Here's a Numberphile video about this news.
https://www.youtube.com/watch?v=zyG8Vlw5aAw
42 is found to be the sum of three cubes
111–120 of 256 posts
Re: 42 is found to be the sum of three cubes
#112What about zero? Can zero be the sum of three cubes? Actually, I read that somebody proved it can't be done, but the proof is too long to fit in this comment. ;-)
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Re: 42 is found to be the sum of three cubes
#113Earlier quoted context omitted.
Mathematicians are interested in which natural numbers k can be expressed as a sum of three cubes. Prior to this year, it had been established that this is possible for all k 33, 42, 114, 165, 390, 579, 627, 633, 732, 795, 906, 921, 975. Earlier this year a solution for k=33 was found [1], so 42 was the next unknown value. [1] Brooker, A., "CRACKING THE PROBLEM WITH 33", https://people.maths.bris.ac.uk/~maarb/papers/…
> Mathematicians are interested in which natural numbers k can be expressed as a sum of three cubes. Why?
That's a question that even Mathematics can not answer
Re: 42 is found to be the sum of three cubes
#114For context, 42 was the only remaining number below 100 where it wasn’t known if this was possible. The general problem of exactly which numbers are the sum of three cubes is unsolved. https://en.m.wikipedia.org/wiki/Sums_of_three_cubes
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…
All integer sets of any length would be enumerable, but that doesn't necessarily mean that one of them can be cubed and added/subtracted to make a given integer.
Re: 42 is found to be the sum of three cubes
#115For context, 42 was the only remaining number below 100 where it wasn’t known if this was possible. The general problem of exactly which numbers are the sum of three cubes is unsolved. https://en.m.wikipedia.org/wiki/Sums_of_three_cubes
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…
The problem is not to enumerate or find the set of numbers that are sums of cubes.
The conjecture is: every integer that is not of the form 9k+4 or 9k+5 can be expressed as the sum of 3 cubes (and, in infinitely many ways).
Re: 42 is found to be the sum of three cubes
#116Clicked hoping for Douglas Adams, got Numberwang
Re: 42 is found to be the sum of three cubes
#117For context, 42 was the only remaining number below 100 where it wasn’t known if this was possible. The general problem of exactly which numbers are the sum of three cubes is unsolved. https://en.m.wikipedia.org/wiki/Sums_of_three_cubes
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…
[deleted]
Re: 42 is found to be the sum of three cubes
#118What about zero? Can zero be the sum of three cubes? Actually, I read that somebody proved it can't be done, but the proof is too long to fit in this comment. ;-)
Yes. 0³+0³+0³. Or x³+(-x)³+0³.
Re: 42 is found to be the sum of three cubes
#119i love bc, just to verify :
bc 1.07.1 Copyright 1991-1994, 1997, 1998, 2000, 2004, 2006, 2008, 2012-2017 Free Software Foundation, Inc. This is free software with ABSOLUTELY NO WARRANTY. For details type `warranty'.
(-80538738812075974)^3 + 80435758145817515^3 + 12602123297335631^3
42
Re: 42 is found to be the sum of three cubes
#120Earlier quoted context omitted.
Mathematicians are interested in which natural numbers k can be expressed as a sum of three cubes. Prior to this year, it had been established that this is possible for all k 33, 42, 114, 165, 390, 579, 627, 633, 732, 795, 906, 921, 975. Earlier this year a solution for k=33 was found [1], so 42 was the next unknown value. [1] Brooker, A., "CRACKING THE PROBLEM WITH 33", https://people.maths.bris.ac.uk/~maarb/papers/…
This isn’t quite right, it’s easy to show that if n is 4 or 5 (mod 9), then there are no integer solutions to a^3 + b^3 + c^3 = n. The conjecture is that the mod 9 obstruction is the only one and all other integers can be represented as the sum of 3 cubes.
Thanks, that was very sloppy of me!