As a mathematician and long-time sufferer of that Douglas Adams meme:
GIVE A FUCK?
42 is found to be the sum of three cubes
171–180 of 256 posts
Re: 42 is found to be the sum of three cubes
#172For 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…
They mean "unsolved" in the sense that for eg we cant prove or disprove claims like "all integers are the sum of three cubes".
Re: 42 is found to be the sum of three cubes
#173Why should I care?
Re: 42 is found to be the sum of three cubes
#174dumb question: What is the practical application of this conjecture?
Re: 42 is found to be the sum of three cubes
#175Earlier quoted context omitted.
It is an open question as to whether every integer not equal to 4 or 5 modulo 9 is the sum of three cubes, although it is suspected to be true. https://en.wikipedia.org/wiki/Sums_of_three_cubes#Computatio... 33 and 42 were known to be exceptions of all sums less than 100 for which solutions were found, until recently. 42 was the most recent to fall.
Thanks, this is helpful. Why do we care if every integer not equal to 4 or 5 modulo 9 is the sum of three cubes? Just for fun?
Some folks are attracted to ways of thinking which are not subject to opinion.
Re: 42 is found to be the sum of three cubes
#176Earlier 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?
Because simple questions with no known simple answers are fun to study and can lead to actually useful techniques sometimes.
Re: 42 is found to be the sum of three cubes
#177Earlier quoted context omitted.
Exactly! In case everyone has forgotten 42 is the "Answer to the Ultimate Question of Life, the Universe, and Everything" [1] [1] https://en.wikipedia.org/wiki/42_(number)#The_Hitchhiker's_G...
This is one of those things that has been mentioned, used and worn out so much that you feel much cooler when you don't acknowledge it.
You come to HN to be cool!?
Re: 42 is found to be the sum of three cubes
#178For 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…
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
Re: 42 is found to be the sum of three cubes
#179Earlier 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…
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).
Re: 42 is found to be the sum of three cubes
#180Why doesn't this evaluate in Excel? =(-80538738812075974)^3 + 80435758145817515^3 + 12602123297335631^3 Returns 1.09785E+36
Sadly `awk/gwak` fails as well returning
198929873392615583518074640613244928
awk -M 'END{print (-80538738812075974)^3 + 80435758145817515^3 + 12602123297335631^3}' /dev/null
42
but `bc` of course works
echo '(-80538738812075974)^3 + 80435758145817515^3 + 12602123297335631^3' | bc
42