查看“︁一般化的士數”︁的源代码

{{noteTA
|T=zh-hans:一般化的士数;zh-hk:一般化的士數;zh-tw:一般化計程車數;
|1=zh-hans:的士;zh-hk:的士;zh-tw:計程車;
}}
{{onesource|time=2019-04-21}}
{{Unsolved|數學|是否存在一正整數能夠用最少兩種方法表示成兩正整數的五次方之和，即 ''a''<sup>5</sup> + ''b''<sup>5</sup> <nowiki>=</nowiki> ''c''<sup>5</sup> + ''d''<sup>5</sup>?}}在[[數學]]中,'''一般化的士數'''''Taxicab''(''k'', ''j'', ''n'') 定義為一最小的數，能夠用n種方法表示成j個[[自然數]]的k次方之和。 若 ''k'' = 3 且 ''j'' = 2, 是為[[的士數]]。
: <math>\mathrm{Taxicab}(1, 2, 2) = 4 = 1 + 3 = 2 + 2.</math>

: <math>\mathrm{Taxicab}(2, 2, 2) = 50 = 1^2 + 7^2 = 5^2 + 5^2.</math>

: <math>\mathrm{Taxicab}(2, 2, 3) = 325 = 1^2 + 18^2 = 6^2 + 17^2 = 10^2 + 15^2.</math>

: <math>\mathrm{Taxicab}(3, 2, 2) = 1729 = 1^3 + 12^3 = 9^3 + 10^3</math>

: <math>\mathrm{Taxicab}(3, 3, 2) = 251 = 1^3 + 5^3 + 5^3= 2^3 + 3^3 + 6^3</math>
[[萊昂哈德·歐拉|歐拉]]證明了
: <math>\mathrm{Taxicab}(4, 2, 2) = 635318657 = 59^4 + 158^4 = 133^4 + 134^4.</math>
然而, ''Taxicab''(5, 2, ''n'')在''n'' &#x2265; 2時尚未被找到; 也就是說，還沒找到任何[[正整數]]可以用多於一種方法表示成2個正整數的5次方之和。<ref>{{Cite book|url=https://books.google.com/books?id=1AP2CEGxTkgC&printsec=frontcover#v=onepage&q=&f=false|title=Unsolved Problems in Number Theory|last=Guy|first=Richard K.|authorlink=Richard K. Guy|date=2004|publisher=Springer-Science+Business Media, Inc.|isbn=0-387-20860-7|edition=Third|location=New York, New York, USA|access-date=2016-12-11|archive-date=2020-12-03|archive-url=https://web.archive.org/web/20201203034631/https://books.google.com/books?id=1AP2CEGxTkgC&printsec=frontcover#v=onepage&q=&f=false|dead-url=no}}</ref>

== 相關條目 ==
* [[士的數]]

== 參見 ==
<references />

== 外部連結 ==
* [https://web.archive.org/web/20150513224105/http://homepage.ntlworld.com/duncan-moore/taxicab/ Generalised Taxicab Numbers and Cabtaxi Numbers]
* [http://euler.free.fr/taxicab4.htm Taxicab Numbers - 4th powers]{{Wayback|url=http://euler.free.fr/taxicab4.htm |date=20200720093011 }}
* [https://web.archive.org/web/20050425023736/http://www.wschnei.de/number-theory/taxicab-numbers.html Taxicab numbers] by Walter Schneider
[[Category:数论]]