含 $\sqrt{2}$ 的四次阶的毕达哥拉斯数
数论
2025-07-24 v3
摘要
设 为一个含 的四次数域， 为一个满足 的阶。我们证明 的毕达哥拉斯数至多为 5。这证实了 Kr\'{a}sensk\'{y}、Ra\v{s}ka 和 Sgallov\'{a} 的一个猜想。证明用到了 Beli 关于二元局部域上二次格的范数生成元基的理论。
引用
@article{arxiv.2204.10468,
title = {Pythagoras number of quartic orders containing $\sqrt{2}$},
author = {Zilong He and Yong Hu},
journal= {arXiv preprint arXiv:2204.10468},
year = {2025}
}
备注
v3: numbering system changed to be consistent with the published version. Chinese version published in Chinese Ann. Math. Ser. A 45 (2024), no. 3, 287--296.; English translation in Chinese Journal of Contemporary Mathematics, 45 (2024) no. 3