
The Man Who Solved the Mysterious Cicada 3301 Puzzle - wallflower
http://www.fastcolabs.com/3025785/meet-the-man-who-solved-the-mysterious-cicada-3301-puzzle
======
spindritf
_It is most likely an underground organization, not related to any government
or intelligence agency_

Or 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.

------
treve
All the links on that website link to the same page :S

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

~~~
ameister14
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
So what was the math problem?

~~~
sillysaurus3
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)

~~~
dil8
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
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
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+...](https://www.google.com/search?q=434437%5E15+%2B+588129%5E15+-+588544%5E15)

~~~
TheLoneWolfling
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
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-...](http://math.stackexchange.com/questions/526330/fermats-last-theorem-
near-misses)

Have fun!

~~~
wbhart
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.

------
anon4
I wish someone would finally report those fuckers to the FBI. Or straight-up
murder them.

Secret societies like that make my skin crawl.

~~~
cLeEOGPw
That's right, they are probably planning on taking all over the world now.

