一个算术定理及其证明
An arithmetic theorem and its demonstration
摘要：
翻译自拉丁原文，“Theorema arithmeticum eiusque demonstratio”，Commentationes arithmeticae collectae 2（1849），588-592。E794在Enestroem指数。对于m个不同的数a，b，c，d，...，\upsilon，x本文计算了\[ \frac{a^n}{(a-b)(a-c)(a-d)...(a-x)}+\frac{b^n}{(b-a)(b-c)(b-d)...(b-x)} +...+\frac{x^n}{(x-a)(x-b)(x-c)...(x-\upsilon)}. \]当$n \leq m-2$，和为0时，欧拉已经在他的Institutiones calculi integralis，vol. I，E366的第1169节中证明了这一点。当为$n=m-1$时，总和为$=1$。当$n \geq m$，然后使用牛顿恒等式，欧拉得到了关于数a，b，c，d，.的幂和的表达式，\upsilon，x.本文中使用的符号令人困惑（ing/ed），事实上，Opera omnia的编辑对原始出版物进行了几次更正。在论文的最后，“对于每一个人，我们都在寻找他的性格。“这是我唯一能想到的句子引用这个定理的作品的例子是J. P. Serre，Corps locaux，第2版，第三章第6节（适用于$n \leq m-1$）和D. E. Knuth，The Art of Computer Programming，vol. 1，sect. 1.2.3，exercise 33.
英文摘要：
Translated from the Latin original, "Theorema arithmeticum eiusque demonstratio", Commentationes arithmeticae collectae 2 (1849), 588-592. E794 in the Enestroem index. For m distinct numbers a,b,c,d,...,\upsilon,x this paper evaluates \[ \frac{a^n}{(a-b)(a-c)(a-d)...(a-x)}+\frac{b^n}{(b-a)(b-c)(b-d)...(b-x)} +...+\frac{x^n}{(x-a)(x-b)(x-c)...(x-\upsilon)}. \] When $n \leq m-2$, the sum is 0, which Euler had already shown in sect. 1169 of his Institutiones calculi integralis, vol. I, E366. When $n=m-1$ the sum is $=1$. When $n \geq m$, then using Newton's identities Euler gets expressions in terms of sums of powers of the numbers a,b,c,d,...,\upsilon,x. The notation used in this paper is confus(ing/ed), and indeed the editors of the Opera omnia made several corrections to the original publication. Near the end of the paper, "... for each of them one searches for what the character will be..." is the only reading of the sentence I can come up with. Examples of a work that refers to this theorem is J.-P. Serre, Corps locaux, 2nd ed., ch. III, sect. 6 (for $n \leq m-1$), and D. E. Knuth, The Art of Computer Programming, vol. 1, sect. 1.2.3, exercise 33.
机构、总结、中文摘要均由AI分析而来，可能存在错误，请以论文正文为准。
