乔·希普曼证明了标记的尺规可以求解一般五次方程。
1 分•作者: jjgreen•大约 3 小时前
我在 FOM 邮件列表中看到了这个，不幸的是存档已失效，所以原文如下：
我刚刚证明了标记的尺和圆规可以解决一般的五次方程。
康威和我经常谈论这个问题。他看到我终于找到了他确信存在的构造，一定会非常高兴。
需要一次 Tschirnhaus 变换来消除 x² 和 x⁴ 项，一次双重 neusis（使用圆规作为分规，半径与尺上的标记相同），以及一些平方根。
Claude Opus 和 ChatGPT Sol 在代数方面提供了很大帮助，如果我是一名终身教授，也许一年就能完成这项工作，而无需他们的帮助，但我从未有过那样的一年。难点在于密集地运用代数几何来弄清楚所有搜索失败的原因，这样我才能最终沿着正确的构造类型进行搜索。对于代数几何学家来说，能够快速获得成千上万个方程的伽罗瓦群、多项式分解以及更高级的算术信息就足够了，但我还需要学习代数几何，而大型语言模型在这方面非常理想，同时也使算法开发速度提高了大约 10 倍。
查看原文
Saw this on the FOM mailing list, unfortunately the archive is down, so the posting verbatim below<p><i>I just proved that marked ruler and compass solve the general quintic equation.<p>Conway and I often talked about this problem. He’d have been so pleased to see I finally found the construction he was sure was there.<p>Needs one Tschirnhaus transformation to remove x^2 and x^4 terms, one double neusis using compass as a divider with same unit radius as the marks on the rules, and a bunch of square roots.<p>Claude Opus and ChatGPT Sol helped a lot with the algebra, if I’d been a tenured professor I maybe could have done it in a year of work without them, but I never had that year. The difficulty was using algebraic geometry intensively to figure out why all the searches were failing, so that I could finally search along the right kinds of constructions. Being able to get Galois groups and factorizations of polynomials and more advanced arithmetical information quickly for thousands of equations would have been enough for an algebraic geometer, but I needed to learn the algebraic geometry too and the LLMs were ideal for that, as well as speeding up the algorithm development by 10x or so.
</i>