 The Man Who Solved the Mysterious Cicada 3301 Puzzle 124 points by wallflower on Nov 2, 2014 | hide | past | web | favorite | 33 comments It is most likely an underground organization, not related to any government or intelligence agencyOr a few bored guys on an IRC channel. A few websites, couple of phone numbers, and some posters: once you have members in the targeted cities, the budget for an operation like this is in tens of dollars. All the links on that website link to the same page :S The technical details of the puzzle described in the article strike me as trivial. 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. 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. So what was the math problem? 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 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. 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.) 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+... 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. 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.pdfHave fun! What you really need is a "solution" that is wrong in just one digit somewhere in the middle that everyone would easily miss when comparing. Python having easy arbitrary-precision integer arithmetic is one of the things I most like about it. Arbitrary-precision integers is definitely one of Python's underrated feature. Well according the the theorem there is no integer solutions when the exponent is greater than 2. His counterexample is incorrect. Umm... About that: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. Also see this same gag on The Simpsons Andrew Wiles would like to have a word with you regarding this claim! Are you sure those were the only restrictions on the problem? Wouldn't [1, 1, 2^(1/20)] also be a solution? Whoops, fixed. I forgot to say that a, b, c must be integers. Thanks! 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. 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 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. This particular solution only works on April 1st. However 0^20 + 1^20 = 1^20 works any day of the year. In general, a^20 + 0^20 = ((+ or -)a)^20, or 0^20 + a^20 = ((+ or -)a)^20. 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. If you think they are trivial why don't you solve them http://uncovering-cicada.wikia.com/wiki/CICADA_3301_2014_PUZ... Wow, aren't you smart. I wish someone would finally report those fuckers to the FBI. Or straight-up murder them.Secret societies like that make my skin crawl. That's right, they are probably planning on taking all over the world now. Seriously? Search: