5 ms·

The technical details of the puzzle described in the article strike me as trivial.

ameister14 12y ago
Well, the last section wasn't. After you got a message on the e-mail account you made using TOR, you were asked a more complex mathematical question, and told not to tell anyone what it was, as they were tired of ircs and skype groups solving everything. I didn't realize I was one of the people selected to move on until a week later, though, and was too late to move forward.
- vex 12y ago
  So what was the math problem?
  - sillysaurus3 12y ago
    Find integers a, b, c such that a^20 + b^20 = c^20. The solution turned out to be 4110^20 + 4693^20 = 4709^20. You can verify this is correct using any old calculator, for example: https://www.google.com/search?q=4110%5E20+%2B+4693%5E20 https://www.google.com/search?q=4110%5E20+%2B+4693%5E20 https://www.google.com/search?q=4709%5E20 https://www.google.com/search?q=4709%5E20
    - bowmessage 12y ago
      Are you sure those were the only restrictions on the problem? Wouldn't [1, 1, 2^(1/20)] also be a solution?
      - sillysaurus3 12y ago
        Whoops, fixed. I forgot to say that a, b, c must be integers. Thanks!
    - wbhart 12y ago
      This particular solution only works on April 1st. However 0^20 + 1^20 = 1^20 works any day of the year.
      - TheLoneWolfling 12y ago
        In general, a^20 + 0^20 = ((+ or -)a)^20, or 0^20 + a^20 = ((+ or -)a)^20.
    - dil8 12y ago
      Am I missing something, doesn't this contradict Fermat's last theorem > In number theory, Fermat's Last Theorem (sometimes called Fermat's conjecture, especially in older texts) states that no three positive integers a, b, and c can satisfy the equation a^n + b^n = c^n for any integer value of n greater than two. https://en.wikipedia.org/wiki/Fermat%27s_Last_Theorem https://en.wikipedia.org/wiki/Fermat%27s_Last_Theorem
      - TheLoneWolfling 12y ago
        You're not missing anything. His numbers do not add up to the same thing. In other words, 4709^20 != (4110^20 + 4693^20). (The difference is ~10^61 or so, whereas the numbers are ~10^73. In other words, they diverge at ~ the 12th digit, whereas many calculators only display 10.)
        sillysaurus3 12y ago
        Hm, well, part 2 of the question was to find a solution for a^15 + b^15 = c^15, where a, b, c are integers > 0. Google verifies the answer is 434437^15 + 588129^15 = 588544^15: https://www.google.com/search?q=434437%5E15+%2B+588129%5E15+-+588544%5E15 https://www.google.com/search?q=434437%5E15+%2B+588129%5E15+...
        TheLoneWolfling 12y ago
        Nope. This is not correct either. > >>> 434437* * 15 + 588129* * 15 - 588544 * *15 > -604550152144288043930860169354171954730939671404246170822386878582482 (Edit: how do I display two asterisks in a row? It's supposed to be number(asterisk)(asterisk)number.) This is using Python, which does arbitrary-precision integer arithmetic. Google's calculator probably uses floating-point numbers internally, and hence starts losing precision.
