p \geqslant 5$, then $p \mid abc$. Without loss of generality, assume $p \mid a$.
Since $p \mid 2(ab + bc + ca)$, it follows that $p \mid bc$.
Without loss of generality, assume $p \mid b$. Since $p \mid (a+b+c)$, it follows that $p \mid c$.
Thus, $p \mid (a, b, c)$, which is a contradiction.
Therefore, $p=2$ or 3.
Hence, $a+b+c = 2^{m} \times 3^{n} (m, n \geqslant 0)$.
If $n \geqslant 2$, then
$3 \mid (a+b+c)$, $3 \mid abc$, and $3 \mid (ab + bc + ca)$.
Following the same reasoning, we can conclude $3 \mid (a, b, c)$, which is a contradiction.
Therefore, $n=0$ or 1.
Let $a+b+c = 2^{m} k (k=1$ or 3$)$, then $2^{m} \mid abc$.
Since $(a, b, c)=1$, without loss of generality, assume $a$ is odd.
Also, $2^{m} \mid (a+b+c)$, without loss of generality, assume $b$ is odd and $c$ is even, so $2^{m} \mid c$.
From $2^{m} \mid (2ab + 2bc + 2ca)$, it follows that
$2^{m-1} \mid ab$.
Thus, $m=0$ or 1.
Since $a+b+c \geqslant 3$, it follows that $a+b+c=3$ or 6.
By verification, we get
$(a, b, c) = (1,1,1)$ or $(1,1,4)$.
(2) It is easy to see that
$$
\begin{array}{l}
a^{n} + b^{n} + c^{n} = (a+b+c)(a^{n-1} + b^{n-1} + c^{n-1}) - (ab + bc + ca)(a^{n-2} + b^{n-2} + c^{n-2}) + \\
abc(a^{n-3} + b^{n-3} + c^{n-3}).
\end{array}
$$
Also, $(a+b+c) \mid (a^{2004} + b^{2004} + c^{2004})$,
$$
(a+b+c) \mid (a^{2005} + b^{2005} + c^{2005}),
$$

Substituting $n=2007$ into equation (1) gives
$(a+b+c) \mid (a^{2007} + b^{2007} + c^{2007})$, which contradicts the given condition.
Therefore, there do not exist $(a, b, c)$ that satisfy the condition.