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Math Technology

Two Mathematicians Solve Old Math Riddle, Possibly the Meaning of Life (livescience.com) 93

Posted by BeauHD from the you-can-rest-easy-now dept.
pgmrdlm shares a report from Live Science: In Douglas Adams' sci-fi series "The Hitchhiker's Guide to the Galaxy," a pair of programmers task the galaxy's largest supercomputer with answering the ultimate question of the meaning of life, the universe and everything. After 7.5 million years of processing, the computer reaches an answer: 42. Only then do the programmers realize that nobody knew the question the program was meant to answer. Now, in this week's most satisfying example of life reflecting art, a pair of mathematicians have used a global network of 500,000 computers to solve a centuries-old math puzzle that just happens to involve that most crucial number: 42.

The question, which goes back to at least 1955 and may have been pondered by Greek thinkers as early as the third century AD, asks, "How can you express every number between 1 and 100 as the sum of three cubes?" Or, put algebraically, how do you solve x^3 + y^3 + z^3 = k, where k equals any whole number from 1 to 100? This deceptively simple stumper is known as a Diophantine equation, named for the ancient mathematician Diophantus of Alexandria, who proposed a similar set of problems about 1,800 years ago. Modern mathematicians who revisited the puzzle in the 1950s quickly found solutions when k equals many of the smaller numbers, but a few particularly stubborn integers soon emerged. The two trickiest numbers, which still had outstanding solutions by the beginning of 2019, were 33 and -- you guessed it -- 42. Using a computer algorithm to look for solutions to the Diophantine equation with x, y and z values that included every number between positive and negative 99 quadrillion, mathematician Andrew Booker, of the University of Bristol in England, found the solution to 33 after several weeks of computing time.

Since his search turned up no solutions for 42, Booker enlisted the help of Massachusetts Institute of Technology mathematician Andrew Sutherland, who helped him book some time with a worldwide computer network called Charity Engine. "Using this crowdsourced supercomputer and 1 million hours of processing time, Booker and Sutherland finally found an answer to the Diophantine equation where k equals 42," reports Live Science. The answer: (-80538738812075974)^3 + (80435758145817515)^3 + (12602123297335631)^3 = 42.
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Two Mathematicians Solve Old Math Riddle, Possibly the Meaning of Life More

Two Mathematicians Solve Old Math Riddle, Possibly the Meaning of Life

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  • What? (Score:5, Funny)

    by NewtonsLaw ( 409638 ) on Wednesday September 11, 2019 @09:41PM (#59183768)

    The answer: (-80538738812075974)^3 + (80435758145817515)^3 + (12602123297335631)^3 = 42.

    Hell, I thought everyone knew that. Bah!

    • 42. Hell, I thought everyone knew that.

      What is interesting is how that number came out in a computerless epoch.

      • The answer: (-80538738812075974)^3 + (80435758145817515)^3 + (12602123297335631)^3 = 42.

        Hell, I thought everyone knew that. Bah!

        Ok Smartypants, the next unsolved number is 114. So what's the solution?

        It is conjectured that any integer can be represented as the sum of three cubes [wikipedia.org].

        • It is conjectured that any integer can be represented as the sum of three cubes

          I'm quite sure that 4 and 5 can't. Well, not as the sums of cubes of integers anyway.

          A proof shouldn't take you more than ten seconds.

          • There is an advanced concept in mathematics called the number line. Integers can go below zero and you get negative numbers; there's one in the solution for 42 even.

          • Re: (Score:2)

            by Rhipf ( 525263 )

            Actually you would be wrong and didn't read the summary. The whole sequence of numbers from 1-100 (this includes 4 and 5 by the way) have already been solved as x^3+y^3+z^3=k (i.e. k equals any number from 1-100 including 4 and 5).

            • Re: (Score:2)

              by Dahan ( 130247 )

              Actually you would be wrong and didn't read the summary. The whole sequence of numbers from 1-100 (this includes 4 and 5 by the way) have already been solved as x^3+y^3+z^3=k (i.e. k equals any number from 1-100 including 4 and 5).