        sillysaurus3 12y ago
        It seems that advancements in technology have made mathematical trolling much more difficult. :) In case anyone is curious, the above "solutions" are called near-misses, since they're almost correct. A clever person came up with an algorithm to generate interesting near-misses for low exponents. See the table on page 15: http://arxiv.org/pdf/math/0005139v1.pdf http://arxiv.org/pdf/math/0005139v1.pdf Related: http://math.stackexchange.com/questions/526330/fermats-last-theorem-near-misses http://math.stackexchange.com/questions/526330/fermats-last-... Have fun!
    - TheLoneWolfling 12y ago
      Umm... About that: https://www.google.com/search?q=4709^20+-+%284110^20+%2B+4693^20%29 https://www.google.com/search?q=4709^20+-+%284110^20+%2B+469... Just because they don't disagree on the significant figures you can see doesn't mean there isn't a difference.
    - diziet 12y ago
      Andrew Wiles would like to have a word with you regarding this claim!
    - IvyMike 12y ago
      Also see this same gag on The Simpsons http://www.npr.org/blogs/krulwich/2014/05/08/310818693/did-homer-simpson-actually-solve-fermat-s-last-theorem-take-a-look http://www.npr.org/blogs/krulwich/2014/05/08/310818693/did-h...
    - lordnacho 12y ago
      Wait a minute, is it somehow clear that you're supposed to be looking for a near-miss? If I came across this in my line of inquiry, I would assume that my previous step was wrong, and I'd have dropped the puzzle eventually.
      - sillysaurus3 12y ago
        Nah, I didn't actually do the Cicada 3301. I was just joking around. It seemed unlikely anyone was going to post an interesting math problem, so I decided to have a little fun. https://news.ycombinator.com/item?id=8549204 https://news.ycombinator.com/item?id=8549204
    - JoshTheGeek 12y ago
      Fermat's last theorem has been proven; there is no solution, people. In this case, you can't verify this using "any old calculator," as most show only ten digits and these diverge at digit 12.
- wbhart 12y ago
  You could at least entertain us with a real mathematical problem, even if not the one you were given, e.g. in how many ways can 54673350319220399841294938973353 be expressed as three times a square plus twice a (generalised) octagonal number. Bonus points for finding one set of explicit values.
na85 12y ago
Wow, aren't you smart.
swiftydev 12y ago
If you think they are trivial why don't you solve them http://uncovering-cicada.wikia.com/wiki/CICADA_3301_2014_PUZZLE_FACTS_PART_9 http://uncovering-cicada.wikia.com/wiki/CICADA_3301_2014_PUZ...
standalone3 12y ago
Are you saying that only based on the technical details presented in a non-technical article? You might be interested to actually check out the actual problems, some of which have remained unsolved (at least by the general public) for years now. http://uncovering-cicada.wikia.com/wiki/Uncovering_Cicada_Wiki http://uncovering-cicada.wikia.com/wiki/Uncovering_Cicada_Wi...

