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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?

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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: $(3-1) + (5-1) + (7-1) + (11-1) + (13-1) + (17-1) = 50$.

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    $\begingroup$ Hint 1b: under that assumption, actually $f(e^{2k\pi i/p})=0$ for $k=1,\ldots,p-1$. $\endgroup$ – Tad Oct 23 '15 at 11:24
  • $\begingroup$ had we but world enough, and time, this coyness, [matey], were no crime. ;) $\endgroup$ – Tad Oct 23 '15 at 22:57
  • $\begingroup$ Are you able to explain the origin of $f(e^\frac{{2k\pi i}}{p})$?... Just roughly. $\endgroup$ – user230283 Oct 25 '15 at 12:58
  • $\begingroup$ That is, I would like you to explain how you got roughly $f(e^\frac{{2k\pi i}}{p}) = 0$ , and where did you get this information (theoretically). $\endgroup$ – user230283 Oct 25 '15 at 15:38
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    $\begingroup$ @J.G, let $z_k = e^{2k\pi i/p}$. Then $z_k^{j+p} = z_k^j$. Therefore, the $p$-balancedness implies $f(z_k) = S\sum_{j=1}^p z_k^j = 0$. $\endgroup$ – zhoraster Oct 25 '15 at 16:22

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