              Or maybe the summary is wrong? I know that's really unlikely on an esteemed site such as /., but there's always a first time, right??

              It is in fact impossible for the sum of the cubes of three integers to total 4 or 5 (or any number that has remainder 4 or 5 when divided by 9). We now have solutions to the equation where k is from 1-100 excluding the ones where k mod 9 is 4 or 5

              • Definitely not. The whole point of the search is that 33 and 42 were the smallest integers for which there is no solution to that specific equation using integers. One point to keep in mind is that nobody said they had to be *positive* integers, so you can add cubes of both positive and negative numbers to get small values such as 4 or 5.

        • So if the ancient question is "How can you express every number between 1 and 100 as the sum of three cubes?"

          The ancient answer should be why the hell do you want to? Now we've figured this out what can we do with it?

          • The ancient answer should be why the hell do you want to? Now we've figured this out what can we do with it?

            Well, knowledge of obscure mathematical trivia is great for impressing chicks at cocktail parties.

          • So if the ancient question is "How can you express every number between 1 and 100 as the sum of three cubes?"

            The ancient answer should be why the hell do you want to? Now we've figured this out what can we do with it?

            Sometimes curiosity-driven studies in mathematics and sciences lead to enormous technological advances. And sometimes they don't. You may not know which until much later. But it's still important to be curious.

        • Re: (Score:2)

          by Dahan ( 130247 )

          It is conjectured that any integer can be represented as the sum of three cubes [wikipedia.org].

          I think you missed an important part of the first paragraph of that Wikipedia article: "A necessary condition for n to equal such a sum is that n cannot equal 4 or 5 modulo 9, because the cubes modulo 9 are 0, 1, and 1, and no three of these numbers can sum to 4 or 5 modulo 9." So no, not any integer; there are an infinite number of integers where we know that there is no possible solution.

        • Ok Smartypants, the next unsolved number is 114. So what's the solution?

          Easy: (15887466312125899211)^3 + (-55514358745699634482)^3 + (567356686756346766)^3 = 114

    • Try to compute this answer in Excel, it gives nothing close to 42.

    • Coincidentally, that's the combination to my luggage.

  • That's one answer. Shouldn't there be an infinite number of correct answers?

  • ...can you find a solution in base13?

  • Author forgets to mention that fact in his article.

  • The mathematical foundation for quantum physics is based on an error.

  • Wow, later than usual. (Score:4, Interesting)

    by Mal-2 ( 675116 ) on Wednesday September 11, 2019 @10:45PM (#59183920) Homepage Journal

    I figured you guys were taking a pass on this one, when the Numberphile video [youtube.com] came out five days ago.

  • Mathematicians? (Score:3)

    by hcs_$reboot ( 1536101 ) on Wednesday September 11, 2019 @11:18PM (#59184008)
    Except for 42 and maybe 33, most of the work was brute force. Half a million computers to the rescue, for a 3 lines algorithm.

  • Not every number (Score:4, Informative)

    by Pretzalzz ( 577309 ) on Wednesday September 11, 2019 @11:33PM (#59184032)

    There are no solutions for k= 4 or 5 mod 9. Every cube is either -1, 0, or 1 mod 9 so the sum of three cubes must be between -3 and +3 mod 9.

  • Should not we have asked the mice? They probably could have told you the answer in rather short order.

    On the gripping hand, I guess this is not the days where computer time worth was measured in dollars per CPU second ...

    • Should not we have asked the mice? They probably could have told you the answer in rather short order.

      That's what they built the earth for, so technically, they are asking us.

