If $ a + b + c \mid a^2 + b^2 + c^2$ then $ a + b + c \mid a^n + b^n + c^n$ for infinitely many $n$ Let $ a,b,c$ positive integer such that $ a + b + c \mid a^2 + b^2 + c^2$.
  Show that $ a + b + c \mid a^n + b^n + c^n$ for infinitely many positive integer $ n$.


(problem composed by Laurentiu Panaitopol)

So far no idea.
Claim. $a+b+c\mid a^{2^n}+b^{2^n}+c^{2^n}$ for all $n\geq0$.


Proof. By induction: True for $n=0,1$ $\checkmark$. Suppose it's true for $0,\ldots,n$. Note that $$a^{2^{n+1}}+b^{2^{n+1}}+c^{2^{n+1}}=(a^{2^n}+b^{2^n}+c^{2^n})^2-2(a^{2^{n-1}}b^{2^{n-1}}+b^{2^{n-1}}c^{2^{n-1}}+c^{2^{n-1}}a^{2^{n-1}})^2+4a^{2^{n-1}}b^{2^{n-1}}c^{2^{n-1}}(a^{2^{n-1}}+b^{2^{n-1}}+c^{2^{n-1}})$$

and that 

$$2(a^{2^{n-1}}b^{2^{n-1}}+b^{2^{n-1}}c^{2^{n-1}}+c^{2^{n-1}}a^{2^{n-1}})=(a^{2^{n-1}}+b^{2^{n-1}}+c^{2^{n-1}})^2-(a^{2^n}+b^{2^n}+c^{2^n})$$

is divisible by $a+b+c$ by the induction hypothesis.
For any positive integer x this is true: $x \leqslant x^2$ (From $1 \leqslant x$ for any positive ineger x ). So $a + b + c \leqslant a^2 + b^2 + c^2$. But for 2 positive integers $x$, $y$ $x$ is divisible by $y$ only if $x \geqslant y$. So $a + b + c \geqslant a^2 + b^2 + c^2$ if $a + b + c \mid a^2 + b^2 + c^2$. 
From these 2 inequalities: 
$a + b + c = a^2 + b^2 + c^2$

So  $a + b - a^2 - b^2 = c^2 - c$

Evaluation for the right part $c^2 - c \geqslant 0$ (because $x \leqslant x^2$ for any positive integer x).
Evaluation for the left part $a + b - a^2 - b^2 \leqslant 0$ (adding 2 inequalities $a - a^2 \leqslant 0$ and $b - b^2 \leqslant 0$)

So the left part is $\leqslant 0$ and the right part $\geqslant 0$. But they are equal so $a + b - a^2 - b^2 = c^2 - c = 0$ and $c^2 = c$. c is positive so we can divide both part of the last equation by c and get $c = 1$.

Similarly $b = 1$ and $a = 1$. So $a + b + c = 3$ and $a^n + b^n + c^n = 3$ for any positive $n$. 3 is divisible by 3 so $a + b + c \mid a^n + b^n + c^n$ for infinitely many positive integer n.