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

Question

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) $.

Answer 1

The answer is indeed $6$. Here is a complete solution.

First, take a prime $p$ such that, $p\mid a+b+c$. Note that, if $p$ divides two of $a$ and $b$, then it must divide the third, contradicting with $(a,b,c)=1$. Thus, $p$ either divides none of them, or only one.

1. If $p\nmid a,b,c$, then by taking $n=p-1$, we have, by Fermat's theorem that $a^{p-1}\equiv b^{p-1}\equiv c^{p-1}\equiv 1 \pmod{p}$. Thus, $p\mid 3$, hence $p=3$.
2. If $p\mid a$, and $p\nmid b,c$, then we obtain $p\mid 2$, by taking $n=p-1$.

Therefore, only prime divisors of $a+b+c$ are $2$ or $3$. Now, for $n=2$, we get using $(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$ that $a+b+c\mid 2(ab+bc+ca)$. Also, for $n=3$, using $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$, we have $a+b+c\mid 3abc$. Next, note that, if $9\mid a+b+c$, then $3\mid abc$. Hence, either $a$ or $b$ or $c$ is divisible by $3$. Now, if $3\mid a$, then using $a+b+c\mid 2(ab+bc+ca)$, we obtain that $3\mid bc$, hence $3\mid b$ or $3\mid c$. However, this, together with $3\mid a+b+c$ contradicts with $(a,b,c)=1$. Hence, $9\nmid a+b+c$.

Similarly, if $4\mid a+b+c$, then note that among $a,b,c$ exactly one is even, suppose it is $a$, that is, $4\mid a$. But this gives, $4\mid 2(ab+bc+ca)$, yielding $2\mid (ab+bc+ca)$, yielding $2\mid bc$. However, since $b$ and $c$ are odd, this is a clear contradiction.

Thus, $a+b+c=3$ or $a+b+c=6$. In former, we get $(1,1,1)$, which is really $n$-good for any $n$. For the latter, we have $(3,2,1)$ or $(4,1,1)$. For the former, $(3,2,1)$ is not $n-$ good for any even $n$, using modulo $3$. For the latter, it is easy to see that it is $n$-good for any $n$.

Hence, we are done.

Note (Alternative) There is an alternative way of proving that $a+b+c$ can only admit $2$ or $3$ as its prime divisors. To see this, suppose $p>3$ divides $a+b+c$. Then, $p\mid 3abc$ implies $p\mid abc$. Using $(a,b,c)=1$, we see that exactly one of $a,b,c$ is divisible by $p$. Suppose, it is $a$. Then, $p\mid ab+bc+ca$, together with $p\mid ab+ac$ implies $p\mid bc$, which clearly is a contradiction. From here, one can finish in exact same way as in above proof, i.e., prove $4,9\nmid a+b+c$, and finish.