42 is found to be the sum of three cubes

Post by [deleted] » Fri, Sep 06, 2019, 7:42 PM UTC

normalhuman wrote: post #135

mombul wrote: post #98

Earlier quoted context omitted.

Why are you so bitter? I asked this genuinely. I doubt PhD+ people do that type of research "just for fun", or do that like someone like me drinks a beer. Someone answered elaborately, and I'm glad they did. Thanks anyway.

> Why are you so bitter? I don't know, it's just how I am. > I doubt PhD+ people do that type of research "just for fun" Well you'd be surprised. All the professional mathematicians I know do math for its own sake. Some amateurs also do it "just for fun", as you say. > Someone answered elaborately, and I'm glad they did. Thanks anyway. A bigger person than me, especially considering the tone of your question.

Post by mombul » Fri, Sep 06, 2019, 7:43 PM UTC

normalhuman wrote: post #135

mombul wrote: post #98

Earlier quoted context omitted.

Why are you so bitter? I asked this genuinely. I doubt PhD+ people do that type of research "just for fun", or do that like someone like me drinks a beer. Someone answered elaborately, and I'm glad they did. Thanks anyway.

> Why are you so bitter? I don't know, it's just how I am. > I doubt PhD+ people do that type of research "just for fun" Well you'd be surprised. All the professional mathematicians I know do math for its own sake. Some amateurs also do it "just for fun", as you say. > Someone answered elaborately, and I'm glad they did. Thanks anyway. A bigger person than me, especially considering the tone of your question.

Post by jjtheblunt » Fri, Sep 06, 2019, 7:44 PM UTC

andrepd wrote: post #198

jjtheblunt wrote: post #103

Earlier quoted context omitted.

Isn't the set of all triples of integers be enumerable, because (inductively) the set of all pairs of integers is enumerable (thanks, Cantor)? Then, if one could enumerate all triples of integers, one could, for each triple, calculate the sum of the cubes. So, integers which are sums of cubes are enumerable. That doesn't mean they're a known recursive set, just recursively enumerable. Am I missing something? Perhaps…

So what you're saying is (with needlessly complicated terms) that we can brute force the search for sums of three cubes?

Post by leecarraher » Fri, Sep 06, 2019, 7:55 PM UTC

erikig wrote: post #186

ColanR wrote: post #178

Earlier quoted context omitted.

It is enumerable, but I think what you may have missed is that negative numbers are included. i.e., > 42 = (-80538738812075974)^3 + 80435758145817515^3 + 12602123297335631^3

Strangely, typing this into Google does NOT result in 42 ((-80538738812075974)^3)+(80435758145817515^3)+(12602123297335631^3)

Post by _cbsz » Fri, Sep 06, 2019, 8:10 PM UTC

robrenaud wrote: post #179

ChrisLomont wrote: post #162

Earlier quoted context omitted.

It is not computationally tractable to enumerate all triples of integers up to the size required to find the values for 42. So certainly any integer N either is or is not the sum of three cubes. Deciding which is the case computationally can be intractable. Each of the integers in the solution are ~53 bits in length. To enumerate all triples of integers (with plus and minus) requires testing around 2^(53+53+53+1+1+1)…

You only need to enumerate pairs of integers, the 3rd is then forced (quickly check if it's a cube).