tag:blogger.com,1999:blog-7147569507271803579.post4706248407875695712..comments2023-02-28T09:33:41.349-06:00Comments on /dev/ragfield: Expressible as the sum of two cubesRagfieldhttp://www.blogger.com/profile/12325362017984149251noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-7147569507271803579.post-35523020051436438902014-02-21T10:25:37.357-06:002014-02-21T10:25:37.357-06:00I had never heard of mathematica, or even seen thi...I had never heard of mathematica, or even seen this kind of computer-friendly notation, before finding this article.<br /><br />Bender's comment mystified me. I tried to figure out the cubed roots myself using estimation and was amazed at how difficult it was.<br /><br />I gave up and Googled, which led me here. I'm so glad it did. I'm downloading the trial version of mathematica as I type this.ceeciltnoreply@blogger.comtag:blogger.com,1999:blog-7147569507271803579.post-49452342008429452102009-04-06T08:23:00.000-05:002009-04-06T08:23:00.000-05:00Even better. Given that I'm more interested in co...Even better. Given that I'm more interested in computer science than math I've never actually used the Reduce function before.Ragfieldhttps://www.blogger.com/profile/12325362017984149251noreply@blogger.comtag:blogger.com,1999:blog-7147569507271803579.post-73900434277359492432009-04-06T01:24:00.000-05:002009-04-06T01:24:00.000-05:00Reduce[i^3 + j^3 == 3370318, Integers]Reduce[i^3 +...Reduce[i^3 + j^3 == 3370318, Integers]<BR/><BR/>Reduce[i^3 + j^3 == 2716057, Integers]RiemannZetaZerohttps://www.blogger.com/profile/01437218362228076238noreply@blogger.com