Divisibility condition implies $a_1=\dotsb=a_k$? Let $a_1, a_2, \dotsc, a_k$ be $k$ positive integers, $k\ge2$. For all $n\ge n_0$ there is a positive integer $f(n)$ such that $n$ and $f(n)$ are relatively prime and $a_{1}^{f(n)}+\dotsb+a_{k}^{f(n)}$ is a multiple of $a_{1}^n+\dotsb+a_{k}^n$.
Is it true that $a_1=a_2=\dotsb=a_k$ or is it possible to construct a counterexample?
 A: As a partial solution,  $a_1=a_2=1$, $a_3=2$ is a "counterexample for $n$ odd" in the sense that in this case there exists $N$ co-prime with $n$ such that $a_1^N+a_2^N+a_3^N$ is a multiple of $a_1^n+a_2^n+a_3^n$.
Specifically, given an odd positive integer $n$, fix arbitrarily an even integer $k>0$ co-prime with $n$ and let $N:=(k+1)(n-1)+1$. Then $\gcd(N,n)=(-k,n)=1$ and $a_1^N+a_2^N+a_2^N = 2+2^N = 2(1+2^{(k+1)(n-1)})$ which is divisible by $2(1+2^{n-1})=a_1^n+a_2^n+a_3^n$.
A: Here's a a tweak of Seva's idea that gives a counterexample. Note that if $r$ is odd, then $2^{n}+1$ divides $2^{rn} + 1$.
Let $k = 6$, $a_{1} = 1$, $a_{2} = a_{3} = a_{4} = 2$, $a_{5} = a_{6} = 4$. Then $a_{1}^{n} + \cdots + a_{6}^{n} = 1 + 3 \cdot 2^{n} + 2 \cdot 4^{n} = (1+2^{n})(1+2^{n+1})$.
If $n$ is any positive integer, let $f(n) = 2n^{2} + n - 1$. We see that $f(n) \equiv -1 \pmod{n}$ and so $\gcd(n,f(n)) = 1$. Now,
$$a_{1}^{f(n)} + \cdots + a_{6}^{f(n)} = (1+2^{2n^{2} + n - 1})(1+2^{2n^{2} + n}).$$
The second factor is $2^{n(2n+1)}+1$ and so is a multiple of $2^{n} + 1$. On the other hand, the first factor is $2^{(n+1)(2n-1)}+1$ and so it is a multiple of $2^{n+1} + 1$. Thus, $a_{1}^{f(n)} + \cdots + a_{6}^{f(n)}$ is a multiple of $a_{1}^{n} + \cdots + a_{6}^{n}$.
