If $a,b$ and $c$ are odd then any postive integer $n$ works since $a+b+c$ is odd. If two of them are even and one is odd the same happens. We can't have that the three are even since $gcd(a,b,c) = 1$ Assume $a$ is even and $b$ and $c$ is odd. Let $a = 2^x \cdot y$, $y$ is an odd integer. If $k$ is even then $b^k \equiv c^k \equiv 1 (mod 4)$, hence $2$ divides $b^k+c^k$ but $4$ doesn't. Clearly, since $x \ge 1$ and $k\ge 2$, we have that $4$ divides $a^k$. Hence $4$ doesn't divide $a^k+b^k+c^k$. Therefore any $n \ge 2$ will work. Consider the case when $k$ is odd. We have $b^k+c^k = (b+c)(b^{k-1}+...+c^{k-1}) = (b+c) \cdot A$ Clearly, since $A$ is the sum of $k$ odd numbers, since k is odd, $A$ is odd. Hence, if $2^{\alpha} || b+c$ then $2^{\alpha} || b^k+c^k$ for all odd $k$. Therefore we can put $a^k+b^k+c^k = 2^{xk} \cdot y^k + 2^{\alpha} \cdot z_k$ where $z_k$ is an odd number that changes as $k$ changes. We take $n$ greater than $\alpha$. If $xk < \alpha$ then $2^{\alpha}$ doesn't divide $a^k$ and $2^{\alpha}$ divides $b^k+c^k$ so $2^{\alpha}$ doesn't divide $a^k+b^k+c^k$, hence $n$ will work. If $xk > \alpha$ then $2^{\alpha+1}$ divides $a^k$ but $2^{\alpha+1}$ doesn't divide $b^k+c^k$ hence $2^{\alpha+1}$ doesn't divide $a^k+b^k+c^k$ and therefore $n$ will work. Finally, if for a fixed $k$ we have that $xk = \alpha$ then $a^k+b^k+c^k = 2^\alpha (y^k+ z_k)$ If $2^{\beta} || y^k + z_k$ then we should take $n$ such that $n> \alpha + \beta$ which will work in all cases.