AI 中文总结
研究最小的非3倍数的次数d，使得存在d次代数数的三个共轭数之和为零，证明该d=20，还推导了关于素数幂次数代数数共轭线性关系系数和的整除性结论等。
AI 中文摘要
设d是最小的正整数，且不是3的倍数，使得存在有理数域上次数为d的代数数α，其三个代数共轭数之和为零。我们证明d=20。这一结论由如下结果推导而来：对于素数p、m≥1，次数d=p^m的代数数的共轭数α_j满足的任意线性关系∑_{j=1}^d a_j α_j=0（其中系数a_j∈ℤ），和∑_{j=1}^d a_j可被p整除；若d=2p^m，p≥3且∑_{j=1}^d |a_j| < p，则∑_{j=1}^d a_j为偶数。
英文摘要
Let $d$ be the smallest positive integer, not a multiple of $3$, for which there exists an algebraic number $\al$ of degree $d$ over $\mathbb{Q}$ whose three algebraic conjugates add to zero. We prove that $d=20$. This is derived from the following result: for any linear relation $\sum_{j=1}^d a_j \al_j=0$ with coefficients $a_j\in\mathbb{Z}$ among the conjugates $\al_j$ of an algebraic number of degree $d=p^m$, where $p$ is a prime number, $m \geq 1$, the sum $\sum_{j=1}a_j$ is divisible by $p$. If $d=2p^m$, $p\geq 3$ and $\sum_{j=1}^d|a_d| < p$, then $\sum_{j=1}a_j$ is an even number.
