A finite sequence $a_1, a_2, ..., a_n$ is called $p$-balanced if any sum of the form $a_k+a_{k+p} + a_{k+2p}+...$ is the same for any $k = 1, 2, 3, ..., p$. For instance the sequence $a_1 = 1$, $a_2 = 2$, $a_3 = 3$, $a_4 = 4$, $a_5 = 3$, $a_6 = 2$ is a $3$-balanced because $a_1 + a_4= a_2 + a_5 = a_3+a_6 = 5$. Prove that if a sequence with $50$ members is $p$-balanced for $p=3,5,7,11,13, 17$, then all its members are equal zero.
Intuitively, I think I have to use a generating function. The problem is that I don't know how to start de problem. Is anyone is able to give me a good hint?
Hint 1: Let $f(x) = \sum_{i=1}^n a_n x^n$. If $a_1,\dots,a_n$ is $p$-balanced, then $f(e^{2\pi i/p}) = 0$.
Hint 2: $(3-1) + (5-1) + (7-1) + (11-1) + (13-1) + (17-1) = 50$.