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.


If $a,b,c,n\in\Bbb Z_{\ge 1}$, $a+b+c\mid a^2+b^2+c^2$, then $$a+b+c\mid a^n+b^n+c^n$$
is true when $n\nmid 3$, but not necessarily when $n\mid 3$.

$$x^2+y^2+z^2+2(xy+yz+zx)=(x+y+z)^2$$
$$\implies x+y+z\mid 2(xy+yz+zx)$$
$$\implies x+y+z\mid (x^k+y^k+z^k)(xy+yz+zx)$$
for all $k\ge 1$ (to see why, check cases when $x+y+z$ is even and when it's odd).
$$x^{n+3}+y^{n+3}+z^{n+3}=(x^{n+2}+y^{n+2}+z^{n+2})(x+y+z)$$
$$-(x^{n+1}+y^{n+1}+z^{n+1})(xy+yz+zx)+(x^n+y^n+z^n)xyz$$
for all $n\ge 1$. We know $$x+y+z\mid (x^{n+2}+y^{n+2}+z^{n+2})(x+y+z)$$
$$-(x^{n+1}+y^{n+1}+z^{n+1})(xy+yz+zx)$$
Now let $(x,y,z)=(x_1,y_1,z_1)=(1,3,9)$. $$x_1+y_1+z_1\nmid x_1^3+y_1^3+z_1^3$$ $$x_1+y_1+z_1\nmid \left(x_1^3+y_1^3+z_1^3\right)x_1y_1z_1$$ $$\implies x_1+y_1+z_1\nmid x_1^6+y_1^6+z_1^6$$
Since $x_1+y_1+z_1$ is coprime to $x_1,y_1,z_1$, we get $$x_1+y_1+z_1\nmid (x_1^6+y_1^6+z_1^6)x_1y_1z_1,$$
and so $x_1+y_1+z_1\nmid x_1^9+y_1^9+z_1^9$, etc.
Therefore $x+y+z$ cannot generally (for all $x,y,z\in\mathbb Z_{\ge 1}$) divide $x^{3m}+y^{3m}+z^{3m}$ for any given $m\ge 1$.
However, we easily get $x+y+z$ always divides $x^n+y^n+z^n$ for $n$ not divisible by $3$,
because $x+y+z\mid (x+y+z)xyz$ and $x+y+z\mid \left(x^2+y^2+z^2\right)xyz$,
because $x+y+z\mid x^2+y^2+z^2$ (given), so $x+y+z\mid x^4+y^4+z^4, x^5+y^5+z^5$,
so $x+y+z\mid \left(x^4+y^4+z^3\right)xyz, \left(x^5+y^5+z^5\right)xyz$,
so $x+y+z\mid x^7+y^7+z^7, x^8+y^8+z^8$, etc.
