Leningrad Mathematical Olympiad $1991$

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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.

Someone gives me those two hints :

Hint 1: Let the generating function $f(x) = \sum_{i=1}^n a_i x^i$. If $a_1,\dots,a_n$ is $p$-balanced, then $f(e^{2\pi i/p}) = 0$.

Hint 2: $(3-1) + (5-1) + (7-1) + (11-1) + (13-1) + (17-1) = 50$.

Hint 3 : Let $z_k = e^{2k\pi i/p}$ . Then $z_k^{j+p} = z_k^j$ . Therefore, the $p$-balancedness implies $f(z_k) = S\sum_{j=1}^p z_k^j = 0$

Here it is probably $e^{2k\pi i/(p+1)}$ instead of $e^{2k\pi i/p}$ cause $k = 1, 2 ,..., p$.

Could anyone register the complete solution? I spent way too much time doing this, and yet I still have not managed to get a justifying answer to this question. I managed to get a partial response, but no global answer.

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We'll let $a_0=0$ for slight simplification. This won't affect any of the proof. As in your hint, let $$ f(x)=a_0+a_1x+\cdots+a_nx^n $$ be the representative generating function. Now suppose $z_p$ is a primitive $p$th root of unity, $z_p=e^{2\pi ik/p}$ (for some $1\le k \le p-1$), for which the sequence is $p$-balanced. Suppose also that $$a_0+a_p+\cdots=a_1+a_{p+1}+\cdots = \cdots = a_{p-1}+a_{2p-1}+\cdots = S$$ for some $S$. Then $$ \begin{align} f(z_p)&=(a_0+a_p+\cdots)+(a_1+a_{p+1}+\cdots)z_p+\cdots + \cdots + (a_{p-1}+a_{2p-1}+\cdots)z_p^{p-1}\\ &=S+Sz_p+\cdots + Sz_p^{p-1}\\&=S(1+z_p+\cdots + z_p^{p-1})\\&=0 \end{align} $$ Now for a given $p$, there are $p-1$ primitive roots of unity, so the polynomial $f$ has at least $$ (3-1)+(5-1)+(7-1)+(11-1)+(13-1)+(17-1)=50 $$ zeroes in these roots of unity. But $0$ is also a root of $f$ since $a_0=0$, so $f$ has $51$ zeroes and degree at most $50$, implying it is the zero polynomial.