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Strangely, typing this into Google does NOT result in 42 ((-80538738812075974)^3)+(80435758145817515^3)+(12602123297335631^3)
python handles big int automatically: sh-4.2$ python -c "print(-80538738812075974 3+80435758145817515 3+12602123297335631 3)" 42
In Python >>> (-80538738812075974)**3 + 80435758145817515**3 + 12602123297335631**3 42
In Google: 1.9892987e+35 It pains me how its 2019 and the fricking Google calculator is less powerful than a solar powered credit-card sized one from the 90s. A single chip made on the earliest imaginable IC processes can effortlessly deal with larger numbers than the lazy ignorant Google implementation. Maybe they can make this an interview question.
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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?
As a prime example, Fermat's Last Theorem was not a particularly applicable mathematical theorem. But when Andrew Wiles found a proof, he ended up proving several other elliptic curve conjectures which were important to the field.
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numerology is not mathematics.
While I do agree with you, I wouldn't be so quick to cast out numerology as a thing of importance. Hippies are pretty decent at sniffing things out, even if they explain them in totally insane ways. EDIT: If you're opposed to this comment, I request that you explain your reasoning. Making the assumption that happiness is of any value, there's actually quite a bit to dig into here.
As for "hippies are pretty decent at sniffing things out", just because someone was correct about something in the past (despite a lack of evidence) does not make them correct today if they still lack evidence.
The burden of proof is on numerologists to show that they are correct, not on everyone else to disprove it.
I’ve just seen that my tweet is here! If anyone wants more details, I wrote a more detailed account, which has just been published at https://aperiodical.com/2019/09/42-is-the-answer-to-the-ques...
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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.
Agreed, though enumeration of triples means you can enumerate sums of cubes of the triples...it just would take literally forever to enumerate them.
For 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 tweet chain gives a type of integer that we've proven cannot be expressed as a sum of three cubes (9k+4 and 9k+5), but that only means we've proven anything for 2/9ths of the integers, and it's only conjectured that the rest can be expressed as such.
There are some trivial other things can we can prove, such as any number that is 3x a cube, or 2x a cube plus/minus another cube. But that doesn't get us to full coverage.
As of yet we haven't proven a result for enough patterns such that every integer belongs to one of those patterns. I suppose to answer your question is that no, we don't have enough small numbers proven and/or we don't have a recursion pattern that covers the integers with the known numbers we have yet.
That's just my layman's understanding though. As it stands now, we do know that there are infinite numbers that cannot be expressed as a sum of three cubes, but we do have a useful finite set of integers that can. So a full solution to the problem will be good to know, but we already know that an answer is "some can, some can't."
Edit: I realize that someone is going to take issue with my changing GP saying enumerable to me saying countable. There's a difference, and I guess it could matter, but I'm too honest to change my original wording. Sorry in advance for getting it wrong.
I’ve just seen that my tweet is here! If anyone wants more details, I wrote a more detailed account, which has just been published at https://aperiodical.com/2019/09/42-is-the-answer-to-the-ques...
Serious question, what is one of the immediate practical applications if this problem (sums of three cubes) is solved? Not a math person so question might sound stupid.
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Please don't use code formatting for quotes. It makes the text unreadable on mobile.
Please don't criticize others for using features of the website you are visiting. If you bring a keyhole to view a landscape you won't get the full experience either. Rendering pages is a matter that should be solved by the client.
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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