10. (2005 Canadian Mathematical Olympiad) If an ordered triplet of positive integers $\{a, b, c\}$ satisfies $a \leqslant b \leqslant c, (a, b, c)=1$, and $a^{n}+b^{n}+c^{n}$ is divisible by $a+b+c$, then $\{a, b, c\}$ is called $n$-powered. For example, $\{1,2,3\}$ is 5-powered.
（1）Find all ordered triplets of positive integers that satisfy: for any $n \geqslant 1$, the ordered triplet of positive integers is $n$-powered;
（2）Find all ordered triplets of positive integers that are both 2004-powered and 2005-powered but not 2007-powered.
10. Solution: (1) Since $(a+b+c)\left|\left(a^{2}+b^{2}+c^{2}\right),(a+b+c)\right|\left(a^{3}+b^{3}+c^{3}\right)$, then $(a+b+c) \mid\left[(a+b+c)^{2}-a^{2}-b^{2}-c^{2}\right]$,
i.e., $(a+b+c) \mid(2 a b+2 b c+2 c a)$.
Also, $(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)=a^{3}+b^{3}+c^{3}-3 a b c$.
Thus, $(a+b+c) \mid 3 a b c$.
Let the prime $p$ satisfy $p^{a} \|(a+b+c)(\alpha \geqslant 1)$ (notation \|\| see page 18 of this book),
If there exists $p \geqslant 5$, then $p \mid a b c$, assume without loss of generality that $p \mid a$.
Since $p \mid 2(a b+b c+c a)$, then $p \mid b c$.
Assume without loss of generality that $p \mid b$, since $p \mid(a+b+c)$, then $p \mid c$.
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 $3|(a+b+c), 3| a b c, 3 \mid(a b+b c+c a)$.
Similarly, it can be deduced that $3 \mid(a, b, c)$, which is a contradiction.
Thus, $n=0$ or 1.
Let $a+b+c=2^{m} \cdot k(k=1$ or 3$)$, then $2^{m} \mid a b c$,
Since $(a, b, c)=1$, assume without loss of generality that $a$ is odd.
Also, $2^{m} \mid(a+b+c)$, assume without loss of generality that $b$ is odd, $c$ is even, so $2^{m} \mid c$.
From $2^{m} \mid(2 a b+2 b c+2 c a)$, then $2^{m-1} \mid a b$.
Thus, $m=0$ or 1.
Also, $a+b+c \geqslant 3$, so, $a+b+c=3$ or 6.
By verification, $(a, b, c)=(1,1,1)$ or $(1,1,4)$.
(2) It is easy to get $a^{n}+b^{n}+c^{n}=(a+b+c)\left(a^{n-1}+b^{n-1}+c^{n-1}\right)-(a b+b c+c a)\left(a^{n-2}+b^{n-2}+\right.$
$$
\begin{aligned}
\left.c^{n-2}\right) & +a b c\left(a^{n-3}+b^{n-3}+c^{n-3}\right) \\
& \text { Also, }(a+b+c)\left|\left(a^{2004}+b^{2004}+c^{2004}\right),(a+b+c)\right|\left(a^{2005}+b^{2005}+c^{2005}\right) .
\end{aligned}
$$

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