Problem A. 917. (November 2025)
A. 917. Let a set \(\displaystyle S\) of complex numbers be called symmetric if the complex conjugate of every element of \(\displaystyle S\) also belongs to \(\displaystyle S\). For each positive integer \(\displaystyle n\), determine the largest positive integer \(\displaystyle K_n\) for which there exists an \(\displaystyle n\)-element symmetric set \(\displaystyle S = \{z_1, z_2, \ldots, z_n\}\) not containing \(\displaystyle 0\), such that \(\displaystyle z_1^k + z_2^k + \ldots + z_n^k \leq 0\) holds true for every integer \(\displaystyle 1 \leq k \leq K_n\).
Proposed by Navid Safaei, Tehran and Géza Kós, Budapest
(7 pont)
Deadline expired on December 10, 2025.
Solution. We will show that \(\displaystyle K_n=2n-1\).
For every integer \(\displaystyle k\), let
\(\displaystyle s_k=r_1^k+\ldots+r_n^k. \)
First we construct a symmetric set \(\displaystyle S=\{r_1,r_2,\ldots,r_n\}\) with \(\displaystyle {s_1,s_2,\ldots,s_{2n-1}\le0}\); this example proves \(\displaystyle K_n\ge 2n-1\).
Let
\(\displaystyle r_j = e^{(2j-1)\pi i/n} =\cos\dfrac{(2j-1)\pi i}{n} +i\sin\dfrac{(2j-1)\pi i}{n} \qquad (j=1,2,\ldots,n), \)
so \(\displaystyle r_1,r_2,\ldots,r_n\) are the complex \(\displaystyle (2n)\)th roots of unity with odd indices.
Notice that for every integer \(\displaystyle k\), the numbers \(\displaystyle r_1^k,r_2^k,\ldots,r_n^k\) form a cyclic geometric progression with quotient \(\displaystyle e^{2k\pi i/n}\). If \(\displaystyle k\) is not divisible by \(\displaystyle n\), then the quotient differs from \(\displaystyle 1\), and the sum is zero. On the other hand, if \(\displaystyle n|k\) then each \(\displaystyle r_j^k\) equals \(\displaystyle (-1)^{k/n}\). Hence,
\(\displaystyle s_k = \begin{cases} 0 & \text{if \(\displaystyle k\not\equiv0\pmod{n}\),} \\ -n & \text{if \(\displaystyle k\equiv n\pmod{2n}\),} \\ +n & \text{if \(\displaystyle k\equiv 0\pmod{2n}\).} \\ \end{cases} \)
Therefore, none of \(\displaystyle s_1,s_2,\ldots,s_{2n-1}\) is positive (but so is \(\displaystyle s_{2n}\)).
Now we show that there is at least one positive among \(\displaystyle s_1,s_2,\ldots,s_{2n}\); that proves \(\displaystyle K_n<2n\).
Let
\(\displaystyle \prod_{j=1}^n(z-r_j)=a_0z^n+a_1z^{n-1}+\ldots+a_{n-1}z+a_n, \quad a_0=1,\text{ and } a_{n+1}=a_{n+2}=\ldots=0. \)
By Newton'identities or Newton–Girard formulas, for every positive integer \(\displaystyle N\) we have
| \(\displaystyle \sum_{k=1}^N a_{N-k}s_k + Na_N = 0. \) | \(\displaystyle (1) \) |
We will consider two cases.
Case 1: At least one of \(\displaystyle a_1,a_2,\ldots,a_n\) is negative.
Let \(\displaystyle 1\le N\le n\) be the first index with \(\displaystyle a_N<0\). By \(\displaystyle (1)\) we have
\(\displaystyle \sum_{k=1}^N a_{N-k}s_k = -Na_N >0, \)
so there is at least one positive term \(\displaystyle a_{N-k}s_k\) on the left-hand side. Due to the minimality of \(\displaystyle N\), we know that \(\displaystyle a_{N-k}\ge0\), forcing \(\displaystyle s_k>0\). Note that \(\displaystyle 1\le k\le N\le n\), so in this case there is at least one positive number among \(\displaystyle s_1,\ldots,s_n\).
Case 2: \(\displaystyle a_0,a_1,\ldots,a_n\ge0\).
By Viète's formulas, \(\displaystyle |a_n|=|r_1\cdots r_n|\ne0\), so \(\displaystyle a_n>0\). Moreover, \(\displaystyle a_{n+1}=a_{n+2}=\dots=0\) by definition.
By applying (1) with \(\displaystyle N=n\) we obtain
\(\displaystyle \sum_{k=1}^n a_{n-k}s_k = -na_n < 0, \)
so there is at least one negative among \(\displaystyle s_1,\ldots,s_n\).
Chose a index \(\displaystyle 1\le\ell\le n\) with \(\displaystyle s_\ell<0\), and apply (1) with \(\displaystyle N=(n+\ell)\):
\(\displaystyle \sum_{k=1}^{n+\ell} a_{n+\ell-k}s_k + (n+\ell)a_{n+\ell} = 0. \)
The last term is \(\displaystyle 0\) because \(\displaystyle a_{n+\ell}=0\). Further, the term \(\displaystyle a_ns_\ell\) is negative. So, the sum must contain a positive term \(\displaystyle a_{n+\ell-k}s_k\). Since \(\displaystyle a_{n+\ell-k}\ge0\), this ensures that \(\displaystyle s_k>0\).
Now we have \(\displaystyle 1\le k\le n+\ell\le 2n\), so found a positive number among \(\displaystyle s_1,\ldots,s_{2n}\).
Statistics:
21 students sent a solution. 7 points: Ali Richárd, Aravin Peter, Bodor Mátyás, Bolla Donát Andor, Diaconescu Tashi, Gyenes Károly, Li Mingdao, Morvai Várkony Albert, Szakács Ábel, Vigh 279 Zalán, Xiaoyi Mo. 6 points: Forrai Boldizsár. 4 points: 1 student. 2 points: 4 students. 0 point: 3 students. Unfair, not evaluated: 1 solutions.
Problems in Mathematics of KöMaL, November 2025