221–230 of 256 posts
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Agreed, though enumeration of triples means you can enumerate sums of cubes of the triples...it just would take literally forever to enumerate them.
You can enumerate all syntactically valid proofs too, and just check each one to see if it proves whatever proposition you’re curious about.
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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)
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python handles big int automatically: sh-4.2$ python -c "print(-80538738812075974 3+80435758145817515 3+12602123297335631 3)" 42
nah, bigint is just another wall you may hit bc/dc arbitrary precision calculators are usually installed by default on *nix platforms and they handle numbers with thousands of digits $ time echo '12345^67890' |bc |wc -c 285941 real 0m2.137s user 0m2.128s sys 0m0.012s
(-80538738812075974)^3 + 80435758145817515^3 + 12602123297335631^3 is 42?
42?!
I always said there was something fundamentally wrong with the universe.
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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.
I think this falls under the mathematical category of "fun things you can do with numbers", and not much else. Just cool.
I must be missing something. Why is this significant?
Same question. People seem to be researching that kind of stuff (numbers that are sums of 3 cubes). Why?
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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)
(-80538738812075974n)**3n + 80435758145817515n**3n + 12602123297335631n**3nEarlier quoted context omitted.
Ok, we've switched the URL from https://math.mit.edu/~drew/ and removed the housing from 42 above.
why did you switch from the mit site to twitter ? (i'm currently blocked on twitter - they want a phone number and are blocking all content till i provide one)
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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…
They mean "unsolved" in the sense that for eg we cant prove or disprove claims like "all integers are the sum of three cubes".
Every cube is within one of a multiple of nine, which means every sum of three cubes is within three of a multiple of nine. So numbers of the form 9k+4 or 9k+5 cannot be expressed as the sum of three cubes.
It’s conjectured that every other whole number can be.
https://twitter.com/robinhouston/status/1169938974246342658?...
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…
You can enumerate a,b pairs and then you need to check whether the "locked in" value of c^3 is a cube.
However imagine it takes 1ns to validate a given pair [a,b].
The eventual solution was [-80538738812075974, 80435758145817515, 12602123297335631].
Since no combination of 3 positive (and therefore small) numbers has worked, we know that one of a,b,c are negative. Let's assume at least one of a,b are negative since it doesn't matter how we allocate them.
To reach the final pair of a = 80435758145817515 (the smaller positive integer) and b = -80538738812075974, you have to increment "a" (starting from 0) 80435758145817515 times and decrement "b" (starting from 0) 80538738812075974 times.
That is 80538738812075974*80435758145817515 possible combinations.
Let's assume each one takes 1 ns (which I believe is fairly optimistic at least for a single machine)
That results in a runtime of 6.5e+24 seconds, aka 2.1e+17 years. No matter how many machines you add, the brute force approach does not appear to be feasible.
I am interested to learn more about how they solved it if not brute force.