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Sum-of-Three-Cubes Problem Solved for ‘Stubborn’ Number 33:
A number theorist with programming prowess has found a solution to 33 = x³ + y³ + z³, a much-studied equation that went unsolved for 64 years.
Mathematicians long wondered whether it’s possible to express the number 33 as the sum of three cubes — that is, whether the equation 33 = x³+ y³+ z³ has a solution. They knew that 29 could be written as 3³ + 1³ + 1³, for instance, whereas 32 is not expressible as the sum of three integers each raised to the third power. But the case of 33 went unsolved for 64 years.
Now, Andrew Booker, a mathematician at the University of Bristol, has finally cracked it: He discovered that (8,866,128,975,287,528)³ + (–8,778,405,442,862,239)³ + (–2,736,111,468,807,040)³ = 33.
Booker found this odd trio of 16-digit integers by devising a new search algorithm to sift them out of quadrillions of possibilities. The algorithm ran on a university supercomputer for three weeks straight. (He says he thought it would take six months, but a solution “popped out before I expected it.”) When the news of his solution hit the internet earlier this month, fellow number theorists and math enthusiasts were feverish with excitement. According to a Numberphile video about the discovery, Booker himself literally jumped for joy in his office when he found out.
Why such elation? Part of it is the sheer difficulty of finding such a solution. Since 1955, mathematicians have used the most powerful computers they can get their hands on to search the number line for trios of integers that satisfy the “sum of three cubes” equation k = x³ + y³ + z³, where k is a whole number. Sometimes solutions are easy, as with k = 29; other times, a solution is known not to exist, as with all whole numbers that leave behind a remainder of 4 or 5 when divided by 9, such as the number 32.
[...]33 was an especially stubborn case: Until Booker found his solution, it was one of only two integers left below 100 (excluding the ones for which solutions definitely don’t exist) that still couldn’t be expressed as a sum of three cubes. With 33 out of the way, the only one left is 42.
Next up is 42? Where have I seen that number before?
(Score: 4, Funny) by driverless on Wednesday March 27 2019, @12:41PM (4 children)
Man, these mathematicians really need to get out more.
(Score: 2) by VLM on Wednesday March 27 2019, @01:22PM
In the dick size war that is academic mathematics, (8,866,128,975,287,528)³ is pr0n star large number material. I can only calculate up to 6 in my head. More caffeine would probably help. I'm not an idiot I figured out 8 pretty easily LOL but I had a mental block or whatever requiring looking up 7 online, which was kinda obvious once I saw it. I think most people could get up to 16 without a computer, 16 is a tough one.
Also there's the primate dominance achievement of bending this monster to your will, regardless how useless the result:
https://www.bristol.ac.uk/acrc/high-performance-computing/bluecrystal-tech-specs/ [bristol.ac.uk]
For some homelab p0rn, here's some images from my basement:
https://www.acrc.bris.ac.uk/gallery/gallery1.htm [bris.ac.uk]
(no actually I'm kidding, thats not my basement, I do have a half rack under my basement stairs harry-potter-bedroom style. I've worked in many facilities vaguely reminiscent of that HPC center. Nice halon plumbing in the background of the pix, that must be like $100K of halon alone.)
(Score: 3, Funny) by slinches on Wednesday March 27 2019, @04:48PM (2 children)
Not sure what took them so long to solve this problem, either. There are an effectively infinite number of solutions and it took me under a minute to come up with this answer:
1^3+2^3+2.8845^3=33.000
That's close enough, at least. Anyone who needs more than five significant digits of precision is measuring the wrong thing.
-- An Engineer
(Score: 2) by opinionated_science on Wednesday March 27 2019, @06:30PM (1 child)
I modded you as funny, as we all know only decimals ".0" count ;-)
(Score: 0) by Anonymous Coward on Thursday March 28 2019, @09:19AM
Easy:
x=y=z=11^(1/3)
See? No decimal places anywhere. ;-)