Divisibility Of Positve Integers Suppose a,b and c are three positive integers which satisfy the condition that ($a$2+$b$2+$c$2) is divisible by $(a+b+c)$.  
Prove that there exists infinitely many positive integers $n$  for which ($a$n+$b$n+$c$n) is also divisible by$(a+b+c)$.

We have that $(a^{2^n}+b^{2^n}+c^{2^n})$ is divisible by $(a+b+c)$ for each $n\geq 0$.
The proof is by induction on $n$. The base cases $n=0$ and $n=1$ are given to us; suppose that it is true for $n$ and $n-1$.  Then $$2(a^{2^{n-1}}b^{2^{n-1}}+a^{2^{n-1}}c^{2^{n-1}}+b^{2^{n-1}}c^{2^{n-1}}) = (a^{2^{n-1}}+b^{2^{n-1}}+c^{2^{n-1}})^2 - (a^{2^n}+b^{2^n}+c^{2^n})$$ is also divisible by $(a+b+c)$.
We then have $$(a^{2^{n+1}}+b^{2^{n+1}} + c^{2^{n+1}}) = (a^{2^n}+b^{2^n}+c^{2^n})^2 - 2(a^{2^{n-1}}b^{2^{n-1}}+a^{2^{n-1}}c^{2^{n-1}}+b^{2^{n-1}}c^{2^{n-1}})^2 + 4(a^{2^{n-1}}+b^{2^{n-1}}+c^{2^{n-1}})(a^{2^{n-1}}b^{2^{n-1}}c^{2^{n-1}}),$$
and each term on the right hand side is divisible by $(a+b+c)$ by the above and the inductive hypothesis.