  • While k is an integer, nowhere in TFS it is said that x, y and z have to be integers. So the solution is x = y = z = sqrt3(k/3)
    sqrt3() being the cubic root function.
    • Mathematics does not give any meaning unless you can apply it to life, the universe and everything, something these mathematicians fail to do. See my comment on this page for an example on how to do it (and please try to prove me wrong): https://science.slashdot.org/c... [slashdot.org]
  • Find k = x^3+y^3+z^3 for any x, y and z being from 1 to 100.
    Like the Greeks, used a good ol pen and a piece of paper. That deserves something!
    x y z => k, 1 1 1 => 3, 1 1 2 => 10, 1 1 3 => 29, 1 1 4 => 66, 1 1 5 => 127, 1 1 6 => 218, 1 1 7 => 345, 1 1 8 => 514, 1 1 9 => 731, 1 1 10 => 1002, 1 1 11 => 1333, 1 1 12 => 1730, 1 1 13 => 2199, 1 1 14 => 2746, 1 1 15 => 3377, 1 1 16 => 4098, 1 1 17 => 4915, 1 1
  • It seems more like a brute force attack rather than mathematical solution. If a computational calculation were to discover that a zero of the Riemann zeta function didn't have a real part 1/2, I don't think we'd learn anything about the distribution of the primes.

    • It seems more like a brute force attack rather than mathematical solution.

      This looks like a very clever mathematical solution, which makes a brute force attack feasible.

      Try finding an algorithm that finds solutions with x, y, z = N faster than O (N^2). O (N^2) is something anyone who is not an absolute beginner can do, but faster is hard, and will need some decent mathematics. This solution however took about 40N nanoseconds. That's astonishing.

      • Re: (Score:3, Informative)

        by Mick McCann ( 6210238 )
        absolutely, you are correct. Here's a link to the paper describing the implementation of the algorithm. https://people.maths.bris.ac.u... [bris.ac.uk]
        • Thanks for that link. Some interesting bits: 1. Their algorithm is specifically designed for the case where k = +/- 3 (modulo 9). On the other hand, that is by for the hardest case; other k have many more solutions. 2. The paper shows the solution for k = 33. The solution for k = 42 uses the exact same algorithm (just needs more time). So they were probably lucky that they picked k = 33 first. 3. For k = 3, there are two single digit solutions known (1 + 1 + 1, and 64 + 64 - 125). For some reason, search

        • Just asking for clarification, but I think you might be retracting your original position because they were not naively brute forcing it, but rather had a sophisticated algorithm that still required a lot of computation to find the answer of 42. Or is it still brute force in the lack of a deep theoretical justification for the algorithm?

          Yes, I know I should read the paper and its references and figure it out rather than bothering you. More so since I suspect you're a serious mathematician (whose time is mor

    • If all else fails, brute force it and see what you get.

      Now that they have solutions for 33 and 42, they don't need to prove that they don't exist. They do. They can now investigate the nature of the solutions. Are there more? How many? Why are they so crazy big?

      ...laura

      • Why are they crazy big? Solutions for x^3 + y^3 + z^3 = k are rare when k = +/- 3 (modulo 9). Heuristically there are less than 0.1 solutions with numbers from 10^k to 10^(k+1). So we have an 18 digit solution, thereâ(TM)s a good chance for another solution with up to 28 digits, and another one with up to 38 digits and so on.

        They took 1 million hours CPU time for this 18 digit solution. Finding a 22 digit solution would take 10,000 times longer. And if some k hasnâ(TM)t got an 18 solution then
  • That was some of the most impressive busy-think I've heard to date. How much carbon did the search add to the atmosphere? Just wondering.

    • Re: (Score:2)

      by shanen ( 462549 )

      That was some of the most impressive busy-think I've heard to date. How much carbon did the search add to the atmosphere? Just wondering.

      Much less than you "think". It was not a naive brute force attack. There's a link to the paper just before your "contribution" to the discussion. Also in the summary of the story.

      Note that I did not call you naive. Draw your own conclusions.

      • Wasn't a contribution.

        • Re: (Score:2)

          by shanen ( 462549 )

          Well, I was considering apologizing for my "snideness". Or should that be "snidity"? Yes, I know I lack gladness (especially around math)--but I would also wager that you can't figure out the reference.

          However, instead I'm going to ask why, if you had nothing to say, did you insist on saying nothing?

          • You may increase your self esteem by assuming the idiocy of those around you if you wish. I don't mind. My self esteem allows that. Thank you for your correspondence. And the answer is: because I wanted to.
  • .. to understand the question.
  • Why is this so hard? 42 / 3 = 14 (oh I love that number, by the way) Now take the cube root of 14 and you get 2.41014226417523 So, another answer is 2.41014226417523^3 + 2.41014226417523^3 + 2.41014226417523^3 = 42 Is there a rule I'm missing?

  • By a curious but not wholly unexpected coincidence, 80538738812075974 is also the telephone number of King's Cross railway station passenger inquiries. Should have started there and saved yourself a million computer hours.

    • Re: (Score:2)

      by shanen ( 462549 )

      By a curious but not wholly unexpected coincidence, 80538738812075974 is also the telephone number of King's Cross railway station passenger inquiries. Should have started there and saved yourself a million computer hours.

      I like the approach to the joke, but... Even if I had the mod point, I'd have to award it on the effort rather than the effect.

      You did get me to try a websearch on the number 80538738812075974 which came back with 7,490 results. However all if them seemed to be about this paper and I didn't want to search all the results to see if any of them were different.

      So I took the opposite tactic. What if your joke were correct? I added "cross" to the search and it was reduced to 425 hits. I even had a hot flash when

      • Fair enough, I realize the attempted joke is not going to win any awards. But for those who completely missed it, it was a reference to the following passage from Life, the Universe, and Everything:

        "That's a pity," said Arthur [upon hearing that a man named Prak who started to tell the Truth, the Whole Truth and Nothing but the Truth, has been left to tell it to himself in a very securely sealed chamber]. "I'd like to hear what he had to say. Presumably he would know what the Ultimate Question to the Ultimate Answer is. It's always bothered me that we never found out."
        "Think of a number," said the computer, "any number."
        Arthur told the computer the telephone number of King's Cross railway station passenger inquiries, on the grounds that it must have some function, and this might turn out to be it.
        The computer injected the number into the ship's reconstituted Improbability Drive.
        In Relativity, Matter tells Space how to curve, and Space tells Matter how to move.
        The Heart of Gold told space to get knotted, and parked itself neatly within the inner steel perimeter of the Argabuthon Chamber of Law.

        • Re: (Score:2)

          by shanen ( 462549 )

          Been a long time since I read his books. Miss 'im. But I think that was the funniest of the series.

          However, you did raise the interesting point of how rare and strange such numbers are. It really would be weird if anyone on earth had ever referenced that particular string of digits before. If it was in any phone directory, then the websearch would have turned it up.

          However my favorite constant may well be the Euler-Mascheroni Constant, which begins 0.5772156649... They can't even prove that it's irrational

  • The answer: (-80538738812075974)^3 + (80435758145817515)^3 + (12602123297335631)^3 = 42.

    No, the answer is 42, the question is What does
      (-80538738812075974)^3 + (80435758145817515)^3 + (12602123297335631)^3 equal?
    They found, after a much heavier calculation than they needed for the other numbers, 42.
    The analogy with THHG is better than suggested!

  • Excel won't work? try 'dc' (on any unix):

    $ echo "_80538738812075974 3 ^ 80435758145817515 3 ^ + 12602123297335631 3 ^ + p" | dc
    42

  • I'm not getting this. The problem is, "Is there a number that is not 4 or 5 modulo 9 and that cannot be expressed as a sum of three cubes?". It is one of the still unsolved problems in number theory,
    https://en.m.wikipedia.org/wik... [wikipedia.org]

    These researchers did not solve this problem. They just did a brute-force calculation for two numbers that had not yet been solved. Maybe that's cool from a computer science perspective (although I'm unclear on that as well), but the Heath-Brown conjecture, the actual math, still

  • Why am I so totally unimpressed? I guess, not being a mathematician, I just don't realize the significance of all this.

  • So (Score:2)

    by longbot ( 789962 )
    How many roads must a man walk down? Still 42, eh?
  • The answer must be Rolling Rock! You know, old number 33!
    The recipe that finally made the cut!

    So, the meaning of life must be ... wait for it ... BEER!

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