What is the maximum value of $(a+ b+c)$ if $(a^n + b^n + c^n)$ is divisible by $(a+ b+c)$ where the remainder is 0? 
The ‘energy’ of an ordered triple $(a, b, c)$ formed by three
positive integers $a$, $b$ and $c$ is said to be n if the following $c$
$\ge b\geq a$, gcd$(a, b, c) = 1$, and $(a^n + b^n + c^n)$ is
divisible (remainder is 0) by  $(a +b+ c) $. There are some
possible ordered triple whose ‘energy’ can be of all values of $n \ge$
$1$. In this case, for which ordered triple, the value of $(a+b+c)$ is
maximum?
Second part (of the original problem) Determine all triples $(a,b,c)$ that are simultaneously $2004$-good and $2005$-good, but not $2007$-good.

Source: Bangladesh Math Olympiad 2017 junior category (Originally from Canada, 2005).
I can't understand the first line of this question. Any 3 consecutive numbers have a gcd of 1. Moreover if $n=1$, then  $(a^n + b^n + c^n) = (a +b+ c) $.

The question is to find the largest sum of $a, b, c$, given they're all relatively prime to each other and divide
$$a^n + b^n + c^n \text{ } \forall \text{ } n \ge 1 \tag{1}\label{eq1}$$
Since $a, b, c \ge 1$, then $a + b + c \ge 3$, so it consists of one or more prime factors.  Call one of these prime factors $d$. Thus,
$$a + b + c \equiv a^n + b^n + c^n \equiv 0 \pmod d \tag{2}\label{eq2}$$
As this must hold for all $n \ge 1$, consider $n = 3$ and substitute
$$c \equiv -a - b \pmod d \tag{3}\label{eq3}$$
into the middle part of the congruence in \eqref{eq2} to get
$$a^3 + b^3 + \left(-a - b\right)^3 \equiv 0 \pmod d \tag{4}\label{eq4}$$
This simplifies to
$$-3ab^2 - 3a^2b = -3ab\left(a + b\right) \equiv 0 \pmod d \tag{5}\label{eq5}$$
Thus, $d$ must divide $3$, $a$, $b$ and/or $a + b \equiv -c \pmod d$. If $d$ is not $3$, then note that if it must divide just one of $a$, $b$ and $-c$. This is because if it divides any $2$, say $a$ and $b$, it must also divide $c$. As the equations are symmetric in $a, b \text{ and } c$, WLOG, assume that $d$ divides $a$, i.e., $a \equiv 0 \pmod d$. Thus, $b + c \equiv 0 \pmod d$, i.e., $b \equiv -c \pmod d$. Now, use $n = 2$ in the congruence in the middle part of \eqref{eq2} to get
$$a^2 + b^2 + c^2 \equiv 0 + \left(-c\right) + c^2 \equiv 2c^2 \equiv 0 \pmod d \tag{6}\label{eq6}$$
As $d$ doesn't divide $c$, this means that $d$ must divide $2$, i.e., it must be $2$. Thus, the only possible prime factors of $a + b + c$ are $3$, as mentioned earlier, and $2$ as shown here.
Based on what I have found here, I believe the maximum value is just the product of those $2$ factors, i.e., $6$, which can be obtained from
$$\left(a, b, c\right) = \left(1, 1, 4\right) \tag{7}\label{eq7}$$
You can easily verify that $a^n + b^n + c^n = 1 + 1 + 4^n \equiv 0 \pmod 6$ for all $n \ge 1$ as $1 + 1 = 2$, so $2$ divides it, and $4^n \equiv 1^n \equiv 1 \pmod 3$, so $3$ divides it as well.  I am posting this partial solution for now so you have something to start from, as I believe I'm on the right track.  I need to go for a while, but I will keep thinking about this and, if I determine the rest of the solution, I will add it later.