Well, the last section wasn't. After you got a message on the e-mail account you made using TOR, you were asked a more complex mathematical question, and told not to tell anyone what it was, as they were tired of ircs and skype groups solving everything. I didn't realize I was one of the people selected to move on until a week later, though, and was too late to move forward.

vex 12y ago
So what was the math problem?
- sillysaurus3 12y ago
  Find integers a, b, c such that a^20 + b^20 = c^20. The solution turned out to be 4110^20 + 4693^20 = 4709^20. You can verify this is correct using any old calculator, for example: https://www.google.com/search?q=4110%5E20+%2B+4693%5E20 https://www.google.com/search?q=4110%5E20+%2B+4693%5E20 https://www.google.com/search?q=4709%5E20 https://www.google.com/search?q=4709%5E20
  - bowmessage 12y ago
    Are you sure those were the only restrictions on the problem? Wouldn't [1, 1, 2^(1/20)] also be a solution?
    - sillysaurus3 12y ago
      Whoops, fixed. I forgot to say that a, b, c must be integers. Thanks!
  - wbhart 12y ago
    This particular solution only works on April 1st. However 0^20 + 1^20 = 1^20 works any day of the year.
    - TheLoneWolfling 12y ago
      In general, a^20 + 0^20 = ((+ or -)a)^20, or 0^20 + a^20 = ((+ or -)a)^20.
  - dil8 12y ago
    Am I missing something, doesn't this contradict Fermat's last theorem > In number theory, Fermat's Last Theorem (sometimes called Fermat's conjecture, especially in older texts) states that no three positive integers a, b, and c can satisfy the equation a^n + b^n = c^n for any integer value of n greater than two. https://en.wikipedia.org/wiki/Fermat%27s_Last_Theorem https://en.wikipedia.org/wiki/Fermat%27s_Last_Theorem
    - TheLoneWolfling 12y ago
      You're not missing anything. His numbers do not add up to the same thing. In other words, 4709^20 != (4110^20 + 4693^20). (The difference is ~10^61 or so, whereas the numbers are ~10^73. In other words, they diverge at ~ the 12th digit, whereas many calculators only display 10.)
      - sillysaurus3 12y ago
        Hm, well, part 2 of the question was to find a solution for a^15 + b^15 = c^15, where a, b, c are integers > 0. Google verifies the answer is 434437^15 + 588129^15 = 588544^15: https://www.google.com/search?q=434437%5E15+%2B+588129%5E15+-+588544%5E15 https://www.google.com/search?q=434437%5E15+%2B+588129%5E15+...
        TheLoneWolfling 12y ago
        Nope. This is not correct either. > >>> 434437* * 15 + 588129* * 15 - 588544 * *15 > -604550152144288043930860169354171954730939671404246170822386878582482 (Edit: how do I display two asterisks in a row? It's supposed to be number(asterisk)(asterisk)number.) This is using Python, which does arbitrary-precision integer arithmetic. Google's calculator probably uses floating-point numbers internally, and hence starts losing precision.
        sillysaurus3 12y ago
        It seems that advancements in technology have made mathematical trolling much more difficult. :) In case anyone is curious, the above "solutions" are called near-misses, since they're almost correct. A clever person came up with an algorithm to generate interesting near-misses for low exponents. See the table on page 15: http://arxiv.org/pdf/math/0005139v1.pdf http://arxiv.org/pdf/math/0005139v1.pdf Related: http://math.stackexchange.com/questions/526330/fermats-last-theorem-near-misses http://math.stackexchange.com/questions/526330/fermats-last-... Have fun!
  - TheLoneWolfling 12y ago
    Umm... About that: https://www.google.com/search?q=4709^20+-+%284110^20+%2B+4693^20%29 https://www.google.com/search?q=4709^20+-+%284110^20+%2B+469... Just because they don't disagree on the significant figures you can see doesn't mean there isn't a difference.
  - diziet 12y ago
    Andrew Wiles would like to have a word with you regarding this claim!
  - IvyMike 12y ago
    Also see this same gag on The Simpsons http://www.npr.org/blogs/krulwich/2014/05/08/310818693/did-homer-simpson-actually-solve-fermat-s-last-theorem-take-a-look http://www.npr.org/blogs/krulwich/2014/05/08/310818693/did-h...
  - lordnacho 12y ago
    Wait a minute, is it somehow clear that you're supposed to be looking for a near-miss? If I came across this in my line of inquiry, I would assume that my previous step was wrong, and I'd have dropped the puzzle eventually.
    - sillysaurus3 12y ago
      Nah, I didn't actually do the Cicada 3301. I was just joking around. It seemed unlikely anyone was going to post an interesting math problem, so I decided to have a little fun. https://news.ycombinator.com/item?id=8549204 https://news.ycombinator.com/item?id=8549204
  - JoshTheGeek 12y ago
    Fermat's last theorem has been proven; there is no solution, people. In this case, you can't verify this using "any old calculator," as most show only ten digits and these diverge at digit 12.
wbhart 12y ago
You could at least entertain us with a real mathematical problem, even if not the one you were given, e.g. in how many ways can 54673350319220399841294938973353 be expressed as three times a square plus twice a (generalised) octagonal number. Bonus points for finding one set of explicit values.

na85 12y ago

Wow, aren't you smart.

swiftydev 12y ago

If you think they are trivial why don't you solve them http://uncovering-cicada.wikia.com/wiki/CICADA_3301_2014_PUZZLE_FACTS_PART_9 http://uncovering-cicada.wikia.com/wiki/CICADA_3301_2014_PUZ...

standalone3 12y ago

Are you saying that only based on the technical details presented in a non-technical article? You might be interested to actually check out the actual problems, some of which have remained unsolved (at least by the general public) for years now. http://uncovering-cicada.wikia.com/wiki/Uncovering_Cicada_Wiki http://uncovering-cicada.wikia.com/wiki/Uncovering_Cicada_Wi